I'm trying to look at all the items we have in a directory, and check whether there are file or directory. Sadly, all of them aren't directory, neither a file.
import io, sys, os, json
path_input = "C:\\Users\\Me\\AppData\\Local\\somewhere\\"
for file in os.listdir(path_input):
print("Looking at " + file)
isFile = os.path.isfile(file) # False
isDir = os.path.isdir(file) # False
I'm pretty sure I'm missing something with how I handle file.
I forgot to add the path in my function to properly locate the file (or the directory)
isDirectory = os.path.isdir(path_input+file)
Someone posted the solution but the post got deleted
Related
I'm having an issue figuring out how can i implement a function to overcome my problem.
My code works fine, but the fault is that sometimes i need more than just one file.
import os
import shutil
target_dir = 'C:\DIST\WR_4.5\Test'
pass_text = 'Test End'
def check_and_copy():
log = open('C:\DIST\WR_4.5\oHistory2.log')
for line in log:
if pass_text in line:
name = '\kp.log'
else:
name = '\kf.log'
if not os.path.exists(target_dir):
os.makedirs(target_dir)
shutil.copy2('C:\DIST\WR_4.5\oHistory2.log', target_dir+name)
else:
os.chdir(target_dir)
shutil.copy2('C:\DIST\WR_4.5\oHistory2.log', target_dir+name)
What i'd like to accomplish is to be able to copy multiple instances of files with or without shutil. The filename is determined by the code itself as you can see.
Thanks for the help in advance.
I've pip-installed the python-dotenv module and am trying to use it as follows. I've created a directory test_dotenv with the following tree:
.
├── config.env
└── test_dotenv_subdir
└── test_dotenv.py
The config.env file is simply an empty file, while test_dotenv.py contains the following:
import dotenv
found_dotenv = dotenv.find_dotenv()
print(found_dotenv)
However, if I run this script using python test_dotenv.py, I find that nothing gets printed; that is, found_dotenv is an empty string ('').
Am I missing something about how to use this method? As far as I can tell, here is the relevant part of the source code:
def _walk_to_root(path):
"""
Yield directories starting from the given directory up to the root
"""
if not os.path.exists(path):
raise IOError('Starting path not found')
if os.path.isfile(path):
path = os.path.dirname(path)
last_dir = None
current_dir = os.path.abspath(path)
while last_dir != current_dir:
yield current_dir
parent_dir = os.path.abspath(os.path.join(current_dir, os.path.pardir))
last_dir, current_dir = current_dir, parent_dir
def find_dotenv(filename='.env', raise_error_if_not_found=False, usecwd=False):
"""
Search in increasingly higher folders for the given file
Returns path to the file if found, or an empty string otherwise
"""
if usecwd or '__file__' not in globals():
# should work without __file__, e.g. in REPL or IPython notebook
path = os.getcwd()
else:
# will work for .py files
frame_filename = sys._getframe().f_back.f_code.co_filename
path = os.path.dirname(os.path.abspath(frame_filename))
for dirname in _walk_to_root(path):
check_path = os.path.join(dirname, filename)
if os.path.exists(check_path):
return check_path
if raise_error_if_not_found:
raise IOError('File not found')
return ''
It seems like os.path.exists(check_path) keeps on returning False, so that the in the end an empty string is returned. Why is this not working as intended?
The dotenv package looks for a file that is literally called .env when you call find_dotenv rather than looking for files like .env. (e.g. config.env)
You can make your code find the desired file by passing the filename to find_dotenv, i.e.
import dotenv
found_dotenv = dotenv.find_dotenv('config.env')
print(found_dotenv)
One possible reason for that can be *.pyc files from another host stored in the project.
i.e. in my case, I copied the content of my repo into the container and start getting this error:
.venv/lib/python3.7/site-packages/dotenv/main.py:257: in _walk_to_root
raise IOError('Starting path not found')
E OSError: Starting path not found
I fixed it by running:
find . -name '*.pyc' | xargs rm -frv {} ;
I need to extract a file called Preview.pdf from a folder called QuickLooks inside of a zip file.
Right now my code looks a little like this:
with ZipFile(newName, 'r') as newName:
newName.extract(\QuickLooks\Preview.pdf)
newName.close()
(In this case, newName has been set equal to the full path to the zip).
It's important to note that the backslash is correct in this case because I'm on Windows.
The code doesn't work; here's the error it gives:
Traceback (most recent call last):
File "C:\Users\Asit\Documents\Evam\Python_Scripts\pageszip.py", line 18, in <module>
ZF.extract("""QuickLooks\Preview.pdf""")
File "C:\Python33\lib\zipfile.py", line 1019, in extract
member = self.getinfo(member)
File "C:\Python33\lib\zipfile.py", line 905, in getinfo
'There is no item named %r in the archive' % name)
KeyError: "There is no item named 'QuickLook/Preview.pdf' in the archive"
I'm running the Python script from inside Notepad++, and taking the output from its console.
How can I accomplish this?
Alternatively, how could I extract the whole QuickLooks folder, move out Preview.pdf, and then delete the folder and the rest of it's contents?
