How can I upload files to a server using JSP/Servlet?
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T

14

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How can I upload files to server using JSP/Servlet?

I tried this:

<form action="upload" method="post">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

However, I only get the file name, not the file content. When I add enctype="multipart/form-data" to the <form>, then request.getParameter() returns null.

During research I stumbled upon Apache Common FileUpload. I tried this:

FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = upload.parseRequest(request); // This line is where it died.

Unfortunately, the servlet threw an exception without a clear message and cause. Here is the stacktrace:

SEVERE: Servlet.service() for servlet UploadServlet threw exception
javax.servlet.ServletException: Servlet execution threw an exception
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:313)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
    at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
    at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
    at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
    at java.lang.Thread.run(Thread.java:637)
Titivate answered 11/3, 2010 at 4:7 Comment(4)
Perhaps this article will be helpful: baeldung.com/upload-file-servletPolyglot
@Adam: They copied from my answer and added a sleuth of advertising on top of it in an attempt to earn money with it. Yeah, great article ..Riposte
No, actually nothing was copied. I wrote the first draft of that article along with the supplemental code. The core reference documentation can be found here: commons.apache.org/proper/commons-fileupload/using.html (and is linked to and cited in the article). Examples are partly reprised from the core reference document (which is the point of reference documentation - i.e. to be a point of reference) but not in their entirety (note that the reference docs don't go into much detail). Thanks!Polyglot
check this sandny.com/2017/05/18/servlet-file-uploadEmmalineemmalyn
R
1258

Introduction

To browse and select a file for upload you need a HTML <input type="file"> field in the form. As stated in the HTML specification you have to use the POST method and the enctype attribute of the form has to be set to "multipart/form-data".

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

After submitting such a form, the binary multipart form data is available in the request body in a different format than when the enctype isn't set.

Before Servlet 3.0 (Dec 2009), the Servlet API didn't natively support multipart/form-data. It supports only the default form enctype of application/x-www-form-urlencoded. The request.getParameter() and consorts would all return null when using multipart form data. This is where the well known Apache Commons FileUpload came into the picture.

Don't manually parse it!

You can in theory parse the request body yourself based on ServletRequest#getInputStream(). However, this is a precise and tedious work which requires precise knowledge of RFC2388. You shouldn't try to do this on your own or copypaste some homegrown library-less code found elsewhere on the Internet. Many online sources have failed hard in this, such as roseindia.net. See also uploading of pdf file. You should rather use a real library which is used (and implicitly tested!) by millions of users for years. Such a library has proven its robustness.

When you're already on Servlet 3.0 or newer, use native API

If you're using at least Servlet 3.0 (Tomcat 7, Jetty 9, JBoss AS 6, GlassFish 3, etc, they exist already since 2010), then you can just use standard API provided HttpServletRequest#getPart() to collect the individual multipart form data items (most Servlet 3.0 implementations actually use Apache Commons FileUpload under the covers for this!). Also, normal form fields are available by getParameter() the usual way.

First annotate your servlet with @MultipartConfig in order to let it recognize and support multipart/form-data requests and thus get getPart() to work:

@WebServlet("/upload")
@MultipartConfig
public class UploadServlet extends HttpServlet {
    // ...
}

Then, implement its doPost() as follows:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String description = request.getParameter("description"); // Retrieves <input type="text" name="description">
    Part filePart = request.getPart("file"); // Retrieves <input type="file" name="file">
    String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
    InputStream fileContent = filePart.getInputStream();
    // ... (do your job here)
}

Note the Path#getFileName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

In case you want to upload multiple files via either multiple="true",

<input type="file" name="files" multiple="true" />

or the old-fashioned way with multiple inputs,

<input type="file" name="files" />
<input type="file" name="files" />
<input type="file" name="files" />
...

then you can collect them as below (unfortunately there is no such method as request.getParts("files")):

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    // ...
    List<Part> fileParts = request.getParts().stream().filter(part -> "files".equals(part.getName()) && part.getSize() > 0).collect(Collectors.toList()); // Retrieves <input type="file" name="files" multiple="true">

    for (Part filePart : fileParts) {
        String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
        InputStream fileContent = filePart.getInputStream();
        // ... (do your job here)
    }
}

When you're not on Servlet 3.1 yet, manually get submitted file name

Note that Part#getSubmittedFileName() was introduced in Servlet 3.1 (Tomcat 8, Jetty 9, WildFly 8, GlassFish 4, etc, they exist since 2013 already). If you're not on Servlet 3.1 yet (really?), then you need an additional utility method to obtain the submitted file name.

