zip file and avoid directory structure

Check out the documentation for Zipfile.write.

ZipFile.write(filename[, arcname[, compress_type]]) Write the file
named filename to the archive, giving it the archive name arcname (by
default, this will be the same as filename, but without a drive letter
and with leading path separators removed)

https://docs.python.org/2/library/zipfile.html#zipfile.ZipFile.write

Try the following:

import zipfile
import os
filename = 'foo.txt'

# Using os.path.join is better than using '/' it is OS agnostic
path = os.path.join(os.path.sep, 'tmp', 'bar', 'baz', filename)
zip_filename = os.path.splitext(filename)[0] + '.zip'
zip_path = os.path.join(os.path.dirname(path), zip_filename)

# If you need exception handling wrap this in a try/except block
with zipfile.ZipFile(zip_path, 'w') as zf:
    zf.write(path, zip_filename)

The bottom line is that if you do not supply an archive name then the filename is used as the archive name and it will contain the full path to the file.

To illustrate most clearly,

directory structure:

/Users
 └── /user
 .    ├── /pixmaps
 .    │    ├── pixmap_00.raw
 .    │    ├── pixmap_01.raw
      │    ├── /jpeg
      │    │    ├── pixmap_00.jpg
      │    │    └── pixmap_01.jpg
      │    └── /png
      │         ├── pixmap_00.png
      │         └── pixmap_01.png
      ├── /docs
      ├── /programs
      ├── /misc
      .
      .
      .

Directory of interest: /Users/user/pixmaps

First attemp

import os
import zipfile

TARGET_DIRECTORY = "/Users/user/pixmaps"
ZIPFILE_NAME = "CompressedDir.zip"

def zip_dir(directory, zipname):
    """
    Compress a directory (ZIP file).
    """
    if os.path.exists(directory):
        outZipFile = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)

        for dirpath, dirnames, filenames in os.walk(directory):
            for filename in filenames:

                filepath   = os.path.join(dirpath, filename)
                outZipFile.write(filepath)

        outZipFile.close()




if __name__ == '__main__':
    zip_dir(TARGET_DIRECTORY, ZIPFILE_NAME)

ZIP file structure:

CompressedDir.zip
.
└── /Users
     └── /user
          └── /pixmaps
               ├── pixmap_00.raw
               ├── pixmap_01.raw
               ├── /jpeg
               │    ├── pixmap_00.jpg
               │    └── pixmap_01.jpg
               └── /png
                    ├── pixmap_00.png
                    └── pixmap_01.png

Avoiding the full directory path

def zip_dir(directory, zipname):
    """
    Compress a directory (ZIP file).
    """
    if os.path.exists(directory):
        outZipFile = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)

        # The root directory within the ZIP file.
        rootdir = os.path.basename(directory)

        for dirpath, dirnames, filenames in os.walk(directory):
            for filename in filenames:

                # Write the file named filename to the archive,
                # giving it the archive name 'arcname'.
                filepath   = os.path.join(dirpath, filename)
                parentpath = os.path.relpath(filepath, directory)
                arcname    = os.path.join(rootdir, parentpath)

                outZipFile.write(filepath, arcname)

    outZipFile.close()




if __name__ == '__main__':
    zip_dir(TARGET_DIRECTORY, ZIPFILE_NAME)

ZIP file structure:

CompressedDir.zip
.
└── /pixmaps
     ├── pixmap_00.raw
     ├── pixmap_01.raw
     ├── /jpeg
     │    ├── pixmap_00.jpg
     │    └── pixmap_01.jpg
     └── /png
          ├── pixmap_00.png
          └── pixmap_01.png

zf.write(tofile)

to change

zf.write(tofile, zipfile_dir)

for example

zf.write("/root/files/result/root/files/result/new.txt", "/root/files/results/new.txt")

You can isolate just the file name of your sources files using:

name_file_only= name_full_path.split(os.sep)[-1]

For example, if name_full_path is /root/files/results/myfile.txt, then name_file_only will be myfile.txt. To zip myfile.txt to the root of the archive zf, you can then use:

zf.write(name_full_path, name_file_only)

I face the same problem and i solve it with writestr. You can use it like this:

zipObject.writestr(<filename> , <file data, bytes or string>)

The arcname parameter in the write method specifies what will be the name of the file inside the zipfile:

import os
import zipfile

# 1. Create a zip file which we will write files to
zip_file = "/home/username/test.zip"
zipf = zipfile.ZipFile(zip_file, 'w', zipfile.ZIP_DEFLATED)

# 2. Write files found in "/home/username/files/" to the test.zip
files_to_zip = "/home/username/files/"
for file_to_zip in os.listdir(files_to_zip):

    file_to_zip_full_path = os.path.join(files_to_zip, file_to_zip)

    # arcname argument specifies what will be the name of the file inside the zipfile
    zipf.write(filename=file_to_zip_full_path, arcname=file_to_zip)

zipf.close()

We can use this

import os
# single File
os.system(f"cd {destinationFolder} && zip fname.zip fname") 
# directory
os.system(f"cd {destinationFolder} && zip -r folder.zip folder") 

For me, This is working.

If you want an elegant way to do it with pathlib you can use it this way:

from pathlib import Path
import zipfile

def zip_dir(path_to_zip: Path):
    zip_file = Path(path_to_zip).with_suffix('.zip')
    z = zipfile.ZipFile(zip_file, 'w', zipfile.ZIP_DEFLATED)
    for f in list(path_to_zip.rglob('*.*')):
        z.write(f, arcname=f.relative_to(path_to_zip))

It is much simpler than expected, I configured the module using the parameter "arcname" as "file_to_be_zipped.txt", so the folders do not appear in my final zipped file:

mmpk_zip_file = zipfile.ZipFile("c:\Destination_folder_namenewzippedfilename.zip", mode='w', compression=zipfile.ZIP_DEFLATED)
mmpk_zip_file.write("c:\Source_folder_namefile_to_be_zipped.txt", "file_to_be_zipped.txt")
mmpk_zip_file.close()

To get rid of the absolute path, I came up with this:

def create_zip(root_path, file_name, ignored=[], storage_path=None):
    """Create a ZIP
    
    This function creates a ZIP file of the provided root path.
    
    Args:
        root_path (str): Root path to start from when picking files and directories.
        file_name (str): File name to save the created ZIP file as.
        ignored (list): A list of files and/or directories that you want to ignore. This
                        selection is applied in root directory only.
        storage_path: If provided, ZIP file will be placed in this location. If None, the
                        ZIP will be created in root_path
    """
    if storage_path is not None:
        zip_root = os.path.join(storage_path, file_name)
    else:
        zip_root = os.path.join(root_path, file_name)
        
    zipf = zipfile.ZipFile(zip_root, 'w', zipfile.ZIP_DEFLATED)
    def iter_subtree(path, layer=0):
        # iter the directory
        path = Path(path)
        for p in path.iterdir():
            if layer == 0 and p.name in ignored:
                continue
            zipf.write(p, str(p).replace(root_path, '').lstrip('/'))
            
            if p.is_dir():
                iter_subtree(p, layer=layer+1)

    iter_subtree(root_path)
    zipf.close()

Maybe it isn’t the most elegant solution, but this works. If we just use p.name when providing the file name to write() method, then it doesn’t create the proper directory structure.

Moreover, if it’s needed to ignore the selected directories or files from the root path, this ignores those selections too.

Specify the arcname input of the write method as following:

tofile =  "/root/files/result/"+file
NewRoot = "files/result/"
zf.write(tofile, arcname=tofile.split(NewRoot)[1])

More info:

ZipFile.write(filename, arcname=None, compress_type=None,
compresslevel=None)
https://docs.python.org/3/library/zipfile.html

This is an example I used. I have one excel file, Treport where I am using python + pandas in my dowork function to create pivot tables, etc. for each of the companies in CompanyNames. I create a zip file of the csv and a non-zip file so I can check as well.
The writer specifies the path where I want my .xlsx to go and for my zip files, I specify that in the zip.write(). I just specify the name of the xlsx file that was recently created, and that is what gets zipped up, not the whole directory. Beforehand I was just specifying ‘writer’ and would zip up the whole directory. This allows me to zip up just the recently created excel file.

Treport = 'TestReportData.csv'
CompanyNames = ['Company1','Company2','Company3']
for CompName in CompanyNames:
    strcomp = str(CompName)
    #Writer Creates pathway to output report to. Each company gets unique file.
    writer = pd.ExcelWriter(f"C:\Users\MyUser\Documents\{strcomp}addReview.xlsx", engine='xlsxwriter')
    DoWorkFunction(CompName, Treport, writer)
    writer.save()
    with ZipFile(f"C:\Users\MyUser\Documents\{strcomp}addR.zip", 'w') as zip:
        zip.write(writer, f"{strcomp}addReview.xlsx")