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?

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

Why does a python program continue to write to the directory in which it was originally located, after I move the file to another directory?

I have 2 directories. I had a python program located in dir_1 writing to a .txt file also in dir_1. I meant to create them in dir_2, but when I move them both to dir_2, the python program, instead of writing to the existing .txt file that is now with it in dir_2, creates a new .txt file in dir_1 and writes to it. How do I fix this? I'm very new to programming and python and googling didn't help me out, probably because I didn't know what exactly to search.
with open('guest_book.txt', 'w') as file:
while True:
name = input('Please enter your name: ')
if name == 'q':
break
else:
print(f"Hello, {name.title()}!\nYou have been added to the guest"
f"book")
file.write(f"{name.title()}\n")

Python writes to the file location you supply it with. If this file location is a relative path, then it will create files relative to the directory of the script. I.e. when you move the script then the .txt file will be created relative to the new directiory.
On the other hand, if you provide an absolute path, then it does not matter where the script is located / where you execute it from. Instead, it will create the file at that location always.
From the sounds of it, you are using an absolute path when you want a relative path.
So change from something like /home/bob/file.txt (Linux) or C:\\Users\Bob\file.txt (Win) to simply file.txt or even ./file.txt.
Update: Since you were using a relative location all along, the problem will lie with the context that you are executing the script from. Your code is not the issue here, it is how you are executing it.
As vlovero suggests, maybe your IDE is not executing the new file in its new location?
One way you can test this robustly is to navigate to dir_2 in a terminal and run
python your_program_name.py
This will execute the script in the dir_2 location.

Since you have not specified an absolute path, your program is then specifying a directory relative to the current working directory (if instead, for example, you had specified a path such as '../guest_book.txt', you would have been specifying a directory one level above the current working directory). So let's imagine your OS is Linux and the Python program resides in /my_home/programs:
cd /my_home/data # this is the current working directory
python ../programs/your_program.py
The current working directory when the program is executed is /home/my_home/data even though the program being executed resides in /my_home/programs, and thus the output file will be created in the /my_home/data directory. os.getcwd() can be called to tell you what the current working directory is.

How to open any program in Python?

Well I searched a lot and found different ways to open program in python,
For example:-
import os
os.startfile(path) # I have to give a whole path that is not possible to give a full path for every program/software in my case.
The second one that I'm currently using
import os
os.system(fileName+'.exe')
In second example problem is:-
If I want to open calculator so its .exe file name is calc.exe and this happen for any other programs too (And i dont know about all the .exe file names of every program).
And assume If I wrote every program name hard coded so, what if user installed any new program. (my program wont able to open that program?)
If there is no other way to open programs in python so Is that possible to get the list of all install program in user's computer.
and there .exe file names (like:- calculator is calc.exe you got the point).
If you want to take a look at code
Note: I want generic solution.

There's always:
from subprocess import call
call(["calc.exe"])
This should allow you to use a dict or list or set to hold your program names and call them at will. This is covered also in this answer by David Cournapeau and chobok.

You can try with os.walk :
import os
exe_list=[]
for root, dirs, files in os.walk("."):
#print (dirs)
for j in dirs:
for i in files:
if i.endswith('.exe'):
#p=os.getcwd()+'/'+j+'/'+i
p=root+'/'+j+'/'+i
#print(p)
exe_list.append(p)
for i in exe_list :
print('index : {} file :{}'.format(exe_list.index(i),i.split('/')[-1]))
ip=int(input('Enter index of file :'))
print('executing {}...'.format(exe_list[ip]))
os.system(exe_list[ip])
os.getcwd()+'/'+i prepends the path of file to the exe file starting from root.
exe_list.index(i),i.split('/')[-1] fetches just the filename.exe
exe_list stores the whole path of an exe file at each index

Can be done with winapps
First install winapps by typing:
pip install winapps
After that use the library:
# This will give you list of installed applications along with some information
import winapps
for app in winapps.list_installed():
print(app)
If you want to search for an app you can simple do:
application = 'chrome'
for app in winapps.search_installed(application):
print(app)

Python3:Save File to Specified Location

I have a rather simple program that writes HTML code ready for use.
It works fine, except that if one were to run the program from the Python command line, as is the default, the HTML file that is created is created where python.exe is, not where the program I wrote is. And that's a problem.
Do you know a way of getting the .write() function to write a file to a specific location on the disc (e.g. C:\Users\User\Desktop)?
Extra cool-points if you know how to open a file browser window.

The first problem is probably that you are not including the full path when you open the file for writing. For details on opening a web browser, read this fine manual.
import os
target_dir = r"C:\full\path\to\where\you\want\it"
fullname = os.path.join(target_dir,filename)
with open(fullname,"w") as f:
f.write("<html>....</html>")
import webbrowser
url = "file://"+fullname.replace("\\","/")
webbrowser.open(url,True,True)
BTW: the code is the same in python 2.6.

I'll admit I don't know Python 3, so I may be wrong, but in Python 2, you can just check the __file__ variable in your module to get the name of the file it was loaded from. Just create your file in that same directory (preferably using os.path.dirname and os.path.join to remain platform-independent).