捕获字符串，无条件结束

Question

I'm trying to capture the middle part of a URL, that contains a conditional ending:

A URL can be of two sorts:

/a/b/(part/needed)
/a/b/(part/needed)/page/#

here's the regexp I use:

preg_match('@/a/b/(.*)(/page/\d)?@i', '/a/b/some/text/page/1', $matches);

returns

0=>"/a/b/some/text/page/1",
1=>"some/text/page/1"

It's ok but it includes the conditional ending which I don't want!

Can someone tell me how to not include the conditional string ending in it but still match when the last segment is present or absent?

Answer 1

By anchoring the expression with ^$ and making the first group non-greedy (.*?), you can get the segment you need. The .* alone is a greedy match, and will eat up everything that follows the .*.

preg_match('@^/a/b/(.*?)(/page/\d)?$@i', '/a/b/some/text/page/1', $matches);
//-----------^-------^^^-----------^
print_r($matches);
Array
(
    [0] => /a/b/some/text/page/1
    [1] => some/text
    [2] => /page/1
)

If you don't need the /page/1, make that a non-capturing group (?:...).

preg_match('@^/a/b/(.*?)(?:/page/\d)?$@i', '/a/b/some/text/more/page/1', $matches);
//----------------------^^^
print_r($matches);
Array
(
    [0] => /a/b/some/text/more/page/4
    [1] => some/text/more
)

regular-expressions.info has good information on character repetition with + and *, and the pitfalls of greediness.