> So, what does (.*?) mean? Well, simply said "any character, occuring 0 or more times" occuring 0 or 1 times.
I don't think so. ((.*)?) would mean that, but in (.*?), the '?' means "make the preceding pattern non-greedy; that is, make it match the minimum number of times. And as the minimum number of matches of (.*) is zero, it ends up meaning 'match no character at all. So it will always be true, wherever it occurs in a match string.
For instance,
$ php -a
Interactive shell
php > $test = "aabbcc";
php > $re = '/.+?(bb?).*/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => bb
)
Note here that the initial pattern piece '.+?' is limited to the minimum match.
The minimum match is a single character, but that is overruled by the attempt to match
capturing sub-expression '(bb?)' so it in fact matches 'aa'. Note that in
this regexp, (bb?) means "...a 'b' char followed by zero or one 'b' chars."
Now change that initial sub-expression.
php > $re = '/.*?(bb?).*/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => bb
)
The minimum match here is no characters, constrained to the minimum. But again,
the minimum match must be extended to accommodate '(bb?)'.
Now remove the minimising constraint.
php > $re = '/.*(bb?).*/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => b
)
Only one 'b'! Which 'b' is matched? It's the second 'b'. The minimum match
for (bb?) is a single 'b' followed by zero 'b's; so the second 'b' satisfies
the capturing expression, and the now-greedy initial subexpression can
gobble up all of the character to that second 'b'.
Don't believe me?
php > $re = '/(.*)(bb?).*/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => aab
[2] => b
)
Let's back up.
php > $re = '/(.*?)(bb?).*/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => aa
[2] => bb
)
As before, with a non-greedy initial sub-expression,
except that we now capture that initial sub-expression.
(bb?) means "...a 'b' followed by zero or one 'b's, greedily.
Can we force that to be non-greedy?
php > $re = '/(.*?)(bb??)(.*)/';
php > preg_match($re, $test, $match);
php > print_r($match);
Array
(
[0] => aabbcc
[1] => aa
[2] => b
[3] => bcc
)
Yes we can, by appending a moderating '?' which curbs the
appetite of the capturing sub-expression: (bb??)
Peter West
"...and behold, something greater than Jonah is here."
> On 23 Feb 2015, at 10:48 pm, Maciek Sokolewicz <maciek.sokolewicz@xxxxxxxxx> wrote:
>
> Secondly, the above two regexp rules are slightly bloated. What they actually mean is:
> ( = start new catchable pattern
> . = any character
> * = 0 or more of the previous pattern
> ? = 0 or 1 of the previous pattern
> ) = end catchable pattern
> \s = any whitespace character
>
> So, what does (.*?) mean? Well, simply said "any character, occuring 0 or more times" occuring 0 or 1 times. But since the any character pattern already occurs 0 or more times, the pattern as a whole will either be matched (1 time) or not (0 times). Making the ? metacharacter useless. Now if it were (.+?) then it would state "one or more of any character, with the entire pattern optional.
> So in practice, the following patterns are equal in what they represent: (.*?), (.*), (.+?)
>
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