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}
String fileName = getSubmittedFileName(filePart);

Note the MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

When you're not on Servlet 3.0 yet, use Apache Commons FileUpload

If you're not on Servlet 3.0 yet (isn't it about time to upgrade? it's released over a decade ago!), the common practice is to make use of Apache Commons FileUpload to parse the multpart form data requests. It has an excellent User Guide and FAQ (carefully go through both). There's also the O'Reilly ("cos") MultipartRequest, but it has some (minor) bugs and isn't actively maintained anymore for years. I wouldn't recommend using it. Apache Commons FileUpload is still actively maintained and currently very mature.

In order to use Apache Commons FileUpload, you need to have at least the following files in your webapp's /WEB-INF/lib:

Your initial attempt failed most likely because you forgot the commons IO.

Here's a kickoff example how the doPost() of your UploadServlet may look like when using Apache Commons FileUpload:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
        for (FileItem item : items) {
            if (item.isFormField()) {
                // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
                String fieldName = item.getFieldName();
                String fieldValue = item.getString();
                // ... (do your job here)
            } else {
                // Process form file field (input type="file").
                String fieldName = item.getFieldName();
                String fileName = FilenameUtils.getName(item.getName());
                InputStream fileContent = item.getInputStream();
                // ... (do your job here)
            }
        }
    } catch (FileUploadException e) {
        throw new ServletException("Cannot parse multipart request.", e);
    }

    // ...
}

It's very important that you don't call getParameter(), getParameterMap(), getParameterValues(), getInputStream(), getReader(), etc on the same request beforehand. Otherwise the servlet container will read and parse the request body and thus Apache Commons FileUpload will get an empty request body. See also a.o. ServletFileUpload#parseRequest(request) returns an empty list.

Note the FilenameUtils#getName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

Alternatively you can also wrap this all in a Filter which parses it all automagically and put the stuff back in the parametermap of the request so that you can continue using request.getParameter() the usual way and retrieve the uploaded file by request.getAttribute(). You can find an example in this blog article.

Workaround for GlassFish3 bug of getParameter() still returning null

Note that Glassfish versions older than 3.1.2 had a bug wherein the getParameter() still returns null. If you are targeting such a container and can't upgrade it, then you need to extract the value from getPart() with help of this utility method:

private static String getValue(Part part) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(part.getInputStream(), "UTF-8"));
    StringBuilder value = new StringBuilder();
    char[] buffer = new char[1024];
    for (int length = 0; (length = reader.read(buffer)) > 0;) {
        value.append(buffer, 0, length);
    }
    return value.toString();
}
String description = getValue(request.getPart("description")); // Retrieves <input type="text" name="description">
    

Saving uploaded file (don't use getRealPath() nor part.write()!)

Head to the following answers for detail on properly saving the obtained InputStream (the fileContent variable as shown in the above code snippets) to disk or database:

Serving uploaded file

Head to the following answers for detail on properly serving the saved file from disk or database back to the client:

Ajaxifying the form

Head to the following answers how to upload using Ajax (and jQuery). Do note that the servlet code to collect the form data does not need to be changed for this! Only the way how you respond may be changed, but this is rather trivial (i.e. instead of forwarding to JSP, just print some JSON or XML or even plain text depending on whatever the script responsible for the Ajax call is expecting).

