zip = zipfile.ZipFile(destination+ff_name,"w")
zip.write(source)
zip.close()
Above is the code that I am using, and here "source" is the path of the directory. But when I run this code it just zips the source folder and not the files and and folders contained in it. I want it to compress the source folder recursively. Using tarfile module I can do this without passing any additional information.
The standard os.path.walk() function will likely be of great use for this.
Alternatively, reading the tarfile module to see how it does its work will certainly be of benefit. Indeed, looking at how pieces of the standard library were written was an invaluable part of my learning Python.
I haven't tested this exactly, but it's something similar to what I use.
zip = zipfile.ZipFile(destination+ff_name, 'w', zipfile.ZIP_DEFLATED)
rootlen = len(source) + 1
for base, dirs, files in os.walk(source):
for file in files:
fn = os.path.join(base, file)
zip.write(fn, fn[rootlen:])
This example is from here:
http://bitbucket.org/jgrigonis/mathfacts/src/ff57afdf07a1/setupmac.py
I'd like to add a "new" python 2.7 feature to this topic: ZipFile can be used as a context manager and therefore you can do things like this:
with zipfile.ZipFile(my_file, 'w') as myzip:
rootlen = len(xxx) #use the sub-part of path which you want to keep in your zip file
for base, dirs, files in os.walk(pfad):
for ifile in files:
fn = os.path.join(base, ifile)
myzip.write(fn, fn[rootlen:])
Related
I got below code from online and I am trying to add a password and I want to change the result directory to be "C:#SFTPDWN" (Final Zip file should be in this folder).
I try to change it like below, it did not work.
with ZipFile('CC-Data.zip', 'w', 'pass word') as zip:
Can anybody please tell how to change this code to add password and change result folder?
One last thing, currently it will zip #SFTPDWN folder, I just want to zip everything inside (Right now it will create two folders (CC-Data.zip and inside it #SFTPDWN )). Can anybody please tell me how to zip everything inside #SFTPDWN folder?
Code
from zipfile import ZipFile
import os
def get_all_file_paths(directory):
file_paths = []
for root, directories, files in os.walk(directory):
for filename in files:
filepath = os.path.join(root, filename)
file_paths.append(filepath)
return file_paths
def main():
# path to folder which needs to be zipped
directory = 'C:\#SFTPDWN'
file_paths = get_all_file_paths(directory)
print('Following files will be zipped:')
for file_name in file_paths:
print(file_name)
with ZipFile('CC-Data.zip', 'w') as zip:
# writing each file one by one
for file in file_paths:
zip.write(file)
print('Zipped successfully!')
if __name__ == "__main__":
main()
For the password question: from the documentation:
This module [...] supports decryption of encrypted files in ZIP archives, but it currently cannot create an encrypted file. Decryption is extremely slow as it is implemented in native Python rather than C.
https://docs.python.org/3/library/zipfile.html
You would need to use a 3rd party library to create an encrypted zip, or encrypt the archive some other way.
For the second part, in ZipFile.write the documentation also mentions:
ZipFile.write(filename, arcname=None, compress_type=None, compresslevel=None)
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). [...]
Note: Archive names should be relative to the archive root, that is, they should not start with a path separator.
https://docs.python.org/3/library/zipfile.html#zipfile.ZipFile.write
So you would need to strip off whatever prefix of your file variable and pass that as the arcname parameter. Using os.path.relpath may help, e.g. (I'm on Linux, but should work with Windows paths under Windows):
>>> os.path.relpath("/folder/subpath/myfile.txt", "/folder/")
'subpath/myfile.txt'
Sidebar: a path like "C:\Something" is an illegal string, as it has the escape \S. Python kinda tolerates this (I think in 3.8 it will error) and rewrites them literally. Either use "C:\\Something", r"C:\Something", or "C:/Something" If you attempted something like "C:\Users" it would actually throw an error, or "C:\nothing" it might silently do something strange...
as the title would imply I am looking to create a script that will allow me to print a list of file names in a directory to a CSV file.
