Develop Reference

A file when you run creates a replica of itself and deletes the original file. (Python)

I tried creating a code where the file, when you run creates a replica of itself and deletes the original file.
Here is my code:
import shutil
import os
loc=os.getcwd()
shutil.move("./aa/test.py", loc, copy_function=shutil.copy2)
But the issue with this is that:
this code is only 1 time usable and to use it again, I need to change the name of the file or delete the newly created file and then run it again.
Also, If I run it inside a folder, It will always create the new file outside the folder (in a dir up from the exceuting program).
How Do I fix this?
Some Notes:
The copy should be made at the exact place where the original file was.
The folder was empty, just having this file. The file doesn't needs to be in a folder but I just used it as a test instance.
Yes, I understand that if I delete the original file it should stop working. I actually have a figure in my mind of how It should work:
First, a new file with the exact same content in it will be made > in the same path as the original file (with a different name probably).
Then, the original file will be deleted and the 2nd file (which is > the copy of the original file) will be renamed as the exact name and > extension as of the original file which got deleted.
This thing above should repeat every time I run the .py file (containing this code) thus making this code portable and suitable for multiple uses.
Maybe the code to be executed after the file deletion can be stored in memory cache (I guess?).

Easiest way (in pseudo code):
Get name of current script.
Read all contents in memory.
Delete current script.
Write memory contents into new file with the same name.

this code is only 1 time usable and to use it again, I need to change the name of the file or delete the newly created file and then run it again.
That is of course because the file is called differently. You could approach this by having no other files in that folder, or always prefixing the filename in the same way, so that you can find the file although it always is called differently.
Also, If I run it inside a folder, It will always create the new file outside the folder (in a dir up from the exceuting program).
That is because you move it from ./aa to ./. You can take the path of the file and reuse it, apart for the filename, and then it would be in the same folder.

Hey TheKaushikGoswami,
I believe your code does exactly what you told him to and as everybody before me stated, surely only works once. :)
I would like to throw in another idea:
First off I'd personally believe that shutil.move is more of a method for actually moving a file into another directory, as you did by accident.
https://docs.python.org/3/library/shutil.html#shutil.move
So why not firstly parameterize your folder (makes it easier for gui/cmd access) and then just copy to a temporary file and then copying from that temporary file. That way you wont get caught in errors raised if you try to create files already existing and have an easy-to-understand code!
Like so:
import shutil
import os
try:
os.mkdir('./aa/')
except:
print('Folder already exists!')
dest= './aa/'
file= 'test.py'
copypath = dest + 'tmp' + file
srcpath = dest + file
shutil.copy2(srcpath, copypath, follow_symlinks=True)
os.remove(srcpath)
shutil.copy2(copypath, srcpath, follow_symlinks=True)
os.remove(copypath)
But may I ask what your use-case is for that since it really doesn't change anything for me other than creating an exact same file?

Python permission error on windows 10 when moving/renaming file

I'm just trying to rename a file or move it to another directory. Either option will work for me. I have the most basic code here that I could find on the internet (first python project here by the way).
import shutil
import os
while True:
for file in os.listdir("C:/Users/lines/OneDrive/Desktop/loc1"):
if file.endswith(".txt"):
shutil.move("C:/Users/lines/OneDrive/Desktop/loc1/" + file, "C:/Users/lines/OneDrive/Desktop/loc2/moved.txt")
# os.rename("C:/Users/lines/OneDrive/Desktop/loc1/" + file, "C:/Users/lines/OneDrive/Desktop/loc2/moved.txt")
But no matter what I try, i ALWAYS get this error in my console when running the file:
PermissionError: [WinError 32] The process cannot access the file because it is being used by another process: 'C:/Users/lines/OneDrive/Desktop/loc1/i000130126543220150615123030.txt' -> 'C:/Users/lines/OneDrive/Desktop/loc2/moved.txt'
I truly have no idea how to fix this. Can anyone help?
Edit: both directories are totally empty and only being used for this test purpose. No other text editors have these files open anywhere.

