Fixing "PermissionError: [Errno 13] Permission denied"

[puredata]

Hello, I'm trying to use Python to automate unzipping of multiple files in a folder. I already have a script that works and unzips my files. But it won't copy contents of the zip files which has subfolders. In order to be able to extract everything into one main folder (disregarding original subfolder structure) I have found this code chunk on stackexchange. A moderator reported that the code is working for him therefore I should be investigating something else, not the code itself.

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import os import shutil import zipfile   my_dir = r"C:\Users\username\My_Dataset\new" my_zip = r"C:\Users\username\My_Dataset"   with zipfile.ZipFile(my_zip) as zip_file:     for member in zip_file.namelist():         filename = os.path.basename(member)         # skip directories         if not filename:             continue           # copy file (taken from zipfile's extract) source = zip_file.open(member) target = open(os.path.join(my_dir, filename), "wb") with source, target:      shutil.copyfileobj(source, target)

Here is the Traceback I get:

Error:
Traceback (most recent call last):
  File "C:/Users/username/PycharmProjects/GeoPandas/mergeAnswer.py", line 8, in <module>
    with zipfile.ZipFile(my_zip) as zip_file:
  File "C:\Users\username\Anaconda3\envs\playground\lib\zipfile.py", line 1207, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\username\\My_Dataset

The only thing that looks suspicious to me is the direction of the slashes on Traceback. Some of them are forwards slashes while some of them are backwards. Could it be related to that? I'm running this code in Windows and haven't tried on a mac or linux.

Would it be possible to get some suggestions in order to make this script work? I don't need to use this specific script, I'm just trying to unzip a lot of files (some are under subfolders some are not) into one main folder.

Thanks in advance!

[stullis]

According to the docs, zipfile.ZipFile() expects a file, not a directory. I'd wager that's the culprit. Try creating the file first and passing that into zipfile.ZipFile().

[jim2007]

Is there a folder called C:\Users\username on your machine? That looks very strange in the error message you posted.

I would expect it to be C:\Users\<your login Id>\My_Dataset

[puredata]

According to the docs, zipfile.ZipFile() expects a file, not a directory. I'd wager that's the culprit. Try creating the file first and passing that into zipfile.ZipFile().

Thanks for looking it up! I checked the documentation as well but I couldn't comprehend to the point to solve my issue.

What do you mean by creating the file first? Do you mean creating a shapefile or do you mean that I need to change the order in the script?

Is there a folder called C:\Users\username on your machine? That looks very strange in the error message you posted.

I would expect it to be C:\Users\<your login Id>\My_Dataset

Thank you, I replaced username before posting since it's my real name in there with the actual path. Do you think backwards/forwards slashes might be creating the issue here?

[puredata]

Wanted to up this post, I still couldn't figure out a way to unzip these files.

[jim2007]

So two things to check:

- Check if you can create a file in the folder with notepad for instance

- Your variable my_zip points to a folder name and not a file. It should be something like: my_zip = r"C:\Users\username\My_Dataset\<name>.zip"

Let us know how you go.

[puredata]

So two things to check:

- Check if you can create a file in the folder with notepad for instance

- Your variable my_zip points to a folder name and not a file. It should be something like: my_zip = r"C:\Users\username\My_Dataset\<name>.zip"

Let us know how you go.

Thank you, I can create a file, actually I have another script, I run from PyCharm and it works on the folder, but this script gives me a permission error.

The reason my_zip points to a folder is I have a folder with zip files and I'm trying to go through them in order to extract their content into a main file. The script I have (which works) does the job but it only extract one layer of zip files, if a zip file has a subfolder in it (not necessarily a zipped folder) it won't extract those to my "main" folder. This is what I'm trying to achieve with the script that gives me an error.

Thank you for your help again.

[TheHacker707]

If permission really is the problem, you could try to run the program with admin user level. Also, I would keep your slashes as backslashes if you're using Windows.
-707

[puredata]

If permission really is the problem, you could try to run the program with admin user level. Also, I would keep your slashes as backslashes if you're using Windows.
-707

Now I'm trying the same on a mac and don't get the permission error but I get this instead

Error:
    self.fp = io.open(file, filemode)
IsADirectoryError: [Errno 21] Is a directory: '/Users/berkaytok/Documents/Boundary_Datasets'

so I imagine stullis seems correct that this script expects a single zip file.

According to the docs, zipfile.ZipFile() expects a file, not a directory. I'd wager that's the culprit. Try creating the file first and passing that into zipfile.ZipFile().

Moving on to my previous script;

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import os import zipfile   user: str = os.getlogin() #to be used in directory names #item = 'item'  # i am not sure if this is necessary dir_name = "/Users/{0}//Documents/Boundary_Datasets".format(user) #folder that contains multiple zipfiles which some of them have subfolders in those zipfiles extract = "/Users/{0}//Documents/Boundary_Datasets/new".format(user) #main folder to be used as an output folder extension = ".zip" #file_name = dir_name + "/" + item #i am not sure if this is necessary   os.chdir(dir_name)  # change directory from working dir to dir with files   for item in os.listdir(dir_name):  # loop through items in dir     if item.endswith(extension):  # check for ".zip" extension         file_name = os.path.abspath(item)  # get full path of files         zip_ref = zipfil e.ZipFile(file_name)  # create zipfile object         zip_ref.extractall(extract)  # extract file to dir         zip_ref.close()  # close file         # os.remove(file_name) # delete zipped file

Can anyone help me to modify this so that it not only extracts the first subfolder content but all subfolders and extract all files into one main folder?

[jim2007]

It is almost certainly an issue with passing the correct file name to the function. Try hard coding one of the file names into the my_zip variable and see if it runs without error.