Python subprocess.call 超时重试

Question

I want to re-run subprocess.call if it's timeout somehow.

subprocess.call('some command', timeout=600)
if timeout:
    subprocess.call('some command')

How do i do something like this?

Answer 1

subprocess.call raises [Python 3.Docs]: exception subprocess.TimeoutExpired when timeout is (given and) reached (just like Popen.communicate ).

Here's a piece of code that keeps launching NotePad with a timeout of 3 seconds, until it runs 2 times, or user manually closes it:

 >>> max_runs = 2 >>> run = 0 >>> while run < max_runs: ... try: ... subprocess.call("notepad", timeout=3) ... except subprocess.TimeoutExpired: ... continue ... else: ... break ... finally: ... run += 1 ... 0

Although this technically answers the question, I don't think it's a good idea to re-launch a process that didn't end, since there's a great chance that the consecutive runs will have the same outcome (will timeout). In that case, you'd have to use Popen and communicate , and if the process times out, kill it via Popen.terminate .

Answer 2

You can use basic error handling to do catch the timeout exception:

try:
    subprocess.call('cmd', timeout=0)
except subprocess.TimeoutExpired:
    print('Expired!')
    # subprocess.call('cmd')

The except block is only run if the specified error is raised. See Python docs tutorial on exceptions , specifically error handling for more information.