I am using shututil to unpack my archive like this:
shutil.unpack_archive(file_path, extract_dir=extract_path)
Unpacking works fine, but I want to be able to store the latest unpacked file name into a variable somehow and then use it later. In the documentation (shutil), I don't see how I can access the filename inside the archive. (note that the unpacked filename might differ from the unpacked archive)
Any ideas on how I can achieve this or even print the status of the extraction?
You can use os.listdir(..) before and after extraction.
Related
Could someone please explain how exactly ZipInfo is supposed to be used? It says that ZipInfo.comment can access "comment for the individual archive member"
I didn't even know archive members can have comments %\ ...
I tried getting it with:
data = zipfile.ZipFile('filename')
info = data.infolist()
but what I'm getting looks like:
[<zipfile.ZipInfo object at 0x0257DBF8>, <zipfile.ZipInfo object at 0x026A7030>, <zipfile.ZipInfo object at 0x026A7098>, ... ]
I don't know what that means :(
Also, i can't seem to call zipinfo.comment at all, but from above it looks like infolist() is the same thing?
So confused...
Calling data.infolist() is giving you a list of ZipInfo objects. These are descriptions of all the individual files and directories stored inside your zip archive (and not the files/directories themselves). To manipulate these individual files/directories, you have to call a method of your ZipFile object data with the name from info. For example if you want to print the first 10 characters in each file you could run
for f in info:
data.read(f)[:10]
I have a folder full of jar, html, css, exe type file. How can I check the file?
I already run "file" command on *NIX and using python-magic. but the result is all like this.
test : Zip archive data, at least v1.0 to extract
How can I get information specifically like test : jar only using using magic number.
How do I do like this?
While not required, most JAR files have a META-INF/MANIFEST.MF file contained within them. You could check for the existence of this file, after checking if it's a zip file:
import zipfile
def zipFileContains(zipFileName, pathName):
f = zipfile.ZipFile(zipFileName, "r")
result = any(x.startswith(pathName.rstrip("/")) for x in f.namelist())
f.close()
return result
print zipFileContains("test.jar", "META-INF/MANIFEST.MF")
However, it might be better to just check if it's a zip file that ends in .jar.
Magic alone won't do it for you, since a JAR is literally just a zip file. Read more about the format here.
Using python 2.7
I have a list of *.tat.gz files on a linux box. Using python, I want to loop through the files and extract those files in a different location, under their respective folders.
For example: if my file name is ~/TargetData/zip/1440198002317590001.tar.gz
then I want to untar and ungzip this file in a different location under its
respective folder name i.e. ~/TargetData/unzip/1440198002317590001.
I have written some code but I am not able to loop through the files. In a command line I am able to untar using $ tar -czf 1440198002317590001.tar.gz 1440198002317590001 command. But I want to be able to loop through the .tar.gz files. The code is mentioned below. Here, I’m not able to loop just the files Or print only the files. Can you please help?
import os
inF = []
inF = str(os.system('ls ~/TargetData/zip/*.tar.gz'))
#print(inF)
if inF is not None:
for files in inF[:-1]:
print files
"""
os.system('tar -czf files /unzip/files[:-7]')
# This is what i am expecting here files = "1440198002317590001.tar.gz" and files[:-7]= "1440198002317590001"
"""
Have you ever worked on this type of use case? Your help is greatly appreciated!! Thank you!
I think you misunderstood the meaning of os.system(), that will do the job, but its return value was not expected by you, it returns 0 for successful done, you can not directly assign its output to a variable. You may consider the module [subprocess], see doc here. However, I DO NOT recommend that way to list files (actually, it returns string instead of list, see doc find the detail by yourself).
The best way I think would be glob module, see doc here. Use glob.glob(pattern), you can put all files match the pattern in a list, then you can loop it easily.
Of course, if you are familiar with os module, you also can use os.listdir(), os.path.join(), or even os.paht.expanduser() to do this. (Unlike glob, it only put filenames without fully path into a list, you need to reconstruct file path).
By the way, for you purpose here, there is no need to declare an empty list first (i.e. inF = [])
For unzip file part, you can do it by os.system, but I also recommend to use subprocess module instead of os.system, you will find the reason in the doc of subprocess.
