renaming the extracted file from zipfile

Question

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 would be holding_march13207.txt.

My problem is that I am not able to change the extracted file's name. I would appreciate if you could advise.

import os 
import sys 
import zipfile
os.chdir("/feeds/lipper/emaxx") 

pwkwd = "/feeds/lipper/emaxx" 

for item in os.listdir(pwkwd): # loop through items in dir
    if item.endswith(".zip"): # check for ".zip" extension
        file_name = os.path.abspath(item) # get full path of files
        fh = open(file_name, "rb")
        zip_ref = zipfile.ZipFile(fh)

        filelist = 'ISSUERS.TXT' , 'SECMAST.TXT' , 'FUND.TXT' , 'HOLDING.TXT'
        for name in filelist :
            try:
                outpath = "/SCRATCH/emaxx" + "/" + os.path.splitext(item)[0]
                zip_ref.extract(name, outpath)

            except KeyError:
                {}

        fh.close()

user_5060185 · Answer

import zipfile

zipdata = zipfile.ZipFile('somefile.zip')
zipinfos = zipdata.infolist()

# iterate through each file
for zipinfo in zipinfos:
    # This will do the renaming
    zipinfo.filename = do_something_to(zipinfo.filename)
    zipdata.extract(zipinfo)

Reference: https://bitdrop.st0w.com/2010/07/23/python-extracting-a-file-from-a-zip-file-with-a-different-name/

user_1343267 · Answer

outpath = '/SCRATCH/emaxx'
suffix = os.path.splitext(item)[0]

for name in filelist :
    index = zip_ref.namelist().find(name)
    if index != -1: # check the file exists in the zipfile
        filename, ext = os.path.splitext(name)
        zip_ref.filelist[index].filename = f'{filename}_{suffix}.{ext}' # rename the extracting file to the suffix file name
        zip_ref.extract(zip_ref.filelist[index], outpath) # use the renamed file descriptor to extract the file

renaming the extracted file from zipfile

5 Answers