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In my Java app I need to get some files and directories.

This is the program structure:

./main.java
./package1/guiclass.java
./package1/resources/resourcesloader.java
./package1/resources/repository/modules/   -> this is the dir I need to get
./package1/resources/repository/SSL-Key/cert.jks    -> this is the file I need to get

guiclass loads the resourcesloader class which will load my resources (directory and file).

As to the file, I tried

resourcesloader.class.getClass().getResource("repository/SSL-Key/cert.jks").toString()

in order to get the real path, but this way does not work.

I have no idea which path to use for the directory.

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13 Answers 13

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187

I had problems with using the getClass().getResource("filename.txt") method. Upon reading the Java docs instructions, if your resource is not in the same package as the class you are trying to access the resource from, then you have to give it relative path starting with '/'. The recommended strategy is to put your resource files under a "resources" folder in the root directory. So for example if you have the structure:

src/main/com/mycompany/myapp

then you can add a resources folder as recommended by maven in:

src/main/resources

furthermore you can add subfolders in the resources folder

src/main/resources/textfiles

and say that your file is called myfile.txt so you have

src/main/resources/textfiles/myfile.txt

Now here is where the stupid path problem comes in. Say you have a class in your com.mycompany.myapp package, and you want to access the myfile.txt file from your resource folder. Some say you need to give the:

"/main/resources/textfiles/myfile.txt" path

or

"/resources/textfiles/myfile.txt"

both of these are wrong. After I ran mvn clean compile, the files and folders are copied in the:

myapp/target/classes

folder. But the resources folder is not there, just the folders in the resources folder. So you have:

myapp/target/classes/textfiles/myfile.txt

myapp/target/classes/com/mycompany/myapp/*

so the correct path to give to the getClass().getResource("") method is:

"/textfiles/myfile.txt"

here it is:

getClass().getResource("/textfiles/myfile.txt")

This will no longer return null, but will return your class. It is strange to me, that the "resources" folder is not copied as well, but only the subfolders and files directly in the "resources" folder. It would seem logical to me that the "resources" folder would also be found under `"myapp/target/classes"

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  • 1
    In my target/classes, I can only see files under main resources and not test resources. Why?
    – Ava
    Commented Aug 10, 2015 at 21:59
  • @Ava Add test folder as your source folder and you are good to go!
    – Aakash Parashar
    Commented Mar 9, 2016 at 7:32
  • @Ava, sure it's late answer for you, but may help someone. You need to execute mvn clean test-compile command, as it goes after compile in Maven lifecycle. That will generate target/test-classes.
    – Alexander Radchenko
    Commented Aug 11, 2020 at 12:15
50

Supply the path relative to the classloader, not the class you're getting the loader from. For instance:

resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
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41

In the hopes of providing additional information for those who don't pick this up as quickly as others, I'd like to provide my scenario as it has a slightly different setup. My project was setup with the following directory structure (using Eclipse):

Project/
  src/                // application source code
    org/
      myproject/
        MyClass.java
  test/               // unit tests
  res/                // resources
    images/           // PNG images for icons
      my-image.png
    xml/              // XSD files for validating XML files with JAXB
      my-schema.xsd
    conf/             // default .conf file for Log4j
      log4j.conf
  lib/                // libraries added to build-path via project settings

I was having issues loading my resources from the res directory. I wanted all my resources separate from my source code (simply for managment/organization purposes). So, what I had to do was add the res directory to the build-path and then access the resource via:

static final ClassLoader loader = MyClass.class.getClassLoader();

// in some function
loader.getResource("images/my-image.png");
loader.getResource("xml/my-schema.xsd");
loader.getResource("conf/log4j.conf");

NOTE: The / is omitted from the beginning of the resource string because I am using ClassLoader.getResource(String) instead of Class.getResource(String).

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  • 2
    .+1 for suggesting ClassLoader
    – pyb
    Commented Nov 24, 2014 at 19:20
  • 1
    Best answer on this subject! The tip about not using / at the beginning of a resource path was key!
    – Kayhadrin
    Commented Sep 15, 2016 at 23:19
  • 1
    How do you add the res directory to the build-path in IntelliJ Idea?
    – Phil
    Commented Apr 5, 2023 at 16:08
10

When you use 'getResource' on a Class, a relative path is resolved based on the package the Class is in. When you use 'getResource' on a ClassLoader, a relative path is resolved based on the root folder.

