I want to capture the text within the square brackets in the html string below. But the regex I have below doesn't get 'image' and imagealt' seperately but returns 'image]" alt="[imagealt' instead. If I take out the alt="[imagealt]" from the string it returns as I would expect/want.
$html = '<h2>[title]</h2>
<div class="content"><img src="[image]" alt="[imagealt]" /></div>
<div class="content">[text]</div>';
preg_match_all("^\[(.*)\]^",$html,$fields, PREG_PATTERN_ORDER);
echo "<pre>";
print_r($fields);
echo "</pre>";
Array
(
[0] => Array
(
[0] => [title]
[1] => [image]" alt="[imagealt]
[2] => [text]
)
[1] => Array
(
[0] => title
[1] => image]" alt="[imagealt
[2] => text
)
)
Square brackets ( “[ ]” ): Any expression within square brackets [ ] is a character set; if any one of the characters matches the search string, the regex will pass the test return true.
[] denotes a character class. () denotes a capturing group. [a-z0-9] -- One character that is in the range of a-z OR 0-9. (a-z0-9) -- Explicit capture of a-z0-9 .
[[\]] will match either bracket. In some regex dialects (e.g. grep) you can omit the backslash before the ] if you place it immediately after the [ (because an empty character class would never be useful): [][] .
Regular expressions can't count brackets.
your regex is being greedy. you need to stop it being greedy to do what you want. Find out a bit more about greediness here.
When a match is greedy it will ignore the first situation which satisfies the regex and will keep trying to match until it consumes as much of the input as it can.
Usually this involves adding a ?
but I'm not certain in php, but you could try:
preg_match_all("^\[(.*?)\]^",$html,$fields, PREG_PATTERN_ORDER);
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