CODEWITHSUNDEEP
javapathjaxb
Dominik Reinert
Dominik Reinert

Reputation: 895

How to convert a ZipPath to File?

In order to work with Jaxb, I need to have a normal java.io.File Object. As I do not want to have legacy code in a quite new project, I want to use java.nio.file.Path objects.

As gradle resolves dependencies in jar files, I need to handle them as com.sun.nio.zipfs.ZipPath. Now the thing is, that I am using these files only for codegeneration, and do not necessarily want to unpack them.

Unfortunately, the ZipPath.toFile() method throws an UnsupportedOperationException, so I cannot convert the Path to a File, which is necessary in order to use the JaxbUnmarshaller, to validate the correct layout of the file and to convert it into an actual runtime object.

I tried:

How can I get a File from a ZipPath without unzipping it?

I suppose it is possible via the ZipFileSystem, but I dont get it.

Upvotes: 6

Views: 4210

Answers (2)

Thomas Eitzenberger
Thomas Eitzenberger

Reputation: 1

With newer Java versions, you can skip IOUtils and directly use the Files class for copying the files out of the ZIP archive to a temp folder. Make sure to replace PREFIX and SUFFIX with some values that make sense for your application of the code.

private Path remapZipPath(Path path) {
    final File tempFile = File.createTempFile("PREFIX" + UUID.randomUUID(), "SUFFIX");
    tempFile.deleteOnExit();
    try (FileOutputStream out = new FileOutputStream(tempFile)) {
      Files.copy(path, out);
    }
    return tempFile.toPath();
  }

Upvotes: 0

wheel.r
wheel.r

Reputation: 211

It's not lovely but I recently solved this problem like so:

InputStream is = Files.newInputStream(zipPath);
final File tempFile = File.createTempFile("PREFIX", "SUFFIX");
tempFile.deleteOnExit();
try (FileOutputStream out = new FileOutputStream(tempFile))
{
    IOUtils.copy(in, out);
}
// do something with tempFile

Upvotes: 8

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