If you use an absolute path, both 'getResource' methods will start at the root folder.

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  • Don't forget the effects of /
    – Gilberto
    Commented Feb 8, 2017 at 17:47
  • @nbeyer Why does using getResource with absolute path start at the root folder? Isn't the purpose of absolute path absolute?
    – atjua
    Commented Jun 7, 2017 at 21:12
10

@GianCarlo: You can try calling System property user.dir that will give you root of your java project and then do append this path to your relative path for example:

String root = System.getProperty("user.dir");
String filepath = "/path/to/yourfile.txt"; // in case of Windows: "\\path \\to\\yourfile.txt
String abspath = root+filepath;



// using above path read your file into byte []
File file = new File(abspath);
FileInputStream fis = new FileInputStream(file);
byte []filebytes = new byte[(int)file.length()];
fis.read(filebytes);
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  • 1
    OP wanted to locate a resource in the classpath (most likely in a jar generated during th build). This method does not provide a way to achieve that goal -- you cannot point a File to a resource inside a jar file, only to a proper entry in the file system (e.g. the jar itself).
    – Attila
    Commented Jan 16, 2013 at 14:47
  • I like the idea of getting "user.dir" to see the desired path. Thanks
    – Indrajeet Gour
    Commented Aug 6, 2019 at 10:46
6

For those using eclipse + maven. Say you try to access the file images/pic.jpg in src/main/resources. Doing it this way : 

ClassLoader loader = MyClass.class.getClassLoader();
File file = new File(loader.getResource("images/pic.jpg").getFile());

is perfectly correct, but may result in a null pointer exception. Seems like eclipse doesn't recognize the folders in the maven directory structure as source folders right away. By removing and the src/main/resources folder from the project's source folders list and putting it back (project>properties>java build path> source>remove/add Folder), I was able to solve this.

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resourcesloader.class.getClass()

Can be broken down to:

Class<resourcesloader> clazz = resourceloader.class;
Class<Class> classClass = clazz.getClass();

Which means you're trying to load the resource using a bootstrap class.

Instead you probably want something like:

resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()

If only javac warned about calling static methods on non-static contexts...

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2

In Order to obtain real path to the file you can try this:

URL fileUrl = Resourceloader.class.getResource("resources/repository/SSL-Key/cert.jks");
String pathToClass = fileUrl.getPath;

Resourceloader is classname here. "resources/repository/SSL-Key/cert.jks" is relative path to the file. If you had your guiclass in ./package1/java with rest of folder structure remaining, you would take "../resources/repository/SSL-Key/cert.jks" as relative path because of rules defining relative path.

This way you can read your file with BufferedReader. DO NOT USE THE STRING to identify the path to the file, because if you have spaces or some characters from not english alphabet in your path, you will get problems and the file will not be found.

BufferedReader bufferedReader = new BufferedReader(
                        new InputStreamReader(fileUrl.openStream()));
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1

Doe the following work?

resourcesloader.class.getClass().getResource("/package1/resources/repository/SSL-Key/cert.jks")

Is there a reason you can't specify the full path including the package?

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Going with the two answers as mentioned above. The first one

resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()

Should be one and same thing?

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0

I made a small modification on @jonathan.cone's one liner ( by adding .getFile() ) to avoid null pointer exception, and setting the path to data directory. Here's what worked for me :

String realmID = new java.util.Scanner(new java.io.File(RandomDataGenerator.class.getClassLoader().getResource("data/aa-qa-id.csv").getFile().toString())).next();
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Use this:

resourcesloader.class.getClassLoader().getResource("/path/to/file").**getPath();**
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One of the stable way to work across all OS would be toget System.getProperty("user.dir")

String filePath = System.getProperty("user.dir") + "/path/to/file.extension";

Path path = Paths.get(filePath);
if (Files.exists(path)) {
    return true;
}
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