I want to insert all .ini files into an archive; it does it well but when I open my .zip, there are the path folders to those files included, too.
Here's my code:
from path import Path
import zipfile
def main():
folderul_cu_demouri = Path('/my/path/bla/bla')
nume_arhiva = 'demoz.zip'
arhiva = zipfile.ZipFile(nume_arhiva, 'w')
for demo in folderul_cu_demouri.files(pattern='*.ini'):
arhiva.write(demo)
arhiva.close()
if __name__ == '__main__':
main()
So when I open my zip file, I gotta browse through /my/path/to/files, and only then I can see my .ini files. How can I make it so only the .ini are inserted in the zip file, without the directories?
Thanks.
PS: I'm using path.py to get their extensions.
if your files are located directly in the archive folder, you could basename your files and pass the name in arcname parameter so the name in the archive is the filename, without the full path:
arhiva.write(demo,arcname=os.path.basename(demo))
else, you could remove the first characters of the full file path so relative paths are preserved:
len_to_strip = len('/my/path/bla/bla')+1
arhiva.write(demo,arcname=demo[:len_to_strip])
Related
I try to run the following code on the folder, which contains only zip files and somehow it does not work. I don't know how to solve it at all. I looked some references on line which said that it may be because some zip files may be damaged itside, but my Python skills are not that good to figure out what may be a solution then. Apparently, it is specifically because some of the zip files are damaged, because I checked extensions and all files inside folder are zip
import os, zipfile
dir_name = "C:/Users/Имя/termpaper/notifications_2020"
extension = ".zip"
os.chdir(dir_name)
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 = zipfile.ZipFile(file_name) # create zipfile object
listOfFileNames = zip_ref.namelist()
for fileName in listOfFileNames:
if fileName.endswith('.xml'): #i choose files only with xml extension, because there are also sig files
if fileName.startswith('fksNotificationEA44'): # code for electronic auction notification
zip_ref.extract(fileName)
I am trying to get full_path of places.sqlite file present in '%APPDATA%\Mozilla\Firefox\Profiles\<random_folder>\places.sqlite' using Python OS module. The issue as you can see that <random_folder> has a random name and there could be multiple folders inside the Profiles folder.
How do I navigate/find the path to the places.sqlite file?
You would ideally want to go through each folder to search for this file. In terminal 'locate file_name' command would do this for you. In python file you can use the following command:
import os
db_path = os.path.join(os.getenv('APPDATA'), r'Mozilla\Firefox\Profiles')
def find_file(file_name, path):
for root_folder, directory, file_names in os.walk(path):
if file_name in file_names:
return os.path.join(root_folder, file_name)
print(find_file('places.sqlite', db_path))
os.walk gives a list of all files in a path recusivly. Use it to search for 'places.sqlite' as follows.
path = ""
for root, dirs, files in os.walk("%APPDATA%\\Mozilla\\Firefox\\Profiles\\"):
if "places.sqlite" in files:
path = os.path.join(root, 'places.sqlite')
break
Use the os module to list out all directories in %APPDATA%\Mozilla\Firefox\Profiles\
loop over the directories until you find places.sqlite file (also using os module)
A glob might be simpler as in this case one expects the file to be there in level below the Profiles folder or not there at all.
import os
import pathlib
profiles = pathlib.Path(os.environ["APPDATA"]) / "Mozilla" / "Firefox" / "Profiles"
# rglob will recursively search as well
if places := list(profiles.rglob("places.sqlite")):
print(places[0]) # will print the sqllite file path
with places[0].open() as f:
# ....
i want a python script that do these three tasks:
check if the path contains word file copied to specific destination
check if the path contains pdf files copied to specific destination
check if the path contains directory and copy the hole folder to
specific destination.
for this reason i am using the os.walk() to list the dirs and files of the path
and i am using shutil library to copy files and dirs.
code
import os
from distutils.dir_util import copy_tree
import shutil
from os import path
import datetime
def main():
src = "C:/Users/LT GM/Desktop/Python_files/"
dst2 = "C:/Users/LT GM/Desktop/"
for root,dirs,files in os.walk(src):
for name in files:
print("files: ",os.path.join(root,name))
for name in dirs:
copieddst = copy_tree(src,dst2)
print("directory: ",os.path.join(root,name))
print(" coppied directory :{0}".format(copieddst) )
# make a duplicate of an existing file
if path.exists(src):
# get the path to the file in the current directory
print("****")
src = path.realpath("pandas.pdf")
#seperate the path from the filter
head, tail = path.split(src)
print("path:" +head)
print("file:" +tail)
dst =str(datetime.date.today()) + tail
# nowuse the shell to make a copy of the file
shutil.copy(src, dst)
if __name__=="__main__":
main()
the problem is that i can copy the files or the content of the directory. not the hole directory and how to check if its pdf or doc files?
