CODEWITHSUNDEEP
pythonbash
Roy van der Valk
Roy van der Valk

Reputation: 23

Problem passing bash output to a python script

I'm fairly new to programming and I searched the internet for a way to pass bash output to a Python script.

I came up with this in bash.

XAS_SVN=`svn info` 
ssh hudson@test "python/runtests.py $XAS_SVN"

And this in python.

import sys
print sys.argv[1]

When I echo $SVN_INFO I get the result.

Path: . URL: //svn/rnd-projects/testing/python Repository Root: //svn/rnd-projects Repository UUID: d07d5450-0a36-4e07-90d2-9411ff75afe9 Revision: 140 Node Kind: directory Schedule: normal Last Changed Author: Roy van der Valk Last Changed Rev: 140 Last Changed Date: 2009-06-09 14:13:29 +0200 (Tue, 09 Jun 2009)

However the python script just prints the following.

Path:

Why is this and how can I solve this? I the $SVN_INFO variable not of string type?

I know about the subprocess module and the Popen function, but I don't think this is a solution since the python script runs on another server.

Upvotes: 2

Views: 1069

Answers (2)

af.
af.

Reputation: 1668

Yes, you need to put in quotes your input to python/runtest.py because otherwise argv[1] gets it only up to the first space. Like this:

ssh hudson@test "python/runtest.py \"$XAS_SVN\""

Upvotes: 0

tonfa
tonfa

Reputation: 24491

Since you have spaces in the variable, you need to escape them or read all the arguments in your script:

print ' '.join(sys.argv[1:])

But it might be better to use stdin/stdout to communicate, especially if there can be some characters susceptible to be interpreted by the shell in the output (like "`$'). In the python script do:

for l in sys.stdin:
     sys.stdout.write(l)

and in shell:

svn info | ssh hudson@test python/runtests.py

Upvotes: 1

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