Windows 任务计划程序无法运行 Python 脚本，错误值：2147944320

Question

I have created a new task, with one action: Start a program. I followed the instructions from https://www.jcchouinard.com/python-automation-using-task-scheduler/

my python script looks like this:

if __name__ == '__main__':
    num = 1

Under general properties, I have 'run whether user is logged on or not' checked, 'run with highest privileges' checked, and myself as the user running the task. I am an administrator.

Action Parameters...

Program/script : C:\Users\myuser\AppData\Local\Microsoft\WindowsApps\python.exe (pasted from command where python )
Add arguments : py_test.py
Start in : C:\Users\myuser\Desktop

I tested from the command line that I can run this command successfully:
C:\Users\myuser\AppData\Local\Microsoft\WindowsApps\python.exe C:\Users\myuser\Desktop\py_test.py

When I click 'run' from the task scheduler library, I get the error 'The file cannot be accessed by the system. (0x80070780)
When I go into the history for the task, I see this error:

Task Scheduler failed to launch action "C:\Users\myuser\AppData\Local\Microsoft\WindowsApps\python.exe" in instance "{6204cea7-bedc-40f9-bc10-ac95b9e02460}" of task "\TestPythonJob". Additional Data: Error Value: 2147944320.

I confirmed under the executable file's properties that I and SYSTEM have access to it. I tried researching this error value but could not find anything. What could be the issue?

Answer 1

Maybe try adding python to your path environment variables. Then make a.bat to run it. Here's how to use a.bat with task scheduler https://www.python.org/ftp/python/3.9.5/python-3.9.5-amd64.exe .