Linux – Skip parameter in Linux bash

bashlinuxshell

I need to write a shell-script which handles 3 parameters ($1 $2 and $3). Within the bash I have to skip parameter 2 (I'm actually writing something like ./script.sh para1 para3).

The script is OK. But I can't skip the second parameter. How do I skip this one?

Best Answer

At last I've found it:

./script.sh "para1" "" "para3"

My teacher should be happy with this. He won't get anything else from me.

Related Solutions

Linux – Does bash have a hook to determine when child shell exits

You were on the right track with trap. What you want is the following:

trap "stty erase ^?" SIGCHLD

You can add that to .bashrc. Note that this will run when any subprocess ends.

This will only work in interactive sessions. For non-interactive sessions, bash will not enable job control by default, so you will need to run set -o monitor first. But I doubt you'd need backspaces in non-interactive scripts.

SIGCHLD is sent to the parent process whenever a subprocess exits.

An alternative method is to wrap your other shell in a script, such as:

#!/bin/sh
tcsh
stty erase ^?

Then, if you launch your shell through the script, the script will run the stty erase command after the shell exits. This is less likely to have side effects than a global trap handler, but of course it'll only work if you launch through the script every time (or create an alias to do so, e.g. alias 'tcsh' '~/launch-tcsh.sh').

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Linux – Does bash have a hook to determine when child shell exits

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