Alright, so here's the problem: I'm trying to make a .pyw file in which the exception handler opens a .py terminal. As far as I understand, you need to use a subroutine. However, I really can't get my head around that. Here's that I tried to do.
except:
subprocess.run('alphareset.py')
Also, I have 2 local variables in a file that have the same name. How would I import them into other files?
You need to specify the executable:
subprocess.run('python alphareset.py')
The association of the extension *.py with the Python executable on Windows typically does not work from within a subprocess.
Related
In a GUI app written in Python and Tkinter, I want to add a menu command which will open the source code of the *.py file in IDLE. I also want it to be platform independent.
I've tried using os.system to open IDLE from Scripts folder of Python, but that would be platform dependent. I couldn't find a way to get the Scripts folder of Python to make it independent. How can I achieve this?
To open a file in an IDLE editor, the command line is
<python> -m idlelib <path>
where <python> is the full path to a python executable or a name that resolves to such. If you want to open IDLE with the same python that is running you python app, which is likely what you want, use the platform-independent sys.executable. The path to the source code for the running app is __file__. This is one of the semi-hidden global variables when python runs a file. Or give the path to any other file.
Whether you use os.system or subprocess is a different issue. The latter is more flexible. subprocess.run replaces os.system. subprocess.Popen should not block. IDLE uses the latter to run user code.
I tried some special way:
import idlelib.pyshell
import sys
sys.argv.clear()
sys.argv.append("")
sys.argv.append(fName)
idlelib.pyshell.main()
It works! But I don't know whether this way will generate some problems.
I want to open a file in LibreOffice Calc on Ubuntu after working on it through Python.
How would I go about this?
I tried:
import subprocess
subprocess.call("explorer path-to-file")
# got error that the path doesn't exist
subprocess.call("calc path-to-file")
# Calc is not executable type error
You could use the xdg-open tool (if you have it) to use the default tool for the file type, or if you really want to always use Libreoffice, the executable for it is libreoffice.
os.system("xdg-open path-to-file")
os.system("libreoffice path-to-file")
(and as always when using os.system(), make sure path-to-file comes from a trusted source.)
My terminal is running python 2 so when I run my python file the program fails.
So basically I am new to programming and trying to create a small python script to help me auto create folders. I also want to give the script to colleges so that they can use it on their systems too.
I know that I can run my file in terminal by using "python3 myfile.py" and it will work, but that's too much off a mission to do for my colleges and as my colleges are not familiar with code or terminal for that matter, I wanted to create an executable file so that they just click to open type a few answers to the promoted question and boom folders created.
This is where I run into a problem, I have "#!/usr/bin/env python3" at the top of my file but when I run the script IDLE opens up and it just shows the code I have written but doesn't seem to run the actual script I wrote. Am I doing something wrong?
I also then though perhaps I could just use the terminal to run the file as it is now executable, so I go into terminal and enter "myfile.py" and the program runs but in python 2 so my script fails as it is in python3. So another question would be is there a way to code into my python file, when running this file make sure you use python3? as I would want this to work on all colleges system without them having to write out anything in terminal?
Sorry for the long explanation but any advice would be greatly appreciated.
Thank you in advance
When you are on windows you can just create a .bat file where you write: python3 myfile.py in it.
Both files have to be in the same directory.
If you want to have a .exe you can also use py2exe.
You could also try just #!/usr/bin/python3 without env.
More here.
I currently have a Python scrip that runs through all Excel files in the current directory and generates a PDF report.
It works fine now but I don't want the users to be anywhere near frozen Python scripts. I created an MSI with cxFreeze which puts the EXE and scripts in the Program Files directory.
What I would like to be able to do is create a shortcut to this executable and pass the directory the shortcut was run from to the Python program so that can be set as the working directory. This would allow the user to move the shortcut to any folder of Excel files and generate a report there.
Does Windows send the location of a opened shortcut to the executable and is there a way to access it from Python?
