Spring getResources（）->获取每个资源的路径会在getFile（）处引发异常

Question

I need to get a list of paths to files for a given pattern.

Resource[] resources = context.getResources("classpath*:**/i18n/**/*.properties"); 
...
File file = resources[i].getFile(); /// throws exception
String path = file.getPath(); 

This works when run from IntelliJ, but fails when run from jar file.

java.io.FileNotFoundException: URL [jar:file:/home/mariusz/.../.../build/libs/app-3.0.0-SNAPSHOT-all.jar!/i18n/test.properties] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/home/mariusz/../../build/libs/app-3.0.0-SNAPSHOT-all.jar!/i18n/test.properties

It returns proper resource, but resource.getFile() throws exception.

How to get path from Resource in a proper way?

Answer 1

File is a file system file on disk. A (read-only) resource is on the class path, possibly packed in a jar/ear/war (zip archives).

The basic java access is via the class loader or class:

URL url = MyApp.class.getResource("..."); // When in a jar: "jar:file:..."
InputStream in = MyApp.class.getResourceAsStream("...");

So use an InputStream instead of a File.

With a Spring Resource :

Resource resource = ...;
InputStream in = resource.getURL().openStream();

Path path = Paths.get("/home/mariusz/test.properties");
Files.copy(in, path); // Copies the resource to a real file.

---

A Path inside a resource:

Path res = Paths.get(resource.getURI());

Answer 2

You are using the getFile() method which return a java.io.File. java.io.File represents a file on a file system. The Jar is a java.io.File. But anything within that file is beyond the reach of java.io.File, unless uncompressed.

Use resource.getInputStream() instead.