DO NOT see the following code, ONLY see them after you really can not solve this by yourself.
import os
import glob
inF = glob.glob('~/TargetData/zip/*.tar.gz')
if inF:
for files in inF:
# consider subprocess.call() instead of os.system
unzip_name = files.replace('zip', 'unzip')[:-7]
# get directory name and make sure it exists, otherwise create it
unzip_dir = os.path.dirname(unzip_name)
if not os.path.exists(unzip_dir):
os.mkdir(unzip_dir)
subprocess.call(['tar -xzf', files, '-C', unzip_name])
# os.system('tar -czf files /unzip/files[:-7]')
I have several very large zip files available to download on a website. I am using Flask microframework (based on Werkzeug) which uses Python.
Is there a way to show the contents of a zip file (i.e. file and folder names) - to someone on a webpage - without actually downloading it? As in doing the working out server side.
Assume that I do not know what are in the zip archives myself.
I apoligize that this post does not include code.
Thank you for helping.
Sure, have a look at zipfile.ZipFile.namelist(). Usage is pretty simple, as you'd expect: you just create a ZipFile object for the file you want, and then namelist() gives you a list of the paths of files stored in the archive.
with ZipFile('foo.zip', 'r') as f:
names = f.namelist()
print names
# ['file1', 'folder1/file2', ...]
http://docs.python.org/library/zipfile.html
Specifically, try using the ZipFile.namelist() method.
The only way I came up for deleting a file from a zipfile was to create a temporary zipfile without the file to be deleted and then rename it to the original filename.
In python 2.4 the ZipInfo class had an attribute file_offset, so it was possible to create a second zip file and copy the data to other file without decompress/recompressing.
This file_offset is missing in python 2.6, so is there another option than creating another zipfile by uncompressing every file and then recompressing it again?
Is there maybe a direct way of deleting a file in the zipfile, I searched and didn't find anything.
The following snippet worked for me (deletes all *.exe files from a Zip archive):
zin = zipfile.ZipFile ('archive.zip', 'r')
zout = zipfile.ZipFile ('archve_new.zip', 'w')
for item in zin.infolist():
buffer = zin.read(item.filename)
if (item.filename[-4:] != '.exe'):
zout.writestr(item, buffer)
zout.close()
zin.close()
If you read everything into memory, you can eliminate the need for a second file. However, this snippet recompresses everything.
After closer inspection the ZipInfo.header_offset is the offset from the file start. The name is misleading, but the main Zip header is actually stored at the end of the file. My hex editor confirms this.
So the problem you'll run into is the following: You need to delete the directory entry in the main header as well or it will point to a file that doesn't exist anymore. Leaving the main header intact might work if you keep the local header of the file you're deleting as well, but I'm not sure about that. How did you do it with the old module?
Without modifying the main header I get an error "missing X bytes in zipfile" when I open it. This might help you to find out how to modify the main header.
Not very elegant but this is how I did it:
import subprocess
import zipfile
z = zipfile.ZipFile(zip_filename)
files_to_del = filter( lambda f: f.endswith('exe'), z.namelist()]
cmd=['zip', '-d', zip_filename] + files_to_del
subprocess.check_call(cmd)
# reload the modified archive
z = zipfile.ZipFile(zip_filename)
The routine delete_from_zip_file from ruamel.std.zipfile¹ allows you to delete a file based on its full path within the ZIP, or based on (re) patterns. E.g. you can delete all of the .exe files from test.zip using
from ruamel.std.zipfile import delete_from_zip_file
delete_from_zip_file('test.zip', pattern='.*.exe')
(please note the dot before the *).
This works similar to mdm's solution (including the need for recompression), but recreates the ZIP file in memory (using the class InMemZipFile()), overwriting the old file after it is fully read.
¹ Disclaimer: I am the author of that package.
Based on Elias Zamaria comment to the question.
Having read through Python-Issue #51067, I want to give update regarding it.
For today, solution already exists, though it is not approved by Python due to missing Contributor Agreement from the author.
Nevertheless, you can take the code from https://github.com/python/cpython/blob/659eb048cc9cac73c46349eb29845bc5cd630f09/Lib/zipfile.py and create a separate file from it. After that just reference it from your project instead of built-in python library: import myproject.zipfile as zipfile.
Usage:
with zipfile.ZipFile(f"archive.zip", "a") as z:
z.remove(f"firstfile.txt")
I believe it will be included in future python versions. For me it works like a charm for given use case.