Issue
I read up on the functions provided by subprocess - call, check_call, check_output, and understand how each works and differs in functionality from one another. I am currently using check_output, so I can have access to the stdout, and used "try block" to catch the exception, as follows:
# "cmnd" is a string that contains the command along with it's arguments.
try:
cmnd_output = check_output(cmnd, stderr=STDOUT, shell=True, timeout=3, universal_newlines=True);
except CalledProcessError:
print("Status : FAIL")
print("Output: \n{}\n".format(cmnd_output))
The issue I am running into is when an exception is thrown, "cmnd_output" is not initialized and don't have access to stderr, and I get the following error message:
print("Output: \n{}\n".format(cmnd_output))
UnboundLocalError: local variable 'cmnd_output' referenced before assignment
I think thats because the exception causes the "check_output" to bail immediately without any further processing, aka assignment to "cmnd_output", in the try block. Please correct me if I am wrong.
Is there any way I can get access to stderr (it's ok if it's sent to stout) and have access to the exit code. I can manually check for pass/fail based on exit code with out the exception being throuwn.
Thank you, Ahmed.
Solution
Try this version:
import subprocess
try:
output = subprocess.check_output(
cmnd, stderr=subprocess.STDOUT, shell=True, timeout=3,
universal_newlines=True)
except subprocess.CalledProcessError as exc:
print("Status : FAIL", exc.returncode, exc.output)
else:
print("Output: \n{}\n".format(output))
This way you will print the output only if the call was successful.
In case of a CalledProcessError
you print the return code and the output.
Answered By - warvariuc
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