Bash Scripting Arrays

Question

I'm trying to read command line arguments through a bash scripts and am having difficulties getting desired results. I want to read in two parameters, one after -var1 and one after var2.

I am passing in these arguments

-var1 hello -var2 world

args=("$@")
x="0"
for (( i=0;i<$#; i++ ))
do
  x=$(($x + 1))
  echo " $i $x "
  if [[ ${args[${i}]} == "-var1" ]] ; then
    if [ $# > $x ] ; then
        var1="${args[${i+1}]}"
    fi
  fi
  echo $i
  if [[ ${args[${i}]} == "-var2" ]] ; then
    if [ $# > $x ] ; then
        var2="${args[${i+1}]}"
    fi
  fi
done

It sets both variables, var1 and var2, equal to hello, rather than var1="hello" and var2="world". Any thoughts? Thanks in advance

What are `-fef` and `-scripts` then? Also, `$x` doesn't seem necessary, especially since it is always just `$i+1`. — chepner, Aug 19 '14 at 18:45
I have previously looked at that code, and I understand that is an alternate way to do it. I am still unsure why the above code does not work, which is why I asked the question. — randydalton411, Aug 19 '14 at 18:48
`[ $var1 > $var2 ]` redirects the output of `[ $var1 ]` to the file named by `$var2`. If you wanted a numeric comparison, that would be `-gt`; if you wanted a string comparison, that would be `'>'` (with the quotes). — Charles Duffy, Aug 19 '14 at 19:26
I strongly recommend (1) using konsolebox's suggestion, and (2) running your original code through http://shellcheck.net/, which will tell you in detail where the original code is broken. — Charles Duffy, Aug 19 '14 at 19:27

konsolebox · Answer 1 · 2014-08-19T19:26:07.733

Try doing it this way instead:

#!/bin/bash

while [[ $# -gt 0 ]]; do
    case "$1" in
    -var1)
         var1=$2
         shift
         ;;
    -var2)
         var2=$2
         shift
         ;;
    *)
         echo "Invalid argument: $1"
         exit 1
    esac
    shift
done

if [[ -z $var1 ]]; then
    echo "Var1 was not specified."
    exit 1
fi

if [[ -z $var2 ]]; then
    echo "Var2 was not specified."
    exit 1
fi

... do something with var1 and var2

glenn jackman · Accepted Answer · 2014-08-19T21:18:06.057

I agree with konsolebox.

Your code is suffering from excessive variable bracing. ${i+1} will not actually perform the addition. The expression within [] in an array expansion is evaluated as an arithmetic expression. Try this:

args=("$@")
for (( i=0; i<$#; i++ ))
do
    [[ ${args[i]} == "-var1" ]] && var1=${args[++i]}
    [[ ${args[i]} == "-var2" ]] && var2=${args[++i]}
done
echo "var1='$var1'"
echo "var2='$var2'"

output

var1='hello'
var2='world'

We could get more dynamic about it:

args=("$@")
for (( i=0; i<$#; i++ )); do
    [[ ${args[i]} == -* ]] && declare ${args[i]#-}=${args[++i]}
done

or even

while (( $# > 0 )); do
    [[ $1 == -* ]] && { declare ${1#-}=$2; shift; }
    shift
done

dganesh2002 · Answer 3 · 2014-08-19T21:55:40.180

0

Use getopts ->

while getopts "a:b:" flag
do
   case "$flag" in
   a) echo "var1=$OPTARG"
   ;;
   b) echo "var2=$OPTARG"
   ;;
   *) echo "Incorrect Usage"
   ;;
   esac
done

Then run it as -a "hello" -b "world"

edited Aug 19 '14 at 21:55

answered Aug 19 '14 at 21:36

dganesh2002

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Bash Scripting Arrays

3 Answers3