如何从 Java jar 文件中读取资源文件？

Question

I'm trying to access an XML file within a jar file, from a separate jar that's running as a desktop application. I can get the URL to the file I need, but when I pass that to a FileReader (as a String) I get a FileNotFoundException saying "The file name, directory name, or volume label syntax is incorrect."

As a point of reference, I have no trouble reading image resources from the same jar, passing the URL to an ImageIcon constructor. This seems to indicate that the method I'm using to get the URL is correct.

URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );

Inside the ServicesLoader class I have

XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));

What's wrong with using this technique to read the XML file?

Answer 1

Looks like you want to use java.lang.Class.getResourceAsStream(String) , see

https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-

Answer 2

You don't say if this is a desktop or web app. I would use the getResourceAsStream() method from an appropriate ClassLoader if it's a desktop or the Context if it's a web app.

Answer 3

It looks as if you are using the URL.toString result as the argument to the FileReader constructor. URL.toString is a bit broken, and instead you should generally use url.toURI().toString() . In any case, the string is not a file path.

Instead, you should either:

• Pass the URL to ServicesLoader and let it call openStream or similar.
• Use Class.getResourceAsStream and just pass the stream over, possibly inside an InputSource . (Remember to check for nulls as the API is a bit messy.)

Answer 4

The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to

xr.parse( new InputSource( filename ));

works just fine.

Answer 5

I'd like to point out that one issues is what if the same resources are in multiple jar files. Let's say you want to read /org/node/foo.txt, but not from one file, but from each and every jar file.

I have run into this same issue several times before. I was hoping in JDK 7 that someone would write a classpath filesystem, but alas not yet.

Spring has the Resource class which allows you to load classpath resources quite nicely.

I wrote a little prototype to solve this very problem of reading resources form multiple jar files. The prototype does not handle every edge case, but it does handle looking for resources in directories that are in the jar files.

I have used Stack Overflow for quite sometime. This is the second answer that I remember answering a question so forgive me if I go too long (it is my nature).

This is a prototype resource reader. The prototype is devoid of robust error checking.

I have two prototype jar files that I have setup.

 <pre>
         <dependency>
              <groupId>invoke</groupId>
              <artifactId>invoke</artifactId>
              <version>1.0-SNAPSHOT</version>
          </dependency>

          <dependency>
               <groupId>node</groupId>
               <artifactId>node</artifactId>
               <version>1.0-SNAPSHOT</version>
          </dependency>

The jar files each have a file under /org/node/ called resource.txt.

This is just a prototype of what a handler would look like with classpath:// I also have a resource.foo.txt in my local resources for this project.

It picks them all up and prints them out.

    package com.foo;

    import java.io.File;
    import java.io.FileReader;
    import java.io.InputStreamReader;
    import java.io.Reader;
    import java.net.URI;
    import java.net.URL;
    import java.util.Enumeration;
    import java.util.zip.ZipEntry;
    import java.util.zip.ZipFile;

    /**
    * Prototype resource reader.
    * This prototype is devoid of error checking.
    *
    *
    * I have two prototype jar files that I have setup.
    * <pre>
    *             <dependency>
    *                  <groupId>invoke</groupId>
    *                  <artifactId>invoke</artifactId>
    *                  <version>1.0-SNAPSHOT</version>
    *              </dependency>
    *
    *              <dependency>
    *                   <groupId>node</groupId>
    *                   <artifactId>node</artifactId>
    *                   <version>1.0-SNAPSHOT</version>
    *              </dependency>
    * </pre>
    * The jar files each have a file under /org/node/ called resource.txt.
    * <br />
    * This is just a prototype of what a handler would look like with classpath://
    * I also have a resource.foo.txt in my local resources for this project.
    * <br />
    */
    public class ClasspathReader {

        public static void main(String[] args) throws Exception {

            /* This project includes two jar files that each have a resource located
               in /org/node/ called resource.txt.
             */


            /* 
              Name space is just a device I am using to see if a file in a dir
              starts with a name space. Think of namespace like a file extension 
              but it is the start of the file not the end.
            */
            String namespace = "resource";

            //someResource is classpath.
            String someResource = args.length > 0 ? args[0] :
                    //"classpath:///org/node/resource.txt";   It works with files
                    "classpath:///org/node/";                 //It also works with directories

            URI someResourceURI = URI.create(someResource);

            System.out.println("URI of resource = " + someResourceURI);

            someResource = someResourceURI.getPath();

            System.out.println("PATH of resource =" + someResource);

            boolean isDir = !someResource.endsWith(".txt");


            /** Classpath resource can never really start with a starting slash.
             * Logically they do, but in reality you have to strip it.
             * This is a known behavior of classpath resources.
             * It works with a slash unless the resource is in a jar file.
             * Bottom line, by stripping it, it always works.
             */
            if (someResource.startsWith("/")) {
                someResource = someResource.substring(1);
            }

              /* Use the ClassLoader to lookup all resources that have this name.
                 Look for all resources that match the location we are looking for. */
            Enumeration resources = null;

            /* Check the context classloader first. Always use this if available. */
            try {
                resources = 
                    Thread.currentThread().getContextClassLoader().getResources(someResource);
            } catch (Exception ex) {
                ex.printStackTrace();
            }

            if (resources == null || !resources.hasMoreElements()) {
                resources = ClasspathReader.class.getClassLoader().getResources(someResource);
            }

            //Now iterate over the URLs of the resources from the classpath
            while (resources.hasMoreElements()) {
                URL resource = resources.nextElement();


