How to compress a single file instead of a directory

[hncc]

i want to such as
ZipUtil.pack(“tmp/demo.txt”), new File("tmp/demo.zip"))
Instead of ZipUtil.pack(“tmp/demo”), new File("tmp/demo.zip"))

can u help me??

[sidgek]

If you need a simple way to zip a single file, you can create a method with something like this:

File zip = new File("tmp/demo.zip");
try (ZipOutputStream out = new ZipOutputStream(Files.newOutputStream(zip.toPath()))) {
  // Pass through to create empty zip archive.
}
ZipUtil.addEntry(zip, "tmp/demo.txt", new File("tmp/demo.txt"));

Don't think the above should be added to the library, just figured I'd offer a possible workaround.

[4asifm]

The ZipUtil.addEntry line fails with
Exception in thread "main" org.zeroturnaround.zip.ZipException: java.util.zip.ZipException: zip file is empty

java 1.8

[github24518]

    //1.create ZipEntrySource array,one or more files add to array.
    List<ZipEntrySource> files = new ArrayList<>();
    files.add(new FileSource("1.txt", new File("/tmp/1.txt")));
    files.add(new FileSource("2.txt", new File("/tmp/2.txt")));
    ZipEntrySource[] array = files.toArray(new ZipEntrySource[files.size()]);
   
    //2.pack
    ZipUtil.pack(array, new File( "/data/zip/demo.zip"));