Riposte answered 11/3, 2010 at 12:27 Comment(17)
Ah sorry, I was seeing request.getParts("file") and was confused x_xAntimere
With Servlet 3.0, if a MultipartConfig condition is violated (eg: maxFileSize), calling request.getParameter() returns null. Is this on purpose? What if I get some regular (text) parameters before calling getPart (and checking for an IllegalStateException)? This causes a NullPointerException to be thrown before I have a chance to check for the IllegalStateException.Rumor
@Riposte I created a post related to this, do you have an idea how I could retrieve extra infos from File API webKitDirectory. More details here stackoverflow.com/questions/45419598/…Woodie
If you aren't on Servlet 3.0 and use the FileUpload trick, I found you can't read the file from the request more than once. If you need this functionality, you might want to look at Spring's MultiPartFilter. This post has a good working example: https://mcmap.net/q/16902/-spring-security-3-2-csrf-support-for-multipart-requestsHelp
Answer is misleading in "When you're already on Servlet 3.0 or newer" section, because according to docs getSubmittedFileName method is available in Part interface since servlet spec 3.1, tomcat 8+ supports 3.1 specification.Koziara
@raviraja: That's correct. See "When you're not on Servlet 3.1 yet, manually get submitted file name" section of the answer.Riposte
Yeah, if someone tries to use the code in 3.0 section with tomcat 7, they might face issue in String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.part similar to meKoziara
In IE/Edge, getSubmittedFileName removes slashes from the full path. How to fix that?Jesse
@Peng: Not. That's the correct behavior. The browser should not send the full path at all but only the file name. All other browsers are doing it correctly. Only IE/Edge incorrectly sends the full client side path along which is a security leak.Riposte
@Riposte Your MSIE fix Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); successfully fix the security issue by removing the full path except the file name. But what I found in IE11 and Edge 44.18362.449 is another issue: getSubmittedFileName itself already removes all slashes in the full path. For example, for the full path C:\temp\file.txt, getSubmittedFileName gives me C:tempfile.txt. So your fix code would give me C:tempfile.txt rather than file.txt.Jesse
We can also disable IE from sending the full path: Internet Options > Security > Internet > Custom Level > disable "Include local directory path when uploading files to server”.Jesse
"getSubmittedFileName itself already removes all slashes in the full path" this is new to me. Apparently it depends on the server make/version being used. "For example, for the full path C:\temp\file.txt, getSubmittedFileName gives me C:tempfile.txt" this is definitely a bug in the server being used. Report it to them. "We can also disable IE from sending the full path" this is not wise to assume. We may not rely on every enduser of your webapp doing that. Ignore that and fix in server side instead.Riposte
after adding reguest.getParameter("password") I am getting a strange FileNotFoundException...Dogooder
@AndroidAnton: Provided that you're using Servlet 3.0 or newer, then you're supposed to obtain the uploaded file via request.getPart() as explained in the answer, not via other ways, and very definitely not via Commons FileUpload. That's only needed when you're not on Servlet 3.0 yet, as explained in the answer.Riposte
I'm working on servlet 2.3(had no choice) so I tried the method mentioned your blog. It works well if the uploaded file type is .txt. When i tried upload pdf & outlook item, they appeared in "C:/upload" folder but unable to open. It was not supported file type or file has been damaged. Any idea how I can solve this?Detrition
@aaa: that can happen when you converted bytes to characters by using a Reader and/or Writer for unclear reason. Do not do that. Use InputStream/OutputStream over all place during reading and writing an uploaded file without massaging bytes into characters. A PDF file is not a character based text file or so. It is a binary file.Riposte
Is this (request.getPart(name)) also valid to use with Spring 2.7 and servlet 3.0?Streaky
L
26

If you happen to use Spring MVC, this is how to (I'm leaving this here in case someone find it useful):

Use a form with enctype attribute set to "multipart/form-data" (the same as BalusC's answer):

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file" />
    <input type="submit" value="Upload"/>
</form>

In your controller, map the request parameter file to MultipartFile type as follows:

@RequestMapping(value = "/upload", method = RequestMethod.POST)
public void handleUpload(@RequestParam("file") MultipartFile file) throws IOException {
    if (!file.isEmpty()) {
            byte[] bytes = file.getBytes(); // alternatively, file.getInputStream();
            // application logic
    }
}

You can get the filename and size using MultipartFile's getOriginalFilename() and getSize().

