well$string = 'something ending in 3'
$n = 3;
if( preg_match( '/'.$n.'$/', $string ){
//do something
}
ought to do it[edit: but im sure there's a more efficient way... i just can't be arsed to think, sorry :)]
($n = 3;
if( preg_match( '/'.$n.'$/', $string ){
//do something
}
ought to do it[edit: but im sure there's a more efficient way... i just can't be arsed to think, sorry :)]
slight yoinks!,
Wed 22 Jan 2003, 22:18,
archived)
actually...dur...
$n=4;
$string = "blahblah4";
if( $string{strlen($string)+1} = $n ){
//do summat
}
that'll be a bit more efficient
($n=4;
$string = "blahblah4";
if( $string{strlen($string)+1} = $n ){
//do summat
}
that'll be a bit more efficient
slight yoinks!,
Wed 22 Jan 2003, 22:29,
archived)
if you'vegot an integer, isn't it easier to just use mod 10 instead of doing complex string thingies with it?
(
freshlegs,
Wed 22 Jan 2003, 22:32,
archived)
.it's liable to go *POOF* when some little turd types in "uraguay"
yes yes i know, validation... but i suspect this is a quick and dirty
(yes yes i know, validation... but i suspect this is a quick and dirty
slight yoinks!,
Wed 22 Jan 2003, 22:45,
archived)