'Run Python Script From Script Directory/Current Directory
[Introduction]
Hi! I'm working on a python script to automate installation of UWP-Apps, it's been a long time i'm not touching Python; until this day. The script uses Depedencies inside the script directory, i've looking up on my older scripts and found this specific code.
os.chdir(os.path.dirname(sys.argv[0]))
[Problematic]
However, using the above code doesn't work on my current script but it's working fine on older scripts. When using above, it shows:
OSError: [WinError 123] The filename, directory name, or volume label syntax is incorrect: ''
Already looking up on Internet about this topic; but most of them was talking about running the script from outer/different directory that leads me to dead end.
Any helps is appreciated :)
Solution 1:[1]
The easiest answer is probably to change your working directory, then call the .py file from where it is:
cd path/to/python/file && python ../.py
Of course you might find it even easier to write a script that does it all for you, like so:
Save this as runPython.sh in the directory where you're running the puthon script from is:
#!/bin/sh
cd path/to/python/file
python ../script.py
Make it executable:
chmod +x ./runPython.sh
Then you can simply enter your directory and run it:
./runPython.sh
If you want to only make changes to the python script:
mydir = os.getcwd() # would be the folder where you're running the file
mydir_tmp = "path/to/python/file" # get the path
mydir_new = os.chdir(mydir_tmp) # change the current working directory
mydir = os.getcwd()
The reason you got an error was because sys.argv[0] is the name of the file itself, instead you can pass the directory as a string and use sys.argv[1] instead.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | kinshukdua |