I've tested this with Spring version 4.1.1.RELEASE.

Limicoline answered 26/8, 2015 at 12:39 Comment(1)
If I'm not mistaken, this requires that you configure a bean in your server's application config...Granicus
S
12

Without components or external libraries in Tomcat 6 or Tomcat 7

Enabling upload in the web.xml file:

Manually Installing PHP, Tomcat and Httpd Lounge.

<servlet>
    <servlet-name>jsp</servlet-name>
    <servlet-class>org.apache.jasper.servlet.JspServlet</servlet-class>
    <multipart-config>
      <max-file-size>3145728</max-file-size>
      <max-request-size>5242880</max-request-size>
    </multipart-config>
    <init-param>
        <param-name>fork</param-name>
        <param-value>false</param-value>
    </init-param>
    <init-param>
        <param-name>xpoweredBy</param-name>
        <param-value>false</param-value>
    </init-param>
    <load-on-startup>3</load-on-startup>
</servlet>

As you can see:

<multipart-config>
  <max-file-size>3145728</max-file-size>
  <max-request-size>5242880</max-request-size>
</multipart-config>

Uploading files using JSP. files:

In the HTML file

<form method="post" enctype="multipart/form-data" name="Form" >

  <input type="file" name="fFoto" id="fFoto" value="" /></td>
  <input type="file" name="fResumen" id="fResumen" value=""/>

In the JSP File or Servlet

InputStream isFoto = request.getPart("fFoto").getInputStream();
InputStream isResu = request.getPart("fResumen").getInputStream();
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte buf[] = new byte[8192];
int qt = 0;
while ((qt = isResu.read(buf)) != -1) {
  baos.write(buf, 0, qt);
}
String sResumen = baos.toString();

Edit your code to servlet requirements, like max-file-size, max-request-size and other options that you can to set...

Soccer answered 25/1, 2014 at 5:44 Comment(0)
O
11

You need the common-io.1.4.jar file to be included in your lib directory, or if you're working in any editor, like NetBeans, then you need to go to project properties and just add the JAR file and you will be done.

To get the common.io.jar file just google it or just go to the Apache Tomcat website where you get the option for a free download of this file. But remember one thing: download the binary ZIP file if you're a Windows user.

Overcritical answered 17/5, 2012 at 11:11 Comment(1)
Can't find .jar but .zip. Do you mean .zip?Nelidanelie
P
9

I am using a common Servlet for every HTML form whether it has attachments or not.

This Servlet returns a TreeMap where the keys are JSP name parameters and values are user inputs and saves all attachments in a fixed directory and later you rename the directory of your choice. Here Connections is our custom interface having a connection object.

public class ServletCommonfunctions extends HttpServlet implements
        Connections {

    private static final long serialVersionUID = 1L;

    public ServletCommonfunctions() {}

    protected void doPost(HttpServletRequest request,
                          HttpServletResponse response) throws ServletException,
                          IOException {}

    public SortedMap<String, String> savefilesindirectory(
            HttpServletRequest request, HttpServletResponse response)
            throws IOException {

        // Map<String, String> key_values = Collections.synchronizedMap(new
        // TreeMap<String, String>());
        SortedMap<String, String> key_values = new TreeMap<String, String>();
        String dist = null, fact = null;
        PrintWriter out = response.getWriter();
        File file;
        String filePath = "E:\\FSPATH1\\2KL06CS048\\";
        System.out.println("Directory Created   ????????????"
            + new File(filePath).mkdir());
        int maxFileSize = 5000 * 1024;
        int maxMemSize = 5000 * 1024;

