Possible Duplicate:
Bash. How compare two strings in “version” format
All,
I need a good algorithm, script to compare "2.0.9" to "2.0.10" 2.0.9 is less than 2.0.10
"2.0.1" is less than "2.0.9" "2.0.9" is less than "2.0.92"
See the picture? this is on the Mac OS 10.6
Take a look at sort -V and ls -v source code.
Also this is a program I wrote before those other programs learned about version sorting.
#!/usr/bin/perl
@S = <>;
print sort byglob @S;
######################################################################
#
# Sorting function which sorts numerically for numerical parts,
# alphabetically otherwise
#
sub byglob
{
my($A) = $a;
my($B) = $b;
my($c,$d,$e);
while ($A && $B)
{
$A =~ s/^([0-9]+|[^0-9]+)//;
$c = $1;
$B =~ s/^([0-9]+|[^0-9]+)//;
$d = $1;
if ($c =~ /\d/ && $d =~ /\d/)
{
$e = $c <=> $d;
}
else
{
$e = $c cmp $d;
}
return $e if ($e);
}
return $a cmp $b;
}
In bash compare them like this using arithmetic operator < inside arithmetic expression brackets [[ and ]] :
x=2.0.9
y=2.0.92
[[ $x < $y ]] && echo "less"
less
[[ 2.0.10 < 2.0.9 ]] && echo "less" wrote "less" on my system.
2.0.10 is indeed less than 2.0.9. However if you compare 2.0.10 with 2.0.09 then it will NOT echo "less";