why does python say it cant find the path specified when it made the path?

Question

I made this program yesterday because I am using py2exe, so what this program does is it zips up the folder created by py2exe and names it to app4export so I can send it to my friends. I also added in where if i already have a zip file called app4export then it deletes it before hand, it worked yesterday but now today I get the error

FileNotFoundError: [WinError 3] The system cannot find the path specified: 'C:\\Users\\severna\\Desktop\\Non_Test_Python_Files\\app4export'

but python made this location so I dont get why it cant find it later?

import os
import zipfile
import shutil

def zip(src, dst):
    zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
    abs_src = os.path.abspath(src)
    for dirname, subdirs, files in os.walk(src):
        for filename in files:
            absname = os.path.abspath(os.path.join(dirname, filename))
            arcname = absname[len(abs_src) + 1:]
            print('zipping %s as %s' % (os.path.join(dirname, filename),
                                        arcname))
            zf.write(absname, arcname)
    zf.close()

source=r"C:\Users\severna\Desktop\Non_Test_Python_Files\dist"
destination=r"C:\Users\severna\Desktop\Non_Test_Python_Files\app4export"

shutil.rmtree(str(destination))

try:
    zip(str(source), str(destination))
    shutil.rmtree(str(source))

except FileNotFoundError:
    print("Source cannot be zipped as it does not exist!")

Are you sure this is the code you're running? At the moment the only line that can throw that error is the shutil.rmtree line (the rest are in the try and the error would be hidden) and that is using a different path (has .zip on the end). — Holloway
– Holloway, Commented Dec 15, 2016 at 13:40
Rather than messing around with r'' or '\\', you can use the fact Windows can handle '/' as a path separator. This will remove one possible problem from the equation. — John Coleman
– John Coleman, Commented Dec 15, 2016 at 13:40

Cleared · Accepted Answer · 2016-12-15 13:50:53Z

1

Your code creates the file C:\Users\severna\Desktop\Non_Test_Python_Files\app4export.zip, but you try to remove the directory C:\Users\severna\Desktop\Non_Test_Python_Files\app4export

So just before the try-block you have

shutil.rmtree(str(destination))

which will throw an FileNotFoundError if the path do not exist. And when you hit that line of code, you still havent created the path. The reason it might have worked yesterday was that you mayby had that path.

edited Dec 15, 2016 at 13:50

answered Dec 15, 2016 at 13:44

Cleared

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8 Comments

user7297278 Over a year ago

the path is there, i am looking right at the zipfile

Cleared Over a year ago

I dont belive you :) Go to the directory using the command-promt, do a dir and post your result.

Cleared Over a year ago

Are you sure that the directory C:\Users\severna\Desktop\Non_Test_Python_Files\app4export exists? (And i am NOT talking about the file C:\Users\severna\Desktop\Non_Test_Python_Files\app4export.zip)

user7297278 Over a year ago

want me to copy and paste the whole thing?

Cleared Over a year ago

You DO NOT have a directory/folder named app4export. You only have a FILE called app4export.zip. Therefore you can not remove the DIRECTORY name app4export, because it does not exist.

|

user7297278 · Accepted Answer · 2016-12-15 14:03:05Z

0

after discussion with Cleared I found out that I needed i file extension because it was a file and shutil.rmtree doesnt remove files it removes directories so I need to use this code instead

os.remove(str(destination)+".zip")

answered Dec 15, 2016 at 14:03

user7297278

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