        // Verify the content type
        String contentType = request.getContentType();
        if ((contentType.indexOf("multipart/form-data") >= 0)) {
            DiskFileItemFactory factory = new DiskFileItemFactory();
            // Maximum size that will be stored in memory
            factory.setSizeThreshold(maxMemSize);
            // Location to save data that is larger than maxMemSize.
            factory.setRepository(new File(filePath));
            // Create a new file upload handler
            ServletFileUpload upload = new ServletFileUpload(factory);
            // maximum file size to be uploaded.
            upload.setSizeMax(maxFileSize);
            try {
                // Parse the request to get file items.
                @SuppressWarnings("unchecked")
                List<FileItem> fileItems = upload.parseRequest(request);
                // Process the uploaded file items
                Iterator<FileItem> i = fileItems.iterator();
                while (i.hasNext()) {
                    FileItem fi = (FileItem) i.next();
                    if (!fi.isFormField()) {
                        // Get the uploaded file parameters
                        String fileName = fi.getName();
                        // Write the file
                        if (fileName.lastIndexOf("\\") >= 0) {
                            file = new File(filePath
                                + fileName.substring(fileName
                                        .lastIndexOf("\\")));
                        } else {
                            file = new File(filePath
                                + fileName.substring(fileName
                                        .lastIndexOf("\\") + 1));
                        }
                        fi.write(file);
                    } else {
                        key_values.put(fi.getFieldName(), fi.getString());
                    }
                }
            } catch (Exception ex) {
                System.out.println(ex);
            }
        }
        return key_values;
    }
}
Periodontal answered 8/1, 2013 at 5:50 Comment(2)
@buhake sindi hey what should be the filepath if i m using live server or i live my project by uploading files to the serverEnforce
Any directory in live server.if you write a code to create a directory in servlet then directory will be created in live srverPeriodontal
T
7

For Spring MVC

I managed to have a simpler version that worked for taking form input, both data and images.

<form action="/handleform" method="post" enctype="multipart/form-data">
    <input type="text" name="name" />
    <input type="text" name="age" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

Controller to handle

@Controller
public class FormController {
    @RequestMapping(value="/handleform",method= RequestMethod.POST)
    ModelAndView register(@RequestParam String name, @RequestParam int age, @RequestParam MultipartFile file)
            throws ServletException, IOException {

        System.out.println(name);
        System.out.println(age);
        if(!file.isEmpty()){
            byte[] bytes = file.getBytes();
            String filename = file.getOriginalFilename();
            BufferedOutputStream stream =new BufferedOutputStream(new FileOutputStream(new File("D:/" + filename)));
            stream.write(bytes);
            stream.flush();
            stream.close();
        }
        return new ModelAndView("index");
    }
}
Trillby answered 15/7, 2017 at 19:42 Comment(1)
Can you please share select image form db mysql and show it on jsp/html?Chayachayote
A
6

Another source of this problem occurs if you are using Geronimo with its embedded Tomcat. In this case, after many iterations of testing Commons IO and commons-fileupload, the problem arises from a parent classloader handling the commons-xxx JAR files. This has to be prevented. The crash always occurred at:

fileItems = uploader.parseRequest(request);

Note that the List type of fileItems has changed with the current version of commons-fileupload to be specifically List<FileItem> as opposed to prior versions where it was generic List.

I added the source code for commons-fileupload and Commons IO into my Eclipse project to trace the actual error and finally got some insight. First, the exception thrown is of type Throwable not the stated FileIOException nor even Exception (these will not be trapped). Second, the error message is obfuscatory in that it stated class not found because axis2 could not find commons-io. Axis2 is not used in my project at all, but it exists as a folder in the Geronimo repository subdirectory as part of standard installation.

Finally, I found one place that posed a working solution which successfully solved my problem. You must hide the JAR files from the parent loader in the deployment plan. This was put into the geronimo-web.xml file with my full file shown below.