I have a folder on my desktop that contains approx 150 pdf's. I'd like to be able to have the file names printed to a csv.
I am brand new to Python and may be jumping out of the frying pan and into the fire with this project.
Can anyone offer some insight to get me started?
First off you will want to start by grabbing all of the files in the directory, then simply by writing them to a file.
from os import listdir
from os.path import isfile, join
import csv
onlyfiles = [f for f in listdir("./") if isfile(join("./", f))]
with open('file_name.csv', 'w') as print_to:
writer = csv.writer(print_to)
writer.writerow(onlyfiles)
Please Note
"./" on line 5 is the directory you want to grab the files from.
Please replace 'file_name.csv' with the name of the file you want to right too.
The following will create a csv file with all *.pdf files:
from glob import glob
with open('/tmp/filelist.csv', 'w') as fout:
# write the csv header -- optional
fout.write("filename\n")
# write each filename with a newline characer
fout.writelines(['%s\n' % fn for fn in glob('/path/to/*.pdf')])
glob() is a nice shortcut to using listdir because it supports wildcards.
import os
csvpath = "csvfile.csv"
dirpath = "."
f = open("csvpath, "wb")
f.write(",".join(os.listdir(dirpath)))
f.close()
This may be improved to present filenames in way that you need, like for getting them back, or something. For instance, this most probably won't include unicode filenames in UTF-8 form but make some mess out of the encoding, but it is easy to fix all that.
If you have very big dir, with many files, you may have to wait some time for os.listdir() to get them all. This also can be fixed by using some other methods instead of os.listdir().
To differentiate between files and subdirectories see Michael's answer.
Also, using os.path.isfile() or os.path.isdir() you can recursively get all subdirectories if you wish.
Like this:
def getall (path):
files = []
for x in os.listdir(path):
x = os.path.join(path, x)
if os.path.isdir(x): files += getall(x)
else: files.append(x)
return files
I'm uploading a zipped folder that contains a folder of text files, but it's not detecting that the folder that is zipped up is a directory. I think it might have something to do with requiring an absolute path in the os.path.isdir call, but can't seem to figure out how to implement that.
zipped = zipfile.ZipFile(request.FILES['content'])
for libitem in zipped.namelist():
if libitem.startswith('__MACOSX/'):
continue
# If it's a directory, open it
if os.path.isdir(libitem):
print "You have hit a directory in the zip folder -- we must open it before continuing"
for item in os.listdir(libitem):
The file you've uploaded is a single zip file which is simply a container for other files and directories. All of the Python os.path functions operate on files on your local file system which means you must first extract the contents of your zip before you can use os.path or os.listdir.
Unfortunately it's not possible to determine from the ZipFile object whether an entry is for a file or directory.
A rewrite or your code which does an extract first may look something like this:
import tempfile
# Create a temporary directory into which we can extract zip contents.
tmpdir = tempfile.mkdtemp()
try:
zipped = zipfile.ZipFile(request.FILES['content'])
zipped.extractall(tmpdir)
# Walk through the extracted directory structure doing what you
# want with each file.
for (dirpath, dirnames, filenames) in os.walk(tmpdir):
# Look into subdirectories?
for dirname in dirnames:
full_dir_path = os.path.join(dirpath, dirname)
# Do stuff in this directory
for filename in filenames:
full_file_path = os.path.join(dirpath, filename)
# Do stuff with this file.
finally:
# ... Clean up temporary diretory recursively here.
Usually to make things handle relative paths etc when running scripts you'd want to use os.path.
It seems to me that you're reading from a Zipfile the items you've not actually unzipped it so why would you expect the file/dirs to exist?
Usually I'd print os.getcwd() to find out where I am and also use os.path.join to join with the root of the data directory, whether that is the same as the directory containing the script I can't tell. Using something like scriptdir = os.path.dirname(os.path.abspath(__file__)).
