Passing multiple arguments in a bash script

Question

The simple script below does not work when, rather than passing a single file name, I want to pass multiple files through expansion characters like *

#!/bin/bash
fgrep -c '$$$$' $1

If I give the command script.sh file.in the script works. If I give the command script.sh *.in it doesn't.

$1 is not set to *.in; it's set to the first item that *.in expands to. — chepner
– chepner, Commented Feb 18, 2014 at 13:49
@chepner which might be *.in, if no matching files exist. ;) — kojiro
– kojiro, Commented Feb 19, 2014 at 2:14

John Kugelman · Accepted Answer · 2014-02-18 13:01:46Z

5

Use "$@" to pass multiple file names to fgrep. $1 only passes the very first file name.

fgrep -c '$$$$' "$@"

answered Feb 18, 2014 at 13:01

John Kugelman

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1 Comment

jpaugh Over a year ago

@mimenico Note that $1 is a subset of $@, referring to exactly the first argument; also, quoting is (usually) better than not quoting:therefore, $@ is better than $*.

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