Java Jar 文件：使用资源错误：URI 不分层

Question

I have deployed my app to jar file. When I need to copy data from one file of resource to outside of jar file, I do this code:

URL resourceUrl = getClass().getResource("/resource/data.sav");
File src = new File(resourceUrl.toURI()); //ERROR HERE
File dst = new File(CurrentPath()+"data.sav");  //CurrentPath: path of jar file don't include jar file name
FileInputStream in = new FileInputStream(src);
FileOutputStream out = new FileOutputStream(dst);
 // some excute code here

The error I have met is: URI is not hierarchical . this error I don't meet when run in IDE.

If I change above code as some help on other post on StackOverFlow:

InputStream in = Model.class.getClassLoader().getResourceAsStream("/resource/data.sav");
File dst = new File(CurrentPath() + "data.sav");
FileOutputStream out = new FileOutputStream(dst);
//....
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) { //NULL POINTER EXCEPTION
  //....
}

Answer 1

You cannot do this

File src = new File(resourceUrl.toURI()); //ERROR HERE

it is not a file! When you run from the ide you don't have any error, because you don't run a jar file. In the IDE classes and resources are extracted on the file system.

But you can open an InputStream in this way:

InputStream in = Model.class.getClassLoader().getResourceAsStream("/data.sav");

Remove "/resource" . Generally the IDEs separates on file system classes and resources. But when the jar is created they are put all together. So the folder level "/resource" is used only for classes and resources separation.

When you get a resource from classloader you have to specify the path that the resource has inside the jar, that is the real package hierarchy.

Answer 2

如果由于某种原因你真的需要创建一个java.io.File对象来指向 Jar 文件中的资源，答案在这里： https : //stackoverflow.com/a/27149287/155167

File f = new File(getClass().getResource("/MyResource").toExternalForm());

Answer 3

Here is a solution for Eclipse RCP / Plugin developers:

Bundle bundle = Platform.getBundle("resource_from_some_plugin");
URL fileURL = bundle.getEntry("files/test.txt");
File file = null;
try {
   URL resolvedFileURL = FileLocator.toFileURL(fileURL);

   // We need to use the 3-arg constructor of URI in order to properly escape file system chars
   URI resolvedURI = new URI(resolvedFileURL.getProtocol(), resolvedFileURL.getPath(), null);
   File file = new File(resolvedURI);
} catch (URISyntaxException e1) {
    e1.printStackTrace();
} catch (IOException e1) {
    e1.printStackTrace();
}

It's very important to use FileLocator.toFileURL(fileURL) rather than resolve(fileURL) , cause when the plugin is packed into a jar this will cause Eclipse to create an unpacked version in a temporary location so that the object can be accessed using File. For instance, I guess Lars Vogel has an error in his article - http://blog.vogella.com/2010/07/06/reading-resources-from-plugin/

Answer 4

While I stumbled upon this problem myself I'd like to add another option (to the otherwise perfect explanation from @dash1e):

Export the plugin as a folder (not a jar) by adding:

Eclipse-BundleShape: dir

to your MANIFEST.MF .

At least when you export your RCP app with the export wizard (based on a *.product ) file this gets respected and will produce a folder.

Answer 5

In addition to the general answers, you can get "URI is not hierarchical" from Unitils library attempting to load a dataset off a .jar file. It may happen when you keep datasets in one maven submodule, but actual tests in another.

There is even a bug UNI-197 filed.

Answer 6

I got a similiar issues before, and I used the code:

new File(new URI(url.toString().replace(" ","%20")).getSchemeSpecificPart());

instead of the code :

new File(new URI(url.toURI())

to solve the problem