Develop Reference

Rename filename on windows enviroment github workflow

I was trying to process some data in a GitHub action.
However, due to a Japanese file name, I can not read the file successfully by:
pd.read_csv('C:\\202204_10エリア計.csv')
And I was trying to rename it before read it by:
for filename in os.listdir(download_path):
if filename.startswith('202204'):
filename = filename.encode('utf-8').decode(locale.getpreferredencoding(False))
print(filename) # this print '202204_10エリア計.csv' on github action
os.rename(os.path.join(download_path, filename), os.path.join(download_path, '202204.csv')
But it gets error:
[WinError 2] The system cannot find the file specified: 'C:\\202204_10エリア計.csv' -> 'C:\\202204.csv'

You should be able to sidestep the encoding/decoding entirely with pathlib:
from pathlib import Path
fn = next(p for p in Path(download_path).glob("*.csv") if p.name.startswith("202204"))
fn.rename(fn.with_stem("202204"))
This is a bit of a workaround to whatever the real issue is, however.
That said I have never needed to meddle with the encoding when using os.path, and a quick search of the docs doesn't turn up anything, so you may be fine if you simple remove your encoding/decoding step. I would expect the os.path api to use the same internal representation throughout.

Writing Data to File then Moving to Desktop

I'm sure this is really simple but I can't figure out how to make a parsed result into its own file then move it to my desktop using python. Here is my code so far. I just want to save the result "names" as its own file then move it to my desktop but I can't find the answer anywhere. Is this an uncommon practice?
from gedcom.element.individual import IndividualElement
from gedcom.parser import Parser
import os
import shutil
import pickle
# Path to your `.ged` file
file_path = '/Users/Justin/Desktop/Lienhard_Line.ged'
# Initialize the parser
gedcom_parser = Parser()
# Parse your file
gedcom_parser.parse_file(file_path, False)
root_child_elements = gedcom_parser.get_root_child_elements()
# Iterate through all root child elements
for element in root_child_elements:
# Is the `element` an actual `IndividualElement`? (Allows usage of extra functions such as `surname_match` and `get_name`.)
if isinstance(element, IndividualElement):
# Get all individuals whose surname matches "Doe"
if element.surname_match('Lienhard'):
# Unpack the name tuple
(first, last) = element.get_name()
names = (first + " " + last)
pickle.dumps(names)

Saving a file to one location and then moving it is to another not how it's usually done, no. Just save to the final location.
from pathlib import Path
pic = Path.home() / 'Desktop' / 'names.pickle'
with pic.open('w') as picklefile:
pickle.dump(names, picklefile)
The pathlib library makes working with file names somewhat easier than the old os.path API, though both are in common use.
Writing and then renaming has some uses; if you need to absolutely make sure your data is saved somewhere, saving it to a temporary file until it can be renamed to its final name is a fairly common technique. But in this case, saving to a different location first seems like it would only introduce brittleness.
The above assumes that you have a directory named Desktop in your home directory, as is commonly the default on beginner-oriented systems. To be perfectly robust, the code should create the directory if it doesn't already exist, or perhaps simply save to your home directory.
Indeed, a much better convention for most purposes is simply to always save in the current directory, and let the user take it from there. Hardcoding paths in another directory just adds confusion and uncertainty.
with open('names.pickle', 'w') as picklefile:
pickle.dump(names, picklefile)
This code creates a file called names.pickle in the invoking user's current working directory. The operating system provides the concept of a current working directory for precisely this purpose. Perhaps see also Difference between ./ and ~/ if you need more details about how this works.

How do I access a file for reading/writing in a different (non-current) directory?

I am working on the listener portion of a backdoor program (for an ETHICAL hacking course) and I would like to be able to read files from any part of my linux system and not just from within the directory where my listener python script is located - however, this has not proven to be as simple as specifying a typical absolute path such as "~/Desktop/test.txt"
So far my code is able to read files and upload them to the virtual machine where my reverse backdoor script is actively running. But this is only when I read and upload files that are in the same directory as my listener script (aptly named listener.py). Code shown below.
def read_file(self, path):
with open(path, "rb") as file:
return base64.b64encode(file.read())
As I've mentioned previously, the above function only works if I try to open and read a file that is in the same directory as the script that the above code belongs to, meaning that path in the above content is a simple file name such as "picture.jpg"
I would like to be able to read a file from any part of my filesystem while maintaining the same functionality.
For example, I would love to be able to specify "~/Desktop/another_picture.jpg" as the path so that the contents of "another_picture.jpg" from my "~/Desktop" directory are base64 encoded for further processing and eventual upload.
Any and all help is much appreciated.
Edit 1:
My script where all the code is contained, "listener.py", is located in /root/PycharmProjects/virus_related/reverse_backdoor/. within this directory is a file that for simplicity's sake we can call "picture.jpg" The same file, "picture.jpg" is also located on my desktop, absolute path = "/root/Desktop/picture.jpg"
When I try read_file("picture.jpg"), there are no problems, the file is read.
When I try read_file("/root/Desktop/picture.jpg"), the file is not read and my terminal becomes stuck.
Edit 2:
I forgot to note that I am using the latest version of Kali Linux and Pycharm.
I have run "realpath picture.jpg" and it has yielded the path "/root/Desktop/picture.jpg"
Upon running read_file("/root/Desktop/picture.jpg"), I encounter the same problem where my terminal becomes stuck.
[FINAL EDIT aka Problem solved]:
Based on the answer suggesting trying to read a file like "../file", I realized that the code was fully functional because read_file("../file") worked without any flaws, indicating that my python script had no trouble locating the given path. Once the file was read, it was uploaded to the machine running my backdoor where, curiously, it uploaded the file to my target machine but in the parent directory of the script. It was then that I realized that problem lied in the handling of paths in the backdoor script rather than my listener.py
Credit is also due to the commentator who pointed out that "~" does not count as a valid path element. Once I reached the conclusion mentioned just above, I attempted read_file("~/Desktop/picture.jpg") which failed. But with a quick modification, read_file("/root/Desktop/picture.jpg") was successfully executed and the file was uploaded in the same directory as my backdoor script on my target machine once I implemented some quick-fix code.
My apologies for not being so specific; efforts to aid were certainly confounded by the unmentioned complexity of my situation and I would like to personally thank everyone who chipped in.
This was my first whole-hearted attempt to reach out to the stackoverflow community for help and I have not been disappointed. Cheers!