                /* if the resource is a file, it just means that we can use normal mechanism
                    to scan the directory.
                */
                if (resource.getProtocol().equals("file")) {
                    //if it is a file then we can handle it the normal way.
                    handleFile(resource, namespace);
                    continue;
                }

                System.out.println("Resource " + resource);

               /*

                 Split up the string that looks like this:
                 jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
                 into
                    this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
                 and this
                     /org/node/
                */
                String[] split = resource.toString().split(":");
                String[] split2 = split[2].split("!");
                String zipFileName = split2[0];
                String sresource = split2[1];

                System.out.printf("After split zip file name = %s," +
                        " \nresource in zip %s \n", zipFileName, sresource);


                /* Open up the zip file. */
                ZipFile zipFile = new ZipFile(zipFileName);


                /*  Iterate through the entries.  */
                Enumeration entries = zipFile.entries();

                while (entries.hasMoreElements()) {
                    ZipEntry entry = entries.nextElement();
                    /* If it is a directory, then skip it. */
                    if (entry.isDirectory()) {
                        continue;
                    }

                    String entryName = entry.getName();
                    System.out.printf("zip entry name %s \n", entryName);

                    /* If it does not start with our someResource String
                       then it is not our resource so continue.
                    */
                    if (!entryName.startsWith(someResource)) {
                        continue;
                    }


                    /* the fileName part from the entry name.
                     * where /foo/bar/foo/bee/bar.txt, bar.txt is the file
                     */
                    String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
                    System.out.printf("fileName %s \n", fileName);

                    /* See if the file starts with our namespace and ends with our extension.        
                     */
                    if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {


                        /* If you found the file, print out 
                           the contents fo the file to System.out.*/
                        try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
                            StringBuilder builder = new StringBuilder();
                            int ch = 0;
                            while ((ch = reader.read()) != -1) {
                                builder.append((char) ch);

                            }
                            System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
                        } catch (Exception ex) {
                            ex.printStackTrace();
                        }
                    }

                    //use the entry to see if it's the file '1.txt'
                    //Read from the byte using file.getInputStream(entry)
                }

            }


        }

        /**
         * The file was on the file system not a zip file,
         * this is here for completeness for this example.
         * otherwise.
         *
         * @param resource
         * @param namespace
         * @throws Exception
         */
        private static void handleFile(URL resource, String namespace) throws Exception {
            System.out.println("Handle this resource as a file " + resource);
            URI uri = resource.toURI();
            File file = new File(uri.getPath());


            if (file.isDirectory()) {
                for (File childFile : file.listFiles()) {
                    if (childFile.isDirectory()) {
                        continue;
                    }
                    String fileName = childFile.getName();
                    if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {

                        try (FileReader reader = new FileReader(childFile)) {
                            StringBuilder builder = new StringBuilder();
                            int ch = 0;
                            while ((ch = reader.read()) != -1) {
                                builder.append((char) ch);

                            }
                            System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
                        } catch (Exception ex) {
                            ex.printStackTrace();
                        }

                    }

                }
            } else {
                String fileName = file.getName();
                if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {

                    try (FileReader reader = new FileReader(file)) {
                        StringBuilder builder = new StringBuilder();
                        int ch = 0;
                        while ((ch = reader.read()) != -1) {
                            builder.append((char) ch);

                        }
                        System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
                    } catch (Exception ex) {
                        ex.printStackTrace();
                    }

                }

            }
        }

    }


   

You can see a fuller example here with the sample output.

Answer 6

Here's a sample code on how to read a file properly inside a jar file (in this case, the current executing jar file)

Just change executable with the path of your jar file if it is not the current running one.

Then change the filePath to the path of the file you want to use inside the jar file. IE if your file is in

someJar.jar\\img\\test.gif

. Set the filePath to "img\\test.gif"

File executable = new File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile jar = new JarFile(executable);
InputStream fileInputStreamReader = jar.getInputStream(jar.getJarEntry(filePath));
byte[] bytes = new byte[fileInputStreamReader.available()];

int sizeOrig = fileInputStreamReader.available();
int size = fileInputStreamReader.available();
int offset = 0;
while (size != 0){
    fileInputStreamReader.read(bytes, offset, size);
    offset = sizeOrig - fileInputStreamReader.available();
    size = fileInputStreamReader.available();
}

Answer 7

I have 2 CSV files that I use to read data. The java program is exported as a runnable jar file. When you export it, you will figure out it doesn't export your resources with it.

I added a folder under project called data in eclipse. In that folder i stored my csv files.

When I need to reference those files I do it like this...

private static final String ZIP_FILE_LOCATION_PRIMARY = "free-zipcode-database-Primary.csv";
private static final String ZIP_FILE_LOCATION = "free-zipcode-database.csv";

private static String getFileLocation(){
    String loc = new File("").getAbsolutePath() + File.separatorChar +
        "data" + File.separatorChar;
    if (usePrimaryZipCodesOnly()){              
        loc = loc.concat(ZIP_FILE_LOCATION_PRIMARY);
    } else {
        loc = loc.concat(ZIP_FILE_LOCATION);
    }
    return loc;
}

Then when you put the jar in a location so it can be ran via commandline, make sure that you add the data folder with the resources into the same location as the jar file.

Answer 8

Outside of your technique, why not use the standard Java JarFile class to get the references you want? From there most of your problems should go away.

Answer 9

If you use resources extensively, you might consider using Commons VFS .

Also supports: * Local Files * FTP, SFTP * HTTP and HTTPS * Temporary Files "normal FS backed) * Zip, Jar and Tar (uncompressed, tgz or tbz2) * gzip and bzip2 * resources * ram - "ramdrive" * mime

There's also JBoss VFS - but it's not much documented.