CODEWITHSUNDEEP
pythonbashsimplehttpserver
fyr91
fyr91

Reputation: 1293

Python running bash script and generate OSError

I have a bash script which helps establish a local SimpleHTTPServer. 

python -m SimpleHTTPServer 8080

I have put this inside my project folder. While I am running the program by using:

subprocess.call('./setup.sh')

an error message comes out:

Traceback (most recent call last):
  File "test.py", line 2, in <module>
    subprocess.call('./setup.sh')
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 522, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 709, in __init__
    errread, errwrite)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 1326, in _execute_child
    raise child_exception
OSError: [Errno 13] Permission denied

I retried this in terminal 

localhost:Desktop XXXX$ sh setup.sh
Serving HTTP on 0.0.0.0 port 8080 ...

It is working fine. I remember there are a few times where the terminal has popped up a window ask me about the permission for python about something related to firewall and I allowed it. Can you help me?

Upvotes: 0

Views: 2059

Answers (2)

mitnk
mitnk

Reputation: 3307

Give subprocess.Popen() a try, with cwd param:

subprocess.Popen(['sh', './setup.sh'], cwd='/dir/contains/setup.sh/')

Upvotes: 1

creichen
creichen

Reputation: 1768

Run it exactly as you would on the shell, i.e., as sh ./setup.sh:

subprocess.call('sh ./setup.sh', shell=True)

That should do the trick. Most likely, your setup.sh is not set to executable or is missing the first #! line that marks its interpreter.

EDIT:

Make sure to set shell=True to execute it via the shell, if you pass it as a single string, or separate the parameters into a list, as you might with execve:

subprocess.call(['sh', './setup.sh'])

Upvotes: 2

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