Pasted from http://osdir.com/ml/user-geronimo-apache/2011-03/msg00026.html:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<web:web-app xmlns:app="http://geronimo.apache.org/xml/ns/j2ee/application-2.0" xmlns:client="http://geronimo.apache.org/xml/ns/j2ee/application-client-2.0" xmlns:conn="http://geronimo.apache.org/xml/ns/j2ee/connector-1.2" xmlns:dep="http://geronimo.apache.org/xml/ns/deployment-1.2" xmlns:ejb="http://openejb.apache.org/xml/ns/openejb-jar-2.2" xmlns:log="http://geronimo.apache.org/xml/ns/loginconfig-2.0" xmlns:name="http://geronimo.apache.org/xml/ns/naming-1.2" xmlns:pers="http://java.sun.com/xml/ns/persistence" xmlns:pkgen="http://openejb.apache.org/xml/ns/pkgen-2.1" xmlns:sec="http://geronimo.apache.org/xml/ns/security-2.0" xmlns:web="http://geronimo.apache.org/xml/ns/j2ee/web-2.0.1">
    <dep:environment>
        <dep:moduleId>
            <dep:groupId>DataStar</dep:groupId>
            <dep:artifactId>DataStar</dep:artifactId>
            <dep:version>1.0</dep:version>
            <dep:type>car</dep:type>
        </dep:moduleId>

        <!-- Don't load commons-io or fileupload from parent classloaders -->
        <dep:hidden-classes>
            <dep:filter>org.apache.commons.io</dep:filter>
            <dep:filter>org.apache.commons.fileupload</dep:filter>
        </dep:hidden-classes>
        <dep:inverse-classloading/>

    </dep:environment>
    <web:context-root>/DataStar</web:context-root>
</web:web-app>
Acidulant answered 10/9, 2013 at 15:15 Comment(1)
The link is (effectively) broken (redirects to https://osdir.com/) - the HTTPS version does as well.Lykins
A
0

Here's an example using apache commons-fileupload:

// apache commons-fileupload to handle file upload
DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setRepository(new File(DataSources.TORRENTS_DIR()));
ServletFileUpload fileUpload = new ServletFileUpload(factory);

List<FileItem> items = fileUpload.parseRequest(req.raw());
FileItem item = items.stream()
  .filter(e ->
  "the_upload_name".equals(e.getFieldName()))
  .findFirst().get();
String fileName = item.getName();

item.write(new File(dir, fileName));
log.info(fileName);
Aegisthus answered 21/5, 2015 at 16:49 Comment(0)
K
0

You first have to set the enctype attribute of the form to "multipart/form-data"

This is shown below.

<form action="Controller" method="post" enctype="multipart/form-data">
     <label class="file-upload"> Click here to upload an Image </label>
     <input type="file" name="file" id="file" required>
</form>

And then, in the Servlet "Controller" add the Annotation for a Multi-part to indicate multipart data is processed in the servlet.

After doing this, retrieve the part sent through the form and then retrieve the file name (with path)of the submitted file. Use this to create a new file in the desired path and write the parts of the file to the newly created file to recreate the file.

As shown below:

@MultipartConfig

public class Controller extends HttpServlet {

    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        insertImage(request, response);
    }

    private void addProduct(HttpServletRequest request, HttpServletResponse response) {
        Part filePart = request.getPart("file");
        String imageName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString();

        String imageSavePath = "specify image path to save image"; //path to save image
        FileOutputStream outputStream = null;
        InputStream fileContent = null;

        try {
            outputStream = new FileOutputStream(new File(imageSavePath + File.separator + imageName));
            // Creating a new file with file path and the file name
            fileContent = filePart.getInputStream();
            // Getting the input stream
            int readBytes = 0;
            byte[] readArray = new byte[1024];
            // Initializing a byte array with size 1024

            while ((readBytes = fileContent.read(readArray)) != -1) {
                outputStream.write(readArray, 0, readBytes);
            } // This loop will write the contents of the byte array unitl the end to the output stream
        } catch (Exception ex) {
            System.out.println("Error Writing File: " + ex);
        } finally {
            if (outputStream != null) {
                outputStream.close();
                // Closing the output stream
            }
            if (fileContent != null) {
                fileContent.close();
                // Closing the input stream
            }
        }
    }
}
Kono answered 24/11, 2020 at 10:49 Comment(2)
This solution is different. Other solution used a library to handle files where as this does tf his with no 3rd party jar files.Kono
This is already covered by the currently accepted answer. Have you read it? The native API already existed since December 2009. Your way of closing the streams is by the way also legacy. Since Java 7 which was introduced in July 2011, you can use try-with-resources statement instead of fiddling with nullchecks in finally.Riposte
A
-1

You can upload a file using JSP /servlet.