Just for context, here's the rest of the script. It's a script to get a PDF of a .pages file. I know there are bonified converters out there; I'm just doing this as an excercise with some sort of real-world application.
import os.path
import zipfile
from zipfile import *
import sys
file = raw_input('Enter the full path to the .pages file in question. Please note that file and directory names cannot contain any spaces.')
dir = os.path.abspath(os.path.join(file, os.pardir))
fileName, fileExtension = os.path.splitext(file)
if fileExtension == ".pages":
os.chdir(dir)
print (dir)
fileExtension = ".zip"
os.rename (file, fileName + ".zip")
newName = fileName + ".zip" #for debugging purposes
print (newName) #for debugging purposes
with ZipFile(newName, 'w') as ZF:
print("I'm about to list names!")
print(ZF.namelist()) #for debugging purposes
ZF.extract("QuickLook/Preview.pdf")
os.rename('Preview.pdf', fileName + '.pdf')
finalPDF = fileName + ".pdf"
print ("Check out the PDF! It's located at" + dir + finalPDF + ".")
else:
print ("Sorry, this is not a valid .pages file.")
sys.exit
I'm not sure if the import of Zipfile is redundant; I read on another SO post that it was better to use from zipfile import * than import zipfile. I wasn't sure, so I used both. =)
EDIT: I've changed the code to reflect the changes suggested by Blckknght.
Here's something that seems to work. There were several issues with your code. As I mentioned in a comment, the zipfile must be opened with mode 'r' in order to read it. Another is that zip archive member names always use forward slash / characters in their path names as separators (see section 4.4.17.1 of the PKZIP Application Note). It's important to be aware that there's no way to extract a nested archive member to a different subdirectory with Python's currentzipfilemodule. You can control the root directory, but nothing below it (i.e. any subfolders within the zip).
Lastly, since it's not necessary to rename the .pages file to .zip — the filename you passZipFile() can have any extension — I removed all that from the code. However, to overcome the limitation on extracting members to a different subdirectory, I had to add code to first extract the target member to a temporary directory, and then copy that to the final destination. Afterwards, of course, this temporary folder needs to deleted. So I'm not sure the net result is much simpler...
import os.path
import shutil
import sys
import tempfile
from zipfile import ZipFile
PREVIEW_PATH = 'QuickLooks/Preview.pdf' # archive member path
pages_file = input('Enter the path to the .pages file in question: ')
#pages_file = r'C:\Stack Overflow\extract_test.pages' # hardcode for testing
pages_file = os.path.abspath(pages_file)
filename, file_extension = os.path.splitext(pages_file)
if file_extension == ".pages":
tempdir = tempfile.gettempdir()
temp_filename = os.path.join(tempdir, PREVIEW_PATH)
with ZipFile(pages_file, 'r') as zipfile:
zipfile.extract(PREVIEW_PATH, tempdir)
if not os.path.isfile(temp_filename): # extract failure?
sys.exit('unable to extract {} from {}'.format(PREVIEW_PATH, pages_file))
final_PDF = filename + '.pdf'
shutil.copy2(temp_filename, final_PDF) # copy and rename extracted file
# delete the temporary subdirectory created (along with pdf file in it)
shutil.rmtree(os.path.join(tempdir, os.path.split(PREVIEW_PATH)[0]))
print('Check out the PDF! It\'s located at "{}".'.format(final_PDF))
#view_file(final_PDF) # see Bonus below
else:
sys.exit('Sorry, that isn\'t a .pages file.')
Bonus: If you'd like to actually view the final pdf file from the script, you can add the following function and use it on the final pdf created (assuming you have a PDF viewer application installed on your system):
import subprocess
def view_file(filepath):
subprocess.Popen(filepath, shell=True).wait()
I would like to edit the file name of several files in a list of folders and export the entire file to a new folder. While I was able to rename the file okay, the contents of the file didn't migrate over. I think I wrote my code to just create a new empty file rather than edit the old one and move it over to a new directory. I feel that the fix should be easy, and that I am missing a couple of important lines of code. Below is what I have so far:
import libraries
import os
import glob
import re
directory
directory = glob.glob('Z:/Stuff/J/extractions/test/*.fsa')
The two files in the directory look like this when printed out
Z:/Stuff/J/extractions/test\c2_D10.fsa
Z:/Stuff/J/extractions/test\c3_E10.fsa
for fn in directory:
print fn
this script was designed to manipulate the file name and export the manipulated file to a another folder
for fn in directory:
output_directory = 'Z:/Stuff/J/extractions/test2'
value = os.path.splitext(os.path.basename(fn))[0]
matchObj = re.match('(.*)_(.*)', value, re.M|re.I)
new_fn = fn.replace(str(matchObj.group(0)), str(matchObj.group(2)) + "_" + str(matchObj.group(1)))
base = os.path.basename(new_fn)
v = open(os.path.join(output_directory, base), 'wb')
v.close()
My end result is the following:
Z:/Stuff/J/extractions/test2\D10_c2.fsa
Z:/Stuff/J/extractions/test2\E10_c3.fsa
But like I said the files are empty (0 kb) in the output_directory
As Stefan mentioned:
import shutil
and replace:
v = open(os.path.join(output_directory, base), 'wb')
v.close()
with:
shutil.copyfile (fn, os.path.join(output_directory, base))
If I'am not wrong, you are only opening the file and then you are immediately closing it again?
With out any writing to the file it is surely empty.
Have a look here:
http://docs.python.org/2/library/shutil.html
shutil.copyfile(src, dst) ;)
How can I tell python to scan the current directory for a file called "filenames.txt" and if that file isn't there, to extract it from a zip file called "files.zip"? I know how to work zipfile, I just don't know how to scan the current directory for that file and use if/then loops with it..
import os.path
try:
os.path.isFile(fname)
# play with the file
except:
# unzip file
import os, zipfile
if 'filenames.txt' in os.listdir('.'):
print 'file is in current dir'
else:
zf = zipfile.ZipFile('files.zip')
zf.extract('filenames.txt')
From the documentation
$ pydoc os.path.exists
Help on function exists in os.path:
os.path.exists = exists(path)
Test whether a path exists. Returns False for broken symbolic links