'ZIp only contents of directory, exclude parent
I'm trying to zip the contents of a directory, without zipping the directory itself, however I can't find an obvious way to do this, and I'm extremely new to python so it's basically german to me. here's the code I'm using which successfully includes the parent as well as the contents:
#!/usr/bin/env python
import os
import zipfile
def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
for file in files:
ziph.write(os.path.join(root, file))
if __name__ == '__main__':
zipf = zipfile.ZipFile('Testing.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('android', zipf)
zipf.close()
I've tried modifying it, but always end up with incomprehensible errors.
Solution 1:[1]
write has second argument - name in archive, ie.
ziph.write(os.path.join(root, file), file)
EDIT:
#!/usr/bin/env python
import os
import zipfile
def zipdir(path, ziph):
length = len(path)
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
folder = root[length:] # path without "parent"
for file in files:
ziph.write(os.path.join(root, file), os.path.join(folder, file))
if __name__ == '__main__':
zipf = zipfile.ZipFile('Testing.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('android', zipf)
zipf.close()
Pathlib
solution
from pathlib import Path
def zipdir(parent_dir : str , ziph : ZipFile) -> None:
for file in Path(parent_dir).rglob('*'): # gets all child items
ziph.write(file, file.name)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Umar.H |