Extraire le fichier sources.zip dans le répertoire "path" ?

Réponse

zipfile.ZipeFile(sources.zip, 'r') as zip_ref:
zip_ref.extractall(path)

Colonne

Explication

with zipfile.ZipFile(zip_file_path, 'r') as zip_ref:
    # Extract all files to the specified directory
    zip_ref.extractall(path_string_extract_dir)