Following is the shell script which iterates over command line arguments and prints the values
for var in "$@"
do
echo $var
done
Now if i want to iterate from the second command line argument (the first argument being used for some other purpose), what is the command to exclude the first argument alone in iteration ?
Use shift:
#!/usr/bin/env bash
shift
for var in "$@"; do
echo "$var"
done
The first argument is in $1, so doing
var1=$1
will store it into var1
You can then use shift to "delete" the first arg and still use your for loop:
~$cat s.sh
var1=$1
shift
echo var1=$var1
for var in "$@"
do
echo $var
done
~$ ./s.sh 1 2 3
var1=1
2
3
From man bash:
shift [n]
The positional parameters from n+1 ... are renamed to $1 .... Parameters represented by the numbers $# down to $#-n+1 are unset. n must be a non-negative number less than or equal to $#. If n is 0, no parameters are changed. If n is not given, it is assumed to be 1. If n is greater than $#, the positional parameters are not changed. The return status is greater than zero if n is greater than $# or less than zero; otherwise 0.
This example behaves more like a standard unix command line utility. Options are processed in any order and can have modifiers. It does not work for every situation (like modifiers with spaces in them), but I've found it to be very useful for scripts that I want to have some default behavior, but occasionally I want to modify one or more parameters.
#!/bin/bash
#defaults
VAR1="Yours"
VAR2="Mine"
VAR3="Ours"
SHARENICE="false"
while [ $# -gt 0 ]; do
case "$1" in
#single parameter
-s) SHARENICE="true"
shift
;;
#modified parameters
-1) VAR1="$2"
shift 2
;;
-2) VAR2="$2"
shift 2
;;
-3) VAR3="$2"
shift 2
;;
#you could put a 'usage' function here,
#or if your last parameter has no modifier,
#like a mandatory input, it will now be at
#$1
*)
break
;;
esac
done
echo -n "$VAR1, $VAR2 and $VAR3. "
if [ "$SHARENICE" != "true" ]; then
echo "Too bad we don't get along."
else
echo "Good thing we play nice!"
fi
