Extract zip to memory, parse contents

Question

I want to read the contents of a zip file into memory rather than extracting them to disc, find a particular file in the archive, open the file and extract a line from it.

Answer 1

IMO just using read is enough:

zfile = ZipFile('name.zip', 'r')
files = []
for name in zfile.namelist():
  if fnmatch.fnmatch(name, '*_readme.xml'):
    files.append(zfile.read(name))

This will make a list with contents of files that match the pattern.

Test: You can then parse contents afterwards by iterating through the list:

for file in files:
  print(file[0:min(35,len(file))].decode()) # "parsing"

Or better use a functor:

import zipfile as zip
import os
import fnmatch

zip_name = os.sys.argv[1]
zfile = zip.ZipFile(zip_name, 'r')

def parse(contents, member_name = ""):
  if len(member_name) > 0:
    print( "Parsed `{}`:".format(member_name) )  
  print(contents[0:min(35, len(contents))].decode()) # "parsing"

for name in zfile.namelist():
  if fnmatch.fnmatch(name, '*.cpp'):
    parse(zfile.read(name), name)

This way there is no data kept in memory for no reason and memory foot print is smaller. It might be important if the files are big.