Bash string comparison w/ leading zeros - regex

Question

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

Answer 1

Strip any leading zeros from the strings, then check if the results are equal. This solution requires extended pattern support from bash.
shopt -s extglob
if [[ "${STR1##*(0)}" = "${STR2##*(0)}" ]]; then
echo "Strings are equal"
fi
You can also use bash's built-in regular expression support, but it might require two comparisons if you don't know which string has more leading 0s. The test works when the string on the left has at least as many leading 0s as the string on the right.
if [[ $STR1 =~ 0*$STR2 || $STR2 =~ 0*$STR1 ]]; then
echo "Strings are equal"

Answer 2

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.

Answer 3

This is how I would make:
#!/bin/bash
STR1=123
STR2=00123
if [ `echo -n ${STR1} | sed 's/^0*//g'` == `echo -n ${STR2} | sed 's/^0*//g'` ]; then
echo "Strings are equal"
else
echo "Strings are NOT equal"
fi
exit 0

Answer 4

The obvious and trivial solution is to explicitly indicate the base of the numbers. Then any leading zeros are insignificant because the numbers will not be interpreted as octal.
if [[ 10#$STR1 -eq 10#$STR2 ]]; then
echo "Numbers are equal"
else
echo "Numbers are NOT equal"
fi
Notice also the switch to -eq for numeric comparison.

Answer 5

This solution uses expr to convert the strings into numeric values. Note an alternative bash method - double parentheses - does NOT work as values with leading zeroes are parsed as octal values. e.g.
STR1=000123
echo $(($STR1)) # outputs 83
SOLUTION
#!/bin/bash
STR1="00123"
STR2="0000123"
n1=$(expr $STR1 + 0)
n2=$(expr $STR2 + 0)
if [ $n1 -eq $n2 ];then
echo "Strings are equal"
else
echo "Strings are NOT equal"
fi
exit 0