Find specific file in an archive

[tester_V]

Hi,
I hope you have better weather than we have tonight in Portland Oregon.
Anyway,
I need to extract a specific file that starts with the word "Debug*.log" from an archive.
An archive may or may not have the file, also the archive is actually a zipped directory with SubDirs and files.
The file I'm looking for could be any of the SubDirs.
I tried to use "zip.namelist" but it only prints files but not recursively,

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zip_filepath = 'C:\\02\\TEST-3.zip' target_dir   = 'C:\\03\\'   zip = zipfile.ZipFile(zip_filepath) print (zip.namelist())

I understand I need to test the Archive first but not sure how to do that.

Thank you.

[BashBedlam]

Try this. I think it will do the trick for you.

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from zipfile import ZipFile   with ZipFile (zip_filepath, 'r') as zip_file :     file_name_list = zip_file.namelist ()     for file_name in file_name_list :         if 'Debug' in file_name :             output_file_name = target_dir + file_name.split ('\')[-1]             with open (output_file_name, 'wb') as out_file :                 out_file.write (zip_file.read (file_name))

[perfringo]

Python has built-in module fnmatch (Unix filename pattern matching) which can be used to directly implement 'Debug*.log' pattern. This module also have filter, so one can write:

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>>> import fnmatch >>> filenames = ['Debug1.log', 'Debug1.txt', 'Debug.log'] >>> fnmatch.filter(filenames, 'Debug*.log') ['Debug1.log', 'Debug.log']

[tester_V]

TO:
BashBedlam
perfringo

Thank you both!

I tried just the BeshBedam code because I'm not sure how to 'perfringo's" snippet.

I'm having an error:
output_file_name = target_dir + file_name.split ('\')[-1]
^
SyntaxError: EOL while scanning string literal

Thank you.

[BashBedlam]

Sorry, I forgot to escape the backslash character. I tested it on a linux system and modified it just before posting to what I thought would be compatible with windows. Try this and if it doesn't work, maybe someone on a windows system can make the output_file_name correct.

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from zipfile import ZipFile    with ZipFile (zip_filepath, 'r') as zip_file :     file_name_list = zip_file.namelist ()     for file_name in file_name_list :         if 'Debug' in file_name :             output_file_name = target_dir + file_name.split ('\\')[-1]             with open (output_file_name, 'wb') as out_file :                 out_file.write (zip_file.read (file_name))

[tester_V]

NO worries!
It works now.

Thank you!

[tester_V]

One more question.
I have a lot of Zip files, I wanted to find ZIp files that have the "Debugxxx.log"
Is it possible to identify which ZIP file has "Debugxxx.log" in it?
I'd like to copy that file too.

Thank you.

[BashBedlam]

Please let me know if this works on your system.

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import os location = 'C:\\02\\' files_in_dir = []   # r=>root, d=>directories, f=>files for r, d, f in os.walk(location):     for item in f:         if '.zip' in item:             files_in_dir.append(os.path.join(r, item))   for item in files_in_dir:     print("file in dir: ", item)

[tester_V]

to BashBedlam:
Thank you!
your help is much appreciated!