[SOLVED] Bash string comparison w/ leading zeros

Issue

Just wondered if there was a simple way in bash scripting to use the 'test' construct to compare two strings matching a given pattern. In this particular case, I want a pair of numeric strings to match if they have leading zeros in front of either of them. Thanks.

#!/bin/bash

STR1=123
STR2=00123

if [[ "0*${STR1}" == "0*${STR2}" ]]; then
    echo "Strings are equal"
else
    echo "Strings are NOT equal"
fi

exit 0

Solution

If you are absolutely sure your strings are numeric, then you should use -eq instead of ==:

if [ $string1 -eq $string2 ]
then
    echo "These are equal"
fi

The -eq doesn't care about leading zeros

The problem is that if neither string is numeric (or one string is equal to zero and the other isn't numeric), this will still work:

string1=foo
string2=bar
if [ $string1 -eq $string2 ]
then
    echo "These are equal"  # This will print, and it shouldn't!
fi

The only way I see getting around this issue is to do something like this:

if expr $string1 + 0 > /dev/null 2&1 && expr $string2 + 0 > /dev/null 2>&1
then  # Both strings are numeric!
    if [ $string1 -eq $string2 ]
    then
        echo "Both strings are numeric and equal."
    else
        echo "Both strings are numeric, but not equal."
elif [ $sring1 = $sring2 ]
then
    echo "Strings aren't numeric, but are the same
else
    echo "Strings aren't numeric or equal to each other"
fi

The expr will return a non-zero exit code if the string isn't numeric. I can use this in my if to test to see if my strings are in fact numeric or not.

If they are both numeric, I can use my second if with the -eq to test for integer equivalency. Leading zeros are no problem.

The elif is used in case my strings are not numeric. In that case, I test with = which tests string equivalency.

Answered By - David W.

Answer Checked By - Mildred Charles (WPSolving Admin)