Not able to parse output of bash from Python correctly

Question

I have the following Python code in which I'm calling a Makefile by embedding the bash command from within the python script. I want to check if a warning is reported on the stdout. I do see the warning on the stdout but my Python code does not seem to be detecting it.

Python Code:

maker = subprocess.Popen(["bash", "-c", "make"], stdout=subprocess.PIPE)
    for line in maker.stdout:
        if "warning:" in line:
            print "Warning(s) detected in make"

Output on stdout which clearly reports a warning:

main.c: In function ‘main’:
main.c:46:14: warning: unused variable ‘options’ [-Wunused-variable]

Compiler error messages get printed to standard error, not standard output. — Barmar
– Barmar, Commented Jan 13, 2014 at 8:51

Michael Helwig · Accepted Answer · 2014-01-13 08:51:46Z

4

Try catching stderr as well:

subprocess.Popen(["bash", "-c", "make"], stdout=subprocess.PIPE, stderr=subprocess.PIPE)

(As user "Barmar" already noted, compiler error messages are sent to stderr.)

answered Jan 13, 2014 at 8:51

Michael Helwig

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