Java 中具有相对路径的开放资源

在我的 Java 应用程序中,我需要获得一些文件和目录。

这是程序结构:

./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加载将加载我的资源(目录和文件)的 resource cesloader 类。

至于文件,我试过了

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

为了得到真正的路径,但这种方式不工作。

我不知道该目录使用哪个路径。

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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?

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();
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...

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.

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"

@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);

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).

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?

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.

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();

Use this:

resourcesloader.class.getClassLoader().getResource("/path/to/file").**getPath();**

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()));

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;
}