open resource with relative path in java

Question

In my java app I need to get some files and dirs.

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`

gui class loads the resourcesloader class which will load my resources (dir and file)

About the file I tried with

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

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

About the dir I have no idea how to do...

Suggestions? Thanks

Answer 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?

Answer 2

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();

Answer 3

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...

Answer 4

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.

Answer 5

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. I hope this helps somebody. 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"

Answer 6

@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);