打开 zip 存档中的特定文件夹并迭代目录结构

Question

We can determine the folder structure of a zip archive using python as follows (we can do the same in Java also):

with zipfile.ZipFile('path to file', 'r') as zipobj:
    for item in zipobj.infolist():
        print(item.filename)

However, is it possible to determine the folder structure of a particular folder inside the zip archive and iterate through all files/folders inside that folder (similar to the path object) only? (instead of iterating all files/folders inside the zip archive as shown in the previous code example)

Answer 1

You can read the file as ZipFile and then iterate over all the entries which are folders. Here is a code snippet in Java:

ZipFile zip = new ZipFile(file);
Enumeration<? extends ZipEntry> entries = zip.entries();
while (entries.hasMoreElements()) {
  ZipEntry entry = entries.nextElement();
  if (entry.isDirectory()) {
      // Code goes here
  }
}

Answer 2

Reading a ZIP structure as Path components is very simple with Java NIO as there are built in handlers for ZIP filesystems - see FileSystems.newFileSystem(zip) , and the Path objects it provides work with other NIO classes such as Files .

For example this scans a zip starting from folder org/apache , and you can substitute any other filters needed treating the ZIP structure just like any other disc filesystem:

try (FileSystem fs = FileSystems.newFileSystem(zip)) {
    Path root = fs.getPath("org/apache");
    try(Stream<Path> stream = Files.find(root, Integer.MAX_VALUE, (p,a) -> true)) {
        stream.forEach(System.out::println);
    }
}