bash script that dynamically calls argument as another bash script

Question

I want to implement a bash script that acts the following way

./script_a.sh script_b.sh $arg1 $arg2 ...

./script_a.sh would invoke script_b.sh on $arg1 $arg2... $argn I don't know how many args there are beforehand. how can this be one?

is script_b.sh already implemented, or is this a fresh design? If it is, why would you want to do it this way? — ghostdog74
– ghostdog74, Commented Oct 30, 2014 at 7:08

Cyrus · Accepted Answer · 2014-10-30 07:50:09Z

4

Try this as a wrapper:

#!/bin/bash
"$@"

edited Oct 30, 2014 at 7:50

answered Oct 30, 2014 at 7:00

Cyrus

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nu11p01n73R · Accepted Answer · 2014-10-30 07:05:33Z

0

The $@ contains the list of all arguments passed to the script.

For example consider:

-sh-3.2$ cat scripta
echo $@
"./$@"

-sh-3.2$ cat scriptb
echo hello world $@

-sh-3.2$ bash scripta scriptb 1 2 3 4
scriptb 1 2 3 4
hello world 1 2 3 4

"./$@" calls the script in the argument, scriptb with the remaining arguments 1 2 3 4

hello world 1 2 3 4 is obtained on running the second sript, scriptb

answered Oct 30, 2014 at 7:05

nu11p01n73R

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