How to pass bash script arguments to a subshell

Posted by Ralf Holly on Super User See other posts from Super User or by Ralf Holly
Published on 2012-03-21T14:56:13Z Indexed on 2012/11/18 5:03 UTC
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I have a wrapper script that does some work and then passes the original parameters on to another tool:

#!/bin/bash
# ...
other_tool -a -b "$@"

This works fine, unless the "other tool" is run in a subshell:

#!/bin/bash
# ...
bash -c "other_tool -a -b $@"

If I call my wrapper script like this:

wrapper.sh -x "blah blup"

then, only the first orginal argument (-x) is handed to "other_tool". In reality, I do not create a subshell, but pass the original arguments to a shell on an Android phone, which shouldn't make any difference:

#!/bin/bash
# ...
adb sh -c "other_tool -a -b $@"

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