<form action="UploadFileServlet" method="post">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

On the other hand, on the server side, use the following code.

package com.abc..servlet;

import java.io.File;
---------
--------


/**
 * Servlet implementation class UploadFileServlet
 */
public class UploadFileServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    public UploadFileServlet() {
        super();
        // TODO Auto-generated constructor stub
    }
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        response.sendRedirect("../jsp/ErrorPage.jsp");
    }

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub

        PrintWriter out = response.getWriter();
        HttpSession httpSession = request.getSession();
        String filePathUpload = (String) httpSession.getAttribute("path") != null ? httpSession.getAttribute("path").toString() : "" ;

        String path1 = filePathUpload;
        String filename = null;
        File path = null;
        FileItem item = null;


        boolean isMultipart = ServletFileUpload.isMultipartContent(request);

        if (isMultipart) {
            FileItemFactory factory = new DiskFileItemFactory();
            ServletFileUpload upload = new ServletFileUpload(factory);
            String FieldName = "";
            try {
                List items = upload.parseRequest(request);
                Iterator iterator = items.iterator();
                while (iterator.hasNext()) {
                     item = (FileItem) iterator.next();

                        if (fieldname.equals("description")) {
                            description = item.getString();
                        }
                    }
                    if (!item.isFormField()) {
                        filename = item.getName();
                        path = new File(path1 + File.separator);
                        if (!path.exists()) {
                            boolean status = path.mkdirs();
                        }
                        /* Start of code fro privilege */

                        File uploadedFile = new File(path + Filename);  // for copy file
                        item.write(uploadedFile);
                        }
                    } else {
                        f1 = item.getName();
                    }

                } // END OF WHILE
                response.sendRedirect("welcome.jsp");
            } catch (FileUploadException e) {
                e.printStackTrace();
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    }
}
Adulterate answered 4/2, 2014 at 10:0 Comment(1)
What do you mean by "Start of code fro privilege" (seems incomprehensible)? Please respond by editing (changing) your answer, not here in comments (without "Edit:", "Update:", or similar - the answer should appear as if it was written today).Lykins
P
-1

Use:

DiskFileUpload upload = new DiskFileUpload();

From this object you have to get the file items and fields, and then you can store into the server like the following:

String loc = "./webapps/prjct name/server folder/" + contentid + extension;
File uploadFile = new File(loc);
item.write(uploadFile);
Pyro answered 23/3, 2015 at 12:37 Comment(0)
C
-1

The simplest way I could come up with for files and input controls, without a billion libraries:

  <%
      if (request.getContentType() == null)
          return;
      // For input type=text controls
      String v_Text =
          (new BufferedReader(new InputStreamReader(request.getPart("Text1").getInputStream()))).readLine();

      // For input type=file controls
      InputStream inStr = request.getPart("File1").getInputStream();
      char charArray[] = new char[inStr.available()];
      new InputStreamReader(inStr).read(charArray);
      String contents = new String(charArray);
  %>
Coenesthesia answered 26/4, 2020 at 5:36 Comment(1)
What is <% for? ASP.NET (C#)? Can you clarify? Please respond by editing (changing) your answer, not here in comments (without "Edit:", "Update:", or similar - the answer should appear as if it was written today).Lykins
W
-2

HTML page

<html>
    <head>
        <title>File Uploading Form</title>
    </head>

    <body>
        <h3>File Upload:</h3>
        Select a file to upload: <br />
        <form action="UploadServlet" method="post"
              enctype="multipart/form-data">

            <input type="file" name="file" size="50" />
            <br />
            <input type="submit" value="Upload File" />
        </form>
    </body>
</html>

