Run file inside a zipfile?

Question

Is it possible to run a .html or .exe for example, that is inside a zipfile? I'm using the Zipfile module.

Here's my sample code:

import zipfile

z = zipfile.ZipFile("c:\\test\\test.zip", "r")
x = ""
g = ""
for filename in z.namelist():
    #print filename
    y = len(filename)
    x = str(filename)[y - 5:]
    if x == ".html":
        g = filename
f = z.open(g)

After f = z.open(g), I don't know what to do next. I tried using the .read() but it only reads whats inside of the html, what I need is for it to run or execute.

Or is there any othere similar ways to do this?

I'm not sure behaviour you want, so it's hard to help you make it happen. Take the zip file out of the equation. Imagine you have a plain ".html" file, e.g. `g = "testfile.html"`. What behaviour do you want to see from the computer? — Jim DeLaHunt, Jan 19 '12 at 03:57
what i would want is, for it to launch the testfile.html on a web browser. like after the g = "testfile.html" theres a code that will open it on a web browser. (but testfile.html is inside a zipfile) — Katherina, Jan 19 '12 at 04:04
I suggest you add to your question the code to open the file in a web browser, run the executable, etc., which you'd use if the file was out of the zip file. Do you already know how to do this part? — Jim DeLaHunt, Jan 19 '12 at 04:18
i got the code on running it outside of the zipfile, the challenge is how do i run it, if it is inside the zipfile, i wish to re create where if you open a zipfile and double clicks on the html file inside the zipfile. it will automatically opens on a web browser. this is under windows btw — Katherina, Jan 19 '12 at 04:33
You "got the code on running it outside of the zipfile". Good. What happens when you give the file handle `f` from `z.open(g)` to this code? What doesn't work? What behaves differently? It would help if you show us this code. — Jim DeLaHunt, Jan 19 '12 at 04:56

Abbas · Answer 1 · 2012-01-20T08:31:40.300

1

The best approach will be to extract the required file to the Windows temp directory and execute it. I have modified your original code to create a temp file and execute it:

import zipfile
import shutil
import os

z = zipfile.ZipFile("c:\\test\\test.zip", "r")
x = ""
g = ""
basename = ""
for filename in z.namelist():
    print filename
    y = len(filename)
    x = str(filename)[y - 5:]
    if x == ".html":
        basename = os.path.basename(filename) #get the file name and extension from the return path
        g = filename
        print basename
        break #found what was needed, no need to run the loop again
f = z.open(g)

temp = os.path.join(os.environ['temp'], basename) #create temp file name
tempfile = open(temp, "wb")
shutil.copyfileobj(f, tempfile) #copy unzipped file to Windows 'temp' folder
tempfile.close()
f.close()
os.system(temp) #run the file

edited Jan 20 '12 at 08:31

answered Jan 20 '12 at 06:30

Abbas

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Don't use `os.system()` use `os.startfile()` on Windows; `subprocess`, `webbrowser` modules on other OSes. – jfs Jan 20 '12 at 08:26
`tempfile` is more robust than `os.environ['temp']` – jfs Jan 20 '12 at 08:27
1

you could use `os.path.splitext()` to extract file extension or just `filename.endswith('.html')` to check it e.g., see [my answer](http://stackoverflow.com/a/8938550/4279). – jfs Jan 20 '12 at 08:30
hi thanks for the teaching let me study it again for a while. i need time for this to sink in my head =) thanks guys – Katherina Jan 24 '12 at 05:37

jfs · Answer 2 · 2012-01-24T18:28:57.183

Run the first .html file in a zip archive specified at the command line:

#!/usr/bin/env python
import os
import shutil
import sys
import tempfile
import webbrowser
import zipfile
from subprocess import check_call
from threading  import Timer

with zipfile.ZipFile(sys.argv[1], 'r') as z:
    # find the first html file in the archive
    member = next(m for m in z.infolist() if m.filename.endswith('.html'))
    # create temporary directory to extract the file to
    tmpdir = tempfile.mkdtemp()
    # remove tmpdir in 5 minutes
    t = Timer(300, shutil.rmtree, args=[tmpdir], kwargs=dict(ignore_errors=True))
    t.start()
    # extract the file
    z.extract(member, path=tmpdir)
    filename = os.path.join(tmpdir, member.filename)

# run the file
if filename.endswith('.exe'):
    check_call([filename]) # run as a program; wait it to complete
else: # open document using default browser
    webbrowser.open_new_tab(filename) #NOTE: returns immediately

Example

T:\> open-from-zip.py file.zip

As an alternative to webbrowser you could use os.startfile(os.path.normpath(filename)) on Windows.

hi thanks for the teaching let me study it again for a while. i need time for this to sink in my head =) thanks guys — Katherina, Jan 24 '12 at 05:38

Run file inside a zipfile?

2 Answers2

Example