Finding a file in zipentry java

问题

I am trying to find a file within a zip file and get it as an InputStream. So this is what I am doing to get it so far and I am not certain if I am doing it correctly.

Here is a sample as the original is slightly longer but this is the main component...

public InputStream Search_Image(String file_located, ZipInputStream zip) 
    throws IOException {
    for (ZipEntry zip_e = zip.getNextEntry(); zip_e != null ; zip_e = zip.getNextEntry()) {
        if (file_located.equals(zip_e.getName())) {
            return zip;
        }
        if (zip_e.isDirectory()) {
            Search_Image(file_located, zip); 
        }
    }
    return null;
}

Now the main problem I am facing is that The ZipInputStream in Search_Image is the same as the original component of the ZipInputStream...

if(zip_e.isDirectory()) {
    //"zip" is the same as the original I need a change here to find folders again.
    Search_Image(file_located, zip); 
}

Now for the question, how do you get the ZipInputStream as the new zip_entry? Also please add in if I did anything wrong in my method as my logic with this class is still lacking.

回答1:

You should use the class ZipFile without worrying yourself with an input stream if you don't need it yet.

ZipFile file = new ZipFile("file.zip");
ZipInputStream zis = searchImage("foo.png", file);

public InputStream searchImage(String name, ZipFile file) {
  for (ZipEntry e : Collections.list(file.entries())) {
    if (e.getName().endsWith(name)) {
      return file.getInputStream(e);
    }
  }
  return null;
}

Some facts:

you should follow conventions for naming methods and variables in your code (Search_Image is not fine, searchImage is)
directories in zip files does not contain any file, they are just entries like everything else so you shouldn't try to recurse into them)
you should compare the name you provide by using endsWith(name) because the file could be inside a folder and a filename inside a zip always contains the path

回答2:

Accessing to a zip entry using ZipInputStream is clearly not the way to do it as you will need to iterate over the entries to find it which is not a scalable approach because the performance will depend on total amount of entries in your zip file.

To get the best possible performances, you need to use a ZipFile in order to access directly to an entry thanks to the method getEntry(name) whatever the size of your archive.

public InputStream searchImage(String name, ZipFile zipFile) throws IOException {
    // Get the entry by its name
    ZipEntry entry = zipFile.getEntry(name);
    if (entry != null) {
        // The entry could be found
        return zipFile.getInputStream(entry);
    }
    // The entry could not be found
    return null;
}

Please note that the name to provide here is the relative path of your image in the archive using / as path separator so if you want to access to foo.png that is in the directory bar, the expected name will be bar/foo.png.

回答3:

Here is my take on this:

ZipFile zipFile = new ZipFile(new File("/path/to/zip/file.zip));
InputStream inputStream = searchWithinZipArchive("findMe.txt", zipFile);

public InputStream searchWithinZipArchive(String name, ZipFile file) throws Exception {
  Enumeration<? extends ZipEntry> entries = file.entries();
  while(entries.hasMoreElements()){
     ZipEntry zipEntry = entries.nextElement();
      if(zipEntry.getName().toLowerCase().endsWith(name)){
             return file.getInputStream(zipEntry);
      }
  }
  return null;
}

来源：https://stackoverflow.com/questions/11123528/finding-a-file-in-zipentry-java

标签

java

zipinputstream