I'd expect you would have to do something like
libitempath = os.path.join(scriptdir, libitem)
if os.path.isdir(libitempath):
....
But I'm guessing at what you're doing as it's a little unclear for me.
I read through the zipfile documentation, but couldn't understand how to unzip a file, only how to zip a file. How do I unzip all the contents of a zip file into the same directory?
import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
zip_ref.extractall(directory_to_extract_to)
That's pretty much it!
If you are using Python 3.2 or later:
import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
zip_ref.extractall("targetdir")
You dont need to use the close or try/catch with this as it uses the
context manager construction.
zipfile is a somewhat low-level library. Unless you need the specifics that it provides, you can get away with shutil's higher-level functions make_archive and unpack_archive.
make_archive is already described in this answer. As for unpack_archive:
import shutil
shutil.unpack_archive(filename, extract_dir)
unpack_archive detects the compression format automatically from the "extension" of filename (.zip, .tar.gz, etc), and so does make_archive. Also, filename and extract_dir can be any path-like objects (e.g. pathlib.Path instances) since Python 3.7.
Use the extractall method, if you're using Python 2.6+
zip = ZipFile('file.zip')
zip.extractall()
You can also import only ZipFile:
from zipfile import ZipFile
zf = ZipFile('path_to_file/file.zip', 'r')
zf.extractall('path_to_extract_folder')
zf.close()
Works in Python 2 and Python 3.
try this :
import zipfile
def un_zipFiles(path):
files=os.listdir(path)
for file in files:
if file.endswith('.zip'):
filePath=path+'/'+file
zip_file = zipfile.ZipFile(filePath)
for names in zip_file.namelist():
zip_file.extract(names,path)
zip_file.close()
path : unzip file's path
If you want to do it in shell, instead of writing code.
python3 -m zipfile -e myfiles.zip myfiles/
myfiles.zip is the zip archive and myfiles is the path to extract the files.
from zipfile import ZipFile
ZipFile("YOURZIP.zip").extractall("YOUR_DESTINATION_DIRECTORY")
The directory where you will extract your files doesn't need to exist before, you name it at this moment
YOURZIP.zip is the name of the zip if your project is in the same directory.
If not, use the PATH i.e : C://....//YOURZIP.zip
Think to escape the / by an other / in the PATH
If you have a permission denied try to launch your ide (i.e: Anaconda) as administrator
YOUR_DESTINATION_DIRECTORY will be created in the same directory than your project
import os
zip_file_path = "C:\AA\BB"
file_list = os.listdir(path)
abs_path = []
for a in file_list:
x = zip_file_path+'\\'+a
print x
abs_path.append(x)
for f in abs_path:
zip=zipfile.ZipFile(f)
zip.extractall(zip_file_path)
This does not contain validation for the file if its not zip. If the folder contains non .zip file it will fail.
I have been able to zip the contents of my folder. But I would like the zipped file to remain in the folder that was just compressed. For example, I've zipped a folder called test in my C: drive. But I would like my "test.zip" file to be contained in C:\test. How can I do this? Thanks in advance.
clarification of question with code example:
Someone kindly pointed out that my question is confusing, but for a python newbie a lot of things are confusing :) - my advance apologies if this question is too basic or the answer is obvious. I don't know how I can ensure that the resulting zip file is inside the folder that has been zipped. In other words, I would like the zip process to take place in 'basedir.' That way the user does not waste time searching for it somewhere on the C drive.
def zip_folder(basedir, zip_file):
z = zipfile.ZipFile(zip_file, 'w', zipfile.ZIP_DEFLATED)
for dirpath, dirnames, filenames in os.walk(basedir):
print "zipping files:"
for fn in filenames:
print fn
absfn = os.path.join(dirpath, fn)
z.write(absfn)
z.close
Whatever you pass as zip_file to your function will be the file that the ZipFile object will write to. So if you pass it a full path, then it will be put there. If you pass it just a filename, then it will be written to that filename under the current working path. It sounds like you just need to make sure that zip_file is an absolute path.