For example I have the text
a1aabca2aa3adefa4a
I want to extract 2 and 3 with a regex between abc and def, so 1 and 4 should be not included in the result.
I tried this
if(preg_match_all('#abc(?:a(\d)a)+def#is', file_get_contents('test.txt'), $m, PREG_SET_ORDER))
print_r($m);
I get this
> Array
(
[0] => Array
(
[0] => abca1aa2adef
[1] => 3
)
)
But I want this
Array
(
[0] => Array
(
[0] => abca1aa2adef
[1] => 2
开发者_如何转开发 [2] => 3
)
)
Is this possible with one preg_match_all call? How can I do it?
Thanks
preg_match_all(
'/\d # match a digit
(?=.*def) # only if followed by <anything> + def
(?!.*abc) # and not followed by <anything> + abc
/x',
$subject, $result, PREG_PATTERN_ORDER);
$result = $result[0];
works on your example. It assumes that there is exactly one instance of abc
and def
per line in your string.
The reason why your attempt didn't work is that your capturing group (\d)
that matches the digit is within another, repeated group (?:a(\d)a)+
. With every repetition, the result of the capture is overwritten. This is how regular expressions work.
In other words - see what's happening during the match:
Current position Current part of regex Capturing group 1
--------------------------------------------------------------
a1a no match, advancing... undefined
abc abc undefined
a2a (?:a(\d)a) 2
a3a (?:a(\d)a) (repeated) 3 (overwrites 2)
def def 3
You ask if it is possible with a single preg_match_all.
Indeed it is. This code outputs exactly what you want.
<?php
$subject='a1aabca2aa3adefa4a';
$pattern='/abc(?:a(\d)a+(\d)a)def/m';
preg_match_all($pattern, $subject, $all_matches,PREG_OFFSET_CAPTURE | PREG_PATTERN_ORDER);
$res[0]=$all_matches[0][0][0];
$res[1]=$all_matches[1][0][0];
$res[2]=$all_matches[2][0][0];
var_dump($res);
?>
Here is the output:
array
0 => string 'abca2aa3adef' (length=12)
1 => string '2' (length=1)
2 => string '3' (length=1)
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