preg_match bug ??

echo10

i have string which can be either '2' or '2 @ 5'
and i match it with
preg_match ( '/[0-9]+|[0-9]+ @ [0-9]+', $s, $m);
BUT, afaik regular expression should match the longest string it works fine with eregi but not with preg_match it always matches just '2' even if string is '2 @ 5'
it works okay if i switch it to '/[0-9]+ @ [0-9]+|[0-9]+/' but that dont makes sense why, does anyone know perl RE so good to give reasonable explanation ?

would be very grateful !

thank you

gammaster

Try this.

   preg_match('/^[0-9]( @ [0-9])?$/', $s, $m);

Houdini

Because you are missing the ending delimiter which in your code is "/" so the regex should be

preg_match ( '/[0-9]+|[0-9]+ @ [0-9]+/', $s, $m);

a regex must have a starting and ending delimiter it could just as well be so long as you have a start and end with the same delimiter, one that is not likely to be used as part of the regex.

preg_match ( '#[0-9]+|[0-9]+ @ [0-9]+#', $s, $m);

BTW $m returns an indexed array you might want to rewrite the regex.

echo10

thank for replies but,

about '/' : no, i just missed it as i was writing this. it works like i described.

gammaster i dont want to use that as i capture the pattern by name by (?P<something>abc)

any other ideas why it behave like it behave ?

xblue

To my knowledge alternation is neither greedy nor ungreedy, the regex machine simply works it way through the branches of the alternation from left to right until it detects a match or the matching fails (Please someone correct me if this is wrong).

Thus,

/[0-9]+ @ [0-9]+|[0-9]+/

should behave more the way you want, but I'd generally prefer

/[0-9]+( @ [0-9]+)?/