Servlet file

// Import required java libraries
import java.io.*;
import java.util.*;

import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.FileUploadException;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.io.output.*;

public class UploadServlet extends HttpServlet {

    private boolean isMultipart;
    private String filePath;
    private int maxFileSize = 50 * 1024;
    private int maxMemSize = 4 * 1024;
    private File file;

    public void init() {
        // Get the file location where it would be stored.
        filePath =
               getServletContext().getInitParameter("file-upload");
    }

    public void doPost(HttpServletRequest request,
                       HttpServletResponse response)
               throws ServletException, java.io.IOException {

        // Check that we have a file upload request
        isMultipart = ServletFileUpload.isMultipartContent(request);
        response.setContentType("text/html");
        java.io.PrintWriter out = response.getWriter();
        if (!isMultipart) {
            out.println("<html>");
            out.println("<head>");
            out.println("<title>Servlet upload</title>");
            out.println("</head>");
            out.println("<body>");
            out.println("<p>No file uploaded</p>");
            out.println("</body>");
            out.println("</html>");
            return;
        }

        DiskFileItemFactory factory = new DiskFileItemFactory();
        // Maximum size that will be stored in memory
        factory.setSizeThreshold(maxMemSize);
        // Location to save data that is larger than maxMemSize.
        factory.setRepository(new File("c:\\temp"));

        // Create a new file upload handler
        ServletFileUpload upload = new ServletFileUpload(factory);
        // maximum file size to be uploaded.
        upload.setSizeMax(maxFileSize);

        try {
            // Parse the request to get file items.
            List fileItems = upload.parseRequest(request);

            // Process the uploaded file items
            Iterator i = fileItems.iterator();

            out.println("<html>");
            out.println("<head>");
            out.println("<title>Servlet upload</title>");
            out.println("</head>");
            out.println("<body>");
            while (i.hasNext())
            {
                FileItem fi = (FileItem)i.next();
                if (!fi.isFormField())
                {
                     // Get the uploaded file parameters
                     String fieldName = fi.getFieldName();
                     String fileName = fi.getName();
                     String contentType = fi.getContentType();
                     boolean isInMemory = fi.isInMemory();
                     long sizeInBytes = fi.getSize();

                     // Write the file
                     if (fileName.lastIndexOf("\\") >= 0) {
                         file = new File(filePath +
                         fileName.substring(fileName.lastIndexOf("\\")));
                     }
                     else {
                         file = new File(filePath +
                         fileName.substring(fileName.lastIndexOf("\\") + 1));
                     }
                     fi.write(file);
                     out.println("Uploaded Filename: " + fileName + "<br>");
                }
            }
            out.println("</body>");
            out.println("</html>");
        }
        catch(Exception ex) {
            System.out.println(ex);
        }
    }

    public void doGet(HttpServletRequest request,
                        HttpServletResponse response)
            throws ServletException, java.io.IOException {

        throw new ServletException("GET method used with " +
                 getClass().getName() + ": POST method required.");
    }
}

File web.xml

Compile the above servlet UploadServlet and create the required entry in the web.xml file as follows.

<servlet>
   <servlet-name>UploadServlet</servlet-name>
   <servlet-class>UploadServlet</servlet-class>
</servlet>

<servlet-mapping>
   <servlet-name>UploadServlet</servlet-name>
   <url-pattern>/UploadServlet</url-pattern>
</servlet-mapping>
Waiver answered 15/7, 2016 at 7:14 Comment(0)
M
-4

Sending multiple files for file, we have to use enctype="multipart/form-data".

And to send multiple files, use multiple="multiple" in the input tag:

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="file" name="fileattachments"  multiple="multiple"/>
    <input type="submit" />
</form>
Mutualism answered 24/10, 2013 at 9:35 Comment(2)
How would we go about doing getPart("fileattachments") so we get an array of Parts instead? I don't think getPart for multiple files will work?Bigmouth
What do you mean by "Sending multiple files for file" (seems incomprehensible)? Please respond by editing (changing) your answer, not here in comments (without "Edit:", "Update:", or similar - the question/answer should appear as if it was written today).Lykins

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