When you launch a shortcut, Windows changes the working directory to the directory specified in the shortcut, in the Start in field. At this point, Windows has no memory of where the shortcut was stored.
You could change the Start in field to point to the directory that the shortcut is in. But you'd have to do that for every single shortcut, and never make a mistake.
The better approach is to use a script, rather than a shortcut. Place your actual Python script (which we'll call doit.py for sake of example) somewhere in your PYTHONPATH. Then create a single-line Python script that imports it:
import doit
Save it (but don't name it doit.py) and copy it to each directory from which you want to be able to invoke the main script. In doit.py you can use os.getcwd() to find out what directory you're being invoked from.
You could also do it with a batch file. This is a little more flexible in that you can specify the exact name of the script and which Python interpreter should be used, and don't need to store the script in a directory in PYTHONPATH. Also, you don't need to worry about the file's name clashing with the name of a Python module. Simply put this line in a file:
C:\path\to\your\python.exe C:\path\to\your\script.py
Save it as (e.g.) doit.bat and copy it into the directories from which you want to invoke it. As before, your Python script can call os.getcwd() to get the directory. Or you can write it so your Python script accepts it as the first argument, and write your batch file like:
C:\path\to\your\python.exe C:\path\to\your\script.py %cd%
Another thing you can do with the batch file approach is add a pause command to the end so that the user is asked to press a key after the script runs, giving them the opportunity to read any output generated by the script. You could even make this conditional so that it only happens if an error occurs (which requires returning a proper exit code from the script). I'll leave that as an exercise. :-)
Is there a problem with modifying the script to take the directory to process as a command line argument?
You could then configure the different shortcuts to pass in the appropriate directory.
Type the following into a batch file (i.e. script.bat):
python \absolute\path\to\your\script.py %~dp0
pause
Then add these imports at the top of your python file script.py (if not already included):
import os
import sys
And add this to the bottom of the python file (or combine it with a similar statement):
if __name__ == "__main__":
# set current working directory:
if len(sys.argv) > 1:
os.chdir(sys.argv[1])
main()
replace main() with whatever function you want to call or code you want to run.
The following is how I came to my answer:
I tried using kindall's answer and had the following issues:
The first suggestion of storing the script somewhere in PYTHONPATH could not be applied to my situation because my script will be used on a server and needs to be independent of the client computer's python environment (besides having the required pip installations).
I tried calling my python script from a Windows Batch File which could be moved to a different location. Instead of the batch file's location being used as the current working directory, it was C:\Windows.
I tried passing %cd% as an argument to my python script, then setting that to be my CWD. This still resulted in a CWD of C:\Windows.
After reviewing the comments, I tried Eryk Sun's suggestion of instead passing %~dp0 as an argument to the python script. This resulted in the CWD being correctly set to the batch file's location.
I hope this helps others facing similar difficulties.
To preface, I've read other threads on the topic, but all their solutions don't work for me.
I have a small .sh file that just runs python3 foo.py. I used a script to turn this file into a .app file, but when I try to open it, I can see the app begin to appear in the dock and then disappear. However, when I open the file inside of the Unix executable itself in Terminal, all is well.
I have tried:
Changing the shebang in the .sh file to both #!/usr/bin and #!/urs/bin/env
Creating an empty .plist file
Making sure every file has execute permissions
Oddly enough, running open appname.app gives the following:
LSOpenURLsWithRole() failed with error -10810 for the file /Users/blah/blah/Lofi.app.
Thanks in advance
When you say you've read other threads on the topic, and the solutions don't work, does that mean you tried every thing in the comments under the sh-to-app script? There are a couple of specific fixes enumerated there.
What OS are using? Looks like higher OS needs you to register the app with the OS.
Check permissions of your personal shell script
Does it open in Terminal with open APP.app? I think you're saying it does. You can look at the last comment under the the sh-to-app script for possibly helpful instructions.
Not so much an answer, more of a workaround. I ended up taking the APP.app/Contents/MacOS/APP executable and creating an application in Automator that simply runs the file, and saving that process as an application in the Applications folder.