fixing #111791: delegating extraction in shutil.unpack_archive to zipfile

shutil.unpack_archive fails, if file name contains '..'; zipfile handles everything correctly, i.e. in the same way than 'unzip'; let zipfile unpack archives, instead of reinventing the wheel here

Author: Sebastian Gassner

Lib/shutil.py

@@ -1252,27 +1252,10 @@ def _unpack_zipfile(filename, extract_dir):
     if not zipfile.is_zipfile(filename):
         raise ReadError("%s is not a zip file" % filename)
 
-    zip = zipfile.ZipFile(filename)
-    try:
-        for info in zip.infolist():
-            name = info.filename
-
-            # don't extract absolute paths or ones with .. in them
-            if name.startswith('/') or '..' in name:
-                continue
-
-            targetpath = os.path.join(extract_dir, *name.split('/'))
-            if not targetpath:
-                continue
-
-            _ensure_directory(targetpath)
-            if not name.endswith('/'):
-                # file
-                with zip.open(name, 'r') as source, \
-                        open(targetpath, 'wb') as target:
-                    copyfileobj(source, target)
-    finally:
-        zip.close()
+    with zipfile.ZipFile(filename) as zf:
+        # delegate extracting the archive to zipfile
+        _ensure_directory(extract_dir)
+        zf.extractall(extract_dir)
 
 def _unpack_tarfile(filename, extract_dir, *, filter=None):
     """Unpack tar/tar.gz/tar.bz2/tar.xz `filename` to `extract_dir`