A solution I found is putting "../" before the filename if the path is right outside of the dictionary.
test.py (in some dictionary right inside dictionary "Desktop" (i.e. /Desktop/test):
with open("../test.txt", "r") as test:
print(test.readlines())
test.txt (in dictionary "/Desktop")
Hi!
Hello!
Result:
["Hi!", "Hello!"]
This is likely the simplest solution. I found this solution because I always use "cd ../" on the terminal.

This not only allows you to modify the current file, but all other files in the same directory as the one you are reading/writing to.
path = os.path.dirname(os.path.abspath(__file__))
dir_ = os.listdir(path)
for filename in dir_:
f = open(dir_ + '/' + filename)
content = f.read()
print filename, len(content)
try:
im = Image.open(filename)
im.show()
except IOError:
print('The following file is not an image type:', filename)

WinError 32 :The process cannot access the file because it is being used by another process

I have written the following code to extract zip files in a directory and a delete a particular excel file in the extracted directory :
def extractZipFiles(dest_directory):
"This function extracts zip files in the destination directory for further processing"
fileFullPath = dest_directory + '\\'
extractedDirList = list()
for file in os.listdir(dest_directory):
dn = fileFullPath+file
dn = re.sub(r'\.zip$', "", fileFullPath+file) #remove the trailing .zip.
extractedDirList.append(dn)
zf = zipfile.ZipFile(fileFullPath+file, mode='r')
zf.extractall(dn) # extract the contents of that zip to the empty directory
zf.close()
return extractedDirList
def removeSelectedReports(extractedDirList):
"This function removes the selected reports from extracted directory"
for i in range(len(extractedDirList)):
for filename in os.listdir(extractedDirList[i]):
if filename.startswith("ABC_8"):
logger.info("File to be removed::"+filename)
fullPathName= "%s/%s" % (extractedDirList[i],filename)
os.remove(fullPathName)
return
extractedDirList = extractZipFiles(attributionRptDestDir)
logger.info("ZIP FILES EXTRACTED:"+str(extractedDirList))
removeSelectedReports(extractedDirList)
I am getting the following intermittent issue even though I have closed the zip file handler.
[WinError 32] The process cannot access the file because it is being used by another process: '\\\\share\\Workingdirectory\\report.20180517.zip'
Can you please help resolve this issue

You should try to figure out what has the file open. Based on your code, it looks like you are on Microsoft Windows.
I would stop all applications on your workstation, including browsers, run with only a minimum number of apps open, and reproduce the problem. Once reproduced you can use a tool to lists all handles open to a particular file.
A handy utility would be handle.exe, but please use any tool with similar functionality.
Once you find the offending application, you can further investigate why the file is open, and take counter measures.
I would be careful not to close any application which has the file open, until you know it is safe to do so.

Get file id in windows with python?

Hello I want to get the file id from files on windows with python. When I searched I could only find how to do it in other languages. Does anybody know how I can achieve this in python?

As far as I have looked and researched, there is no such file id available. But instead, you can have the creation date on Windows and Mac, and the last modified on Linux. These two are usually sufficient to find unique files, even if they are renamed, altered, or whatever.
Here's how to do it, along with the source SO thread I found the solution.
import os
import platform
def creation_date(path_to_file):
"""
Try to get the date that a file was created, falling back to when it was
last modified if that isn't possible.
See http://stackoverflow.com/a/39501288/1709587 for explanation.
"""
if platform.system() == 'Windows':
return os.path.getctime(path_to_file)
else:
stat = os.stat(path_to_file)
try:
return stat.st_birthtime
except AttributeError:
# We're probably on Linux. No easy way to get creation dates here,
# so we'll settle for when its content was last modified.
return stat.st_mtime

import os
path_to_file = r"path_to_your_file"
file_id = os.stat(path_to_file, follow_symlinks=False).st_ino
print(hex(file_id))
to check the result from the commandline:
c:\> fsutil file queryfileid path_to_your_file
so in Python you can also use
print(os.popen(fr"fsutil file queryfileid path_to_your_file").read())
or when you have hardlinks:
print(os.popen(fr"fsutil hardlink list path_to_your_file").read())
to find the filename with an id:
print(os.popen(fr'fsutil file queryFileNameById c:\ the_file_id').read())