Can i shorten a file location to the .py files location

first post here so sorry if it's hard to understand. Is it possible to shorten the directory in python to the location of the .py file?. For example, if I wanted to grab something from the directory "C:\Users\Person\Desktop\Code\Data\test.txt", and if the .py was located in the Code folder, could I shorten it to "\data\test.txt". I'm new to python so sorry if this is something really basic and I just didn't understand it correctly.
I forgot to add i plan to use this with multiple files, for example: "\data\test.txt" and \data\test2.txt

import os
CUR_FILE = os.path.abspath(__file__)
TARGET_FILE = "./data/test.txt"
print(os.path.join(CUR_FILE, TARGET_FILE))
With this, you can move around your Code directory anywhere and not have to worry about getting the full path to the file.
Also, you can run the script from anywhere and it will work (you don't have to move to Code's location to run the script.

You can import os and get current working directory ,this will give you the location of python file and then you can add the location of folder data and the file stored in that ,code is given below
import os
path=os.getcwd()
full_path1=path+"\data\test.txt"
full_path2=path+"\data\test2.txt"
print(full_path1)
print(full_path2)
I think this will work for your case and if it doesn't work then add a comment

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)

Permission denied when pandas dataframe to tempfile csv

I'm trying to store a pandas dataframe to a tempfile in csv format (in windows), but am being hit by:
[Errno 13] Permission denied: 'C:\Users\Username\AppData\Local\Temp\tmpweymbkye'
import tempfile
import pandas
with tempfile.NamedTemporaryFile() as temp:
df.to_csv(temp.name)
Where df is the dataframe. I've also tried changing the temp directory to one I am sure I have write permissions:
tempfile.tempdir='D:/Username/Temp/'
This gives me the same error message
Edit:
The tempfile appears to be locked for editing as when I change the loop to:
with tempfile.NamedTemporaryFile() as temp:
df.to_csv(temp.name + '.csv')
I can write the file in the temp directory, but then it is not automatically deleted at the end of the loop, as it is no longer a temp file.
However, if I change the code to:
with tempfile.NamedTemporaryFile(suffix='.csv') as temp:
training_data.to_csv(temp.name)
I get the same error message as before. The file is not open anywhere else.

I encountered the same error message and the issue was resolved after adding "/df.csv" to file_path.
df.to_csv('C:/Users/../df.csv', index = False)

Check your permissions and, according to this post, you can run your program as an administrator by right click and run as administrator.
We can use the to_csv command to do export a DataFrame in CSV format. Note that the code below will by default save the data into the current working directory. We can save it to a different folder by adding the foldername and a slash to the file
verticalStack.to_csv('foldername/out.csv').
Check out your working directory to make sure the CSV wrote out properly, and that you can open it! If you want, try to bring it back into python to make sure it imports properly.
newOutput = pd.read_csv('out.csv', keep_default_na=False, na_values=[""])
ref
Unlike TemporaryFile(), the user of mkstemp() is responsible for deleting the temporary file when done with it.
With the use of this function may introduce a security hole in your program. By the time you get around to doing anything with the file name it returns, someone else may have beaten you to the punch. mktemp() usage can be replaced easily with NamedTemporaryFile(), passing it the delete=False paramete.
Read more.
After export to CSV you can close your file with temp.close().
with tempfile.NamedTemporaryFile(delete=False) as temp:
df.to_csv(temp.name + '.csv')
temp.close()

Sometimes，you need check the file path that if you have right permission to read and write file. Especially when you use relative path.
xxx.to_csv('%s/file.csv'%(file_path), index = False)

Sometimes, it gives that error simply because there is another file with the same name and it has no permission to delete the earlier file and replace it with the new file.
So either name the file differently while saving it,
or
If you are working on Jupyter Notebook or a other similar environment, delete the file after executing the cell that reads it into memory. So that when you execute the cell which writes it to the machine, there is no other file that exists with that name.

I encountered the same error. I simply had not yet saved my entire python file. Once I saved my python file in VS code as "insertyourfilenamehere".py to documents(which is in my path), I ran my code again and I was able to save my data frame as a csv file.

As per my knowledge, this error pops up when one attempt to save the file that have been saved already and currently open in the background.
You may try closing those files first and then rerun the code.

Just give a valid path and a file name
e.g:
final_df.to_csv('D:\Study\Data Science\data sets\MNIST\sample.csv')