Java Jar file: use resource errors: URI is not hierarchical
JavaDeploymentResourcesExecutable JarJava Problem Overview
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
//....
}
Java Solutions
Solution 1 - Java
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.
Solution 2 - Java
If for some reason you really need to create a java.io.File
object to point to a resource inside of a Jar file, the answer is here: https://stackoverflow.com/a/27149287/155167
File f = new File(getClass().getResource("/MyResource").toExternalForm());
Solution 3 - Java
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/
Solution 4 - Java
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
Solution 5 - Java
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.
Solution 6 - Java
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.