Converting MultipartFile to java.io.File without copying to local machine
Asked Answered
L

4

14

I have a Java Spring MVC web application. From client, through AngularJS, I am uploading a file and posting it to Controller as webservice.

In my Controller, I am gettinfg it as MultipartFile and I can copy it to local machine.

But I want to upload the file to Amazone S3 bucket. So I have to convert it to java.io.File. Right now what I am doing is, I am copying it to local machine and then uploading to S3 using jets3t.

Here is my way of converting in controller

MultipartHttpServletRequest mRequest=(MultipartHttpServletRequest)request;

Iterator<String> itr=mRequest.getFileNames();
        while(itr.hasNext()){
            MultipartFile mFile=mRequest.getFile(itr.next());
            String fileName=mFile.getOriginalFilename();

            fileLoc="/home/mydocs/my-uploads/"+date+"_"+fileName; //date is String form of current date.

Then I am using FIleCopyUtils of SpringFramework

File newFile = new File(fileLoc);

                  // if the directory does not exist, create it
                  if (!newFile.getParentFile().exists()) {
                    newFile.getParentFile().mkdirs();  
                  }
                FileCopyUtils.copy(mFile.getBytes(), newFile);

So it will create a new file in the local machine. That file I am uplaoding in S3

S3Object fileObject = new S3Object(newFile);

s3Service.putObject("myBucket", fileObject);

It creates file in my local system. I don't want to create.

Without creating a file in local system, how to convert a MultipartFIle to java.io.File?

Leer answered 13/1, 2014 at 10:34 Comment(0)
H
9

MultipartFile, by default, is already saved on your server as a file when user uploaded it. From that point - you can do anything you want with this file. There is a method that moves that temp file to any destination you want. http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartFile.html#transferTo(java.io.File)

But MultipartFile is just API, you can implement any other MultipartResolver http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartResolver.html

This API accepts input stream and you can do anything you want with it. Default implementation (usually commons-multipart) saves it to temp dir as a file.

But other problem stays here - if S3 API accepts a file as a parameter - you cannot do anything with this - you need a real file. If you want to avoid creating files at all - create you own S3 API.

Hashim answered 13/1, 2014 at 10:53 Comment(3)
Only problem is creating the file in local machine. What I am doing now is, after uploading in amazon, I am deleting from my local machine.Leer
What inspires that problem ? S3 java-API ? Or what ? Do you have ability to send input stream to S3 ?Hashim
This is incorrect in regard to S3 API accepting only file parameter, it accepts also InputStreamStream
M
5

The question is already more than one year old, so I'm not sure if the jets35 link provided by the OP had the following snippet at that time.

If your data isn't a File or String you can use any input stream as a data source, but you must manually set the Content-Length.

// Create an object containing a greeting string as input stream data.
String greeting = "Hello World!";

S3Object helloWorldObject = new S3Object("HelloWorld2.txt");

ByteArrayInputStream greetingIS = new ByteArrayInputStream(greeting.getBytes());

helloWorldObject.setDataInputStream(greetingIS);
helloWorldObject.setContentLength(
    greeting.getBytes(Constants.DEFAULT_ENCODING).length);
helloWorldObject.setContentType("text/plain");

s3Service.putObject(testBucket, helloWorldObject);

It turns out you don't have to create a local file first. As @Boris suggests you can feed the S3Object with the Data Input Stream, Content Type and Content Length you'll get from MultipartFile.getInputStream(), MultipartFile.getContentType() and MultipartFile.getSize() respectively.

Monteverdi answered 9/4, 2015 at 11:25 Comment(0)
M
0

Instead of copying it to your local machine, you can just do this and replace the file name with this:

File newFile = new File(multipartFile.getOriginalName());

This way, you don't have to have a local destination create your file

Mixologist answered 17/4, 2022 at 8:29 Comment(0)
F
-3

if you are try to use in httpentity check my answer here

https://mcmap.net/q/186416/-how-can-i-make-a-multipart-form-data-post-request-using-java
Fustian answered 17/6, 2021 at 15:57 Comment(0)

© 2022 - 2024 — McMap. All rights reserved.