Can't Access Resources In Executable Jar

Question

Can someone please point out what I'm doing wrong here.

I have a small weather app that generates and sends an HTML email. With my code below, everything works fine when I run it from Eclipse. My email gets generated, it's able to access my image resources and it sends the email with the included attachment.

However, when I build the executable jar by running mvn install and run the jar using java -jar NameOfMyJar.jar I get java.io.FileNotFound Exceptions for my image resource.

I know that I have to be doing something wrong with how I'm accessing my image resources, I just don't understand why it works fine when it's not packaged, but bombs out whenever I package it into a jar.

Any advice is very much appreciated it.

---

My project layout

---

How I'm accessing my image resource

//Setup the ATTACHMENTS
        MimeBodyPart attachmentsPart = new MimeBodyPart();
        try {
            attachmentsPart.attachFile("resources/Cloudy_Day.png");
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }   

---

The StackTrace

    Exception in thread "main" java.lang.RuntimeException: javax.mail.MessagingException: IOException while sending message;
  nested exception is:
    java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
    at Utilities.SendEmailUsingGmailSMTP.SendTheEmail(SendEmailUsingGmailSMTP.java:139)
    at Utilities.SendEmailUsingGmailSMTP.SendWeatherEmail(SendEmailUsingGmailSMTP.java:66)
    at Weather.Main.start(Main.java:43)
    at Weather.Main.main(Main.java:23)
Caused by: javax.mail.MessagingException: IOException while sending message;
  nested exception is:
    java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
    at com.sun.mail.smtp.SMTPTransport.sendMessage(SMTPTransport.java:1167)
    at javax.mail.Transport.send0(Transport.java:195)
    at javax.mail.Transport.send(Transport.java:124)
    at Utilities.SendEmailUsingGmailSMTP.SendTheEmail(SendEmailUsingGmailSMTP.java:134)
    ... 3 more
Caused by: java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
    at java.io.FileInputStream.open(Native Method)
    at java.io.FileInputStream.<init>(FileInputStream.java:146)
    at javax.activation.FileDataSource.getInputStream(FileDataSource.java:97)
    at javax.activation.DataHandler.writeTo(DataHandler.java:305)
    at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:1485)
    at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:865)
    at javax.mail.internet.MimeMultipart.writeTo(MimeMultipart.java:462)
    at com.sun.mail.handlers.multipart_mixed.writeTo(multipart_mixed.java:103)
    at javax.activation.ObjectDataContentHandler.writeTo(DataHandler.java:889)
    at javax.activation.DataHandler.writeTo(DataHandler.java:317)
    at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:1485)
    at javax.mail.internet.MimeMessage.writeTo(MimeMessage.java:1773)
    at com.sun.mail.smtp.SMTPTransport.sendMessage(SMTPTransport.java:1119)
    ... 6 more

Enugu

try to use /resources/Cloudy_Day.png (/ in front of the path) – Gladisgladney

what is your current working directory ? – Contradictory

Thanks everyone for the suggestions! When I get home tonight, I'll give them a shot and let y'all know how it goes. – Enugu

Answer 1

Others are correct with the use of getResourceAsStream, but the path is a little tricky. You see the little package icon in the resources folder? That signifies that all the files in the resource folder will be put into the root of the classpath. Just like all the packages in src/main/java are put in the root. So you would take out the resources from the path

InputStream is = getClass().getResourceAsStream("/Cloudy_Day.png");

An aside: Maven has a file structure conventions. Class path resources are usually put into src/main/resources. If you create a resources dir in the src/main, Eclipse should automatically pick it up, and create the little package icon for a path src/main/resource that you should see in the project explorer. These files would also go to the root and could be accessed the same way. I would fix the file structure to follow this convention.

Note: A MimeBodyPart, can be Constructed from an InputStream (As suggested by Bill Shannon, this is incorrect). As mentioned in his comment below

"You can also attach the data using"

mbp.setDataHandler(new DataHandler(new ByteArrayDataSource(
          this.getClass().getResourceAsStream("/Cloudy_Day.png", "image/png"))));

Faradize

Note: constructing a MimeBodyPart from a stream doesn't do what you think it does. It reads an entire MIME part - headers and data - not just the data you want in an attachment. – Para

@BillShannon Gotcha. Thanks for the heads up. I'll scratch that out. I'm not familiar with this API. How would you suggest handling the OPs use case, using an InputStream? – Faradize

@peeskillet and BillShannon thank you both so much!!! I modified my project structure and used the setDataHandler() method to access my resource file and now everything is working. Sorry it took me a couple days to get back to you, but I've been swamped lately. Anyway, thank you both very much for you help. – Enugu

Answer 2

You can't access resources inside a JAR file as a File, only read them as an InputStream: getResourceAsStream().

As the MimeBodyPart has no attach() method for an InputStream, the easiest way should be to read your resources and write them to temp files, then attach these files.

Seigniorage

You can also attach the data using mbp.setDataHandler(new DataHandler(new ByteArrayDataSource(this.getClass().getResourceAsStream("/Cloudy_Day.png", "image/png")))); – Para

Answer 3

Try this

new MimeBodyPart().attachFile(new File(this.getClass().getClassLoader().getResource("resources/Cloudy_Day.png").toURI());

Bathulda

Answer 4

I don't know if this will help anyone or not. But, I have a similar case as the OP and I solved the case by finding the file in the classpath using recursive function. The idea is so that when another developer decided to move the resources into another folder/path. It will still be found as long as the name is still the same.

For example, in my work we usually put our resources outside the jar, and then we add said resources path into our classpath, so here the classpath of the resources will be different depending on where it is located.

That's where my code comes to work, no matter where the file is put, as long as it's in the classpath it will be found.

Here is an example of my code in action:

import java.io.File;

public class FindResourcesRecursive {

    public File findConfigFile(String paths, String configFilename) {
        for (String p : paths.split(File.pathSeparator)) {
            File result = findConfigFile(new File(p), configFilename);
            if (result != null) {
                return result;
            }
        }
        return null;
    }

    private File findConfigFile(File path, String configFilename) {
        if (path.isDirectory()) {
            String[] subPaths = path.list();
            if (subPaths == null) {
                return null;
            }
            for (String sp : subPaths) {
                File subPath = new File(path.getAbsoluteFile() + "/" + sp);
                File result = findConfigFile(subPath, configFilename);
                if (result != null && result.getName().equalsIgnoreCase(configFilename)) {
                    return result;
                }
            }
            return null;
        } else {
            File file = path;
            if (file.getName().equalsIgnoreCase(configFilename)) {
                return file;
            }
            return null;
        }
    }

}

Here I have a test case that is coupled with a file "test.txt" in my test/resources folder. The content of said file is:

A sample file

Now, here is my test case:

import org.junit.Test;

import java.io.*;

import static org.junit.Assert.fail;

public class FindResourcesRecursiveTest {

    @Test
    public void testFindFile() {
        // Here in the test resources I have a file "test.txt"
        // Inside it is a string "A sample file"
        // My Unit Test will use the class FindResourcesRecursive to find the file and print out the results.
        File testFile = new FindResourcesRecursive().findConfigFile(
                System.getProperty("java.class.path"),
                "test.txt"
        );

        try (FileInputStream is = new FileInputStream(testFile)) {
            int i;
            while ((i = is.read()) != -1) {
                System.out.print((char) i);
            }
            System.out.println();
        } catch (IOException e) {
            fail();
        }

    }
}

Now, if you run this test, it will print out "A sample file" and the test will be green.

Catamenia