Develop Reference

How to open and load a JSON file inside a .zip archive?

So I have a python script I run in a directory which contains .zip files and these zip files all have JSON files which I want to load. With my current method I get an error of 'No such file or directory' when I try to 'open(filename)' I assume this is because namelist() doesn't actually enter the .zip directory. How can I load the JSON file inside this zip archive once I confirm that it is indeed a .zip archive?
import zipfile
import json
import os
myList = []
files = [f for f in os.listdir('.') if os.path.isfile(f)]
for f in files:
if f.endswith('.zip'):
with zipfile.ZipFile(f) as myzip:
for filename in myzip.namelist():
if filename.endswith('.json'):
g = open(filename)
data = json.load(g)
#do stuff with g and myList

Compress all files in a folder with python?

this code takes a bunch of files in a folder (based on the file name), zips them into bz2 and adds them into a tar file. Is there a way I can modify this to only compress the files into bz2 (or gzip)? I do not want to have to deal with having them packaged into a tar. I just want to go through each file in a directory and compress it.
import os
from glob import glob
import tarfile
os.chdir(r'C:\Documents\FTP\\')
compression = "w:bz2"
extension = '.tar.bz2'
filename = 'survey_'
filetype = 'survey_report_*.csv'
tarname = saveloc+filename+extension
files = glob(filetype)
tar = tarfile.open(tarname, compression)
for file in files:
if file not in tarname:
print('Packaging file:', file)
tar.add(file)
tar.close()
EDIT:
This code seems to work for some files, but for other ones it makes them 1kb and when I open it there are just some random characters. Any suggestions?
import bz2
import os
location = r'C:\Users\Documents\FTP\\'
os.chdir(location)
filelist = os.listdir(location)
for file in filelist:
data = open(file).read()
try:
output = bz2.BZ2File(file + '.bz2', 'wb')
output.write(data)
finally:
output.close()

PYTHON - Zip each file in directory independently

I've been looking through and have tried a few different codes without results. What I'm trying to do is zip each file in a subdirectory/folder independently.
Ex:
FileName.prj
FileName.dwg
FileName.mp3
Each as it's own .zip
Thanks!

Try this
import os
import zipfile
folder = "/tmp/in"
dest_folder = "/tmp/out"
l = [os.path.join(folder, fname) for fname in os.listdir(folder)]
os.chdir(folder)
for f in l:
f_name = f[f.rfind("/")+1:]+".zip"
z = zipfile.ZipFile(f_name, 'w')
z.write(f_name[:f_name.rfind(".zip")])
os.rename(folder+"/"+f_name, dest_folder+"/"+f_name)
where folder is you folder that contains files you want to zip and dest_folder is folder where the zip files will be written.

How to compress a file with shutil.make_archive in python?

I want to compress one text file using shutil.make_archive command. I am using the following command:
shutil.make_archive('gzipped'+fname, 'gztar', os.path.join(os.getcwd(), fname))
OSError: [Errno 20] Not a directory: '/home/user/file.txt'
I tried several variants but it keeps trying to compress the whole folders. How to do it correctly?

Actually shutil.make_archive can make one-file archive! Just pass path to target directory as root_dir and target filename as base_dir.
Try this:
import shutil
file_to_zip = 'test.txt' # file to zip
target_path = 'C:\\test_yard\\' # dir, where file is
try:
shutil.make_archive(target_path + 'archive', 'zip', target_path, file_to_zip)
except OSError:
pass

shutil can't create an archive from one file. You can use tarfile, instead:
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
os.chdir('/home/user')
tar.add("file.txt")
tar.close()
or
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
tar.addfile(tarfile.TarInfo("/home/user/file.txt"), "/home/user/file.txt")
tar.close()

Try this and Check shutil
copy your file to a directory.
cd directory
shutil.make_archive('gzipped', 'gztar', os.getcwd())

#CommonSense had a good answer, but the file will always be created zipped inside its parent directories. If you need to create a zipfile without the extra directories, just use the zipfile module directly
import os, zipfile
inpath = "test.txt"
outpath = "test.zip"
with zipfile.ZipFile(outpath, "w", compression=zipfile.ZIP_DEFLATED) as zf:
zf.write(inpath, os.path.basename(inpath))

If you don't mind doing a file copy op:
def single_file_to_archive(full_path, archive_name_no_ext):
tmp_dir = tempfile.mkdtemp()
shutil.copy2(full_path, tmp_dir)
shutil.make_archive(archive_name_no_ext, "zip", tmp_dir, '.')
shutil.rmtree(tmp_dir)

Archiving a directory to another destination was a pickle for me but shutil.make_archive not zipping to correct destination helped a lot.
from shutil import make_archive
make_archive(
base_name=path_to_directory_to_archive},
format="gztar",
root_dir=destination_path,
base_dir=destination_path)

copying files from one folder to another

I'm trying to make a script that copies all the files other than the zipped files from a source folder to another destination folder and extract zipped files from the source folder to the destination, this is what i have come till this far:
import os
import zipfile
import shutil
myPath = "Source dir"
for root, dirs, files in os.walk(myPath):
for file in files:
if file.endswith('.zip'):
fh = open(file, 'rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
outpath = "destination dir"#Put the name of the destination folder
z.extract(name, outpath)
fh.close()
else:
fileList = os.listdir('source dir')
for f in fileList:
shutil.copy2(f, 'destination directory')
The code shows no error but there is no output too.

From Python Standard Library To get a full path (which begins with top) to a file or directory in dirpath, do os.path.join(dirpath, name) so you should write
fh = open(so.path.join(root,file))
to have the correct path.