zipfile

问题 I have lots of zipped files on a Linux server and each file includes multiple text files. what I want is to extract some of those text files, which have the same name across zipped files and save it a folder; I am creating one folder for each zipped file and extract the text file to it. I need to add the parent zipped folder name to the end of file names and save all text files in one directory. For example, if the zipped folder was March132017.zip and I extracted holding.txt, my filename

问题 I'm trying to create a zip archive of a possibly huge folder. For this purpose I'm using the python zipfile module, but as far as I can see there is no option to split the created archive into multiple chunks with a max size. The zipped archive is supposed to be sent via Telegram, which has a size limitation of 1.5 GB per file. Thereby I need to split the resulting zip archive. I would really like to not use a subprocess and shell commands for creating this archive. My current code looks like

问题 I'm trying to create a zip archive of a possibly huge folder. For this purpose I'm using the python zipfile module, but as far as I can see there is no option to split the created archive into multiple chunks with a max size. The zipped archive is supposed to be sent via Telegram, which has a size limitation of 1.5 GB per file. Thereby I need to split the resulting zip archive. I would really like to not use a subprocess and shell commands for creating this archive. My current code looks like

问题 To rename the one file in a ZipFile that I'm downloading, I do the following: for item in zipfile.infolist(): old_name = item.filename match = re.search(r'(.*)(.mdb)', item.filename) item.filename = "%s_friend%s" % (match.group(1),, match.group(2)) # I should probably be using replace here zipfile.extract(old_name, save_dir) However when I want to extract that file and save it into a specific directory, I need to reference the "old_name" and cannot reference the new. Is there a "cleaner" way

问题 i used zipfile lib to extract file from zip and now after unzip the directory i found the permission of my file has been corrupted , import zipfile fh = open('sample.zip', 'rb') z = zipfile.ZipFile(fh) print z.namelist() for name in z.namelist(): z.extract(name, '/tmp/') fh.close() but when i use linux unzip tools this issue don't happen i try to use os.system('unzip sample.zip') but i still want to do this with zipfile 回答1: The related Python issues provide some insight as to why the issue

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf

问题 I'm trying to read ZipFile data from a URL and via StringIO parse the data inside the ZipFile as csv using pandas.read_csv r = req.get("http://seanlahman.com/files/database/lahman-csv_2014-02-14.zip").content file = ZipFile(StringIO(r)) salaries_csv = file.open("Salaries.csv") salaries = pd.read_csv(salaries_csv) The last line gave me an error: CParserError: Error tokenizing data. C error: Calling read(nbytes) on source failed. Try engine='python'. However if i try using salaries = pd.read

来源： https://stackoverflow.com/questions/52484821/python-zipfile-with-no-parent-folder