bash script

indikabandara19 · 05-21-2007, 07:07 AM

hope this is the correct place to post...otherwise please redirect me

how can i pass all the arguments of a shell script inside? i.e instead of using $1,$2 etc

eg. say i want to wrap grep(as i have nothing better to do)

----a.sh---------
#!/bin/bash

[you fill this]
grep [you fill this]

----a.sh---------

here is how i want to use it
./a.sh -Rn pattern file

a.sh is expected to pass all arguments to grep
how can u do it?

druuna · 05-21-2007, 07:45 AM

Hi,

Take a look at bash' getopt.

Here are 2 sites dealing with just that:

More Power with Bash Getopts
Getopt and getopts

Also take a look at Advanced Bash Scripting

Hope this helps.

marozsas · 05-21-2007, 08:43 AM

druuna give you the right directions to learn more about how to handle arguments in shell.

However, a short answer is to use "$*" which expand for all arguments in cmd line.

Code:

grep $*

theYinYeti · 05-21-2007, 09:26 AM

Not quite. You'll use "$@" (WITH the quotes). See bash man page for details.

Yves.

indikabandara19 · 05-21-2007, 10:40 PM

thanks.. u've pointed me in correct direction