php regex find part of string

Question

I'm trying to find a string inside curly braces that can appear in many variations.

i.e. In the following string i'm searching for the word Link

asdasd{Link1111:TAG}\r\n {TAG:Link2sds}

I'm using the following command: $pattern = "/{(Link.*?)}|{(.*Link.*?)}/";

and the result is:

{array} [3]
 0 => {array} [2]
     0 = "{Link1:TAG}"
     1 = "{TAG:Link2}"
 1 = {array} [2]
     0 = "Link1:TAG"
     1 = ""
 2