If you want to copy directory rather than file, then use shutil.copytree. In usage it is similiar to shutil.copy2, that is:
import shutil
shutil.copytree('mydir', 'mydircopy')
Note that by default dirs_exist_ok is False meaning that destination should not exist, when shutil.copytree is launched.
I am trying to get the most recent file added to a directory using python 2.7 os and glob modules.
import os
import glob
path = "files/"
newestFile = max(glob.iglob(path + '*.txt'), key=os.path.getctime)
print newestFile
When I print the newestFile variable I get the path included i.e.
files\file.txt
I just want the filename but my .txt file and .py script are not in the same directory. The text file is one directory down under the files directory. How do I refer to the directory and get the newest .txt file added to that directory.
You can use os.path.basename to just get the filename:
newestFile = os.path.basename(max(glob.iglob(path + '*.txt'), key=os.path.getctime))
os.path.getctime is going to need the full path so one way or another you would have to use the full path.
Once I have all the files I require in a particular folder, I would like my python script to zip the folder contents.
Is this possible?
And how could I go about doing it?
On python 2.7 you might use: shutil.make_archive(base_name, format[, root_dir[, base_dir[, verbose[, dry_run[, owner[, group[, logger]]]]]]]).
base_name archive name minus extension
format format of the archive
root_dir directory to compress.
For example
shutil.make_archive(target_file, format="bztar", root_dir=compress_me)
Adapted version of the script is:
#!/usr/bin/env python
from __future__ import with_statement
from contextlib import closing
from zipfile import ZipFile, ZIP_DEFLATED
import os
def zipdir(basedir, archivename):
assert os.path.isdir(basedir)
with closing(ZipFile(archivename, "w", ZIP_DEFLATED)) as z:
for root, dirs, files in os.walk(basedir):
#NOTE: ignore empty directories
for fn in files:
absfn = os.path.join(root, fn)
zfn = absfn[len(basedir)+len(os.sep):] #XXX: relative path
z.write(absfn, zfn)
if __name__ == '__main__':
import sys
basedir = sys.argv[1]
archivename = sys.argv[2]
zipdir(basedir, archivename)
Example:
C:\zipdir> python -mzipdir c:\tmp\test test.zip
It creates 'C:\zipdir\test.zip' archive with the contents of the 'c:\tmp\test' directory.
Here is a recursive version
def zipfolder(path, relname, archive):
paths = os.listdir(path)
for p in paths:
p1 = os.path.join(path, p)
p2 = os.path.join(relname, p)
if os.path.isdir(p1):
zipfolder(p1, p2, archive)
else:
archive.write(p1, p2)
def create_zip(path, relname, archname):
archive = zipfile.ZipFile(archname, "w", zipfile.ZIP_DEFLATED)
if os.path.isdir(path):
zipfolder(path, relname, archive)
else:
archive.write(path, relname)
archive.close()
Both jfs's solution and Kozyarchuk's solution could work for the OP's use case, however:
jfs's solution zips all of the files in a source folder and stores them in the zip at the root level (not preserving the original source folder within the structure of the zip).
Kozyarchuk's solution inadvertently puts the newly-created zip file into itself since it is a recursive solution (e.g. creating new zip file "myzip.zip" with this code will result in the archive "myzip.zip" itself containing an empty file "myzip.zip")
Thus, here is a solution that will simply add a source folder (and any subfolders to any depth) to a zip archive. This is motivated by the fact that you cannot pass a folder name to the built-in method ZipFile.write() -- the function below, add_folder_to_zip(), offers a simple method to add a folder and all of its contents to a zip archive. Below code works for Python2 and Python3.
import zipfile
import os
def add_folder_to_zip(src_folder_name, dst_zip_archive):
""" Adds a folder and its contents to a zip archive
Args:
src_folder_name (str): Source folder name to add to the archive
dst_zip_archive (ZipFile): Destination zip archive
Returns:
None
"""
for walk_item in os.walk(src_folder_name):
for file_item in walk_item[2]:
# walk_item[2] is a list of files in the folder entry
# walk_item[0] is the folder entry full path
fn_to_add = os.path.join(walk_item[0], file_item)
dst_zip_archive.write(fn_to_add)
if __name__ == '__main__':
zf = zipfile.ZipFile('myzip.zip', mode='w')
add_folder_to_zip('zip_this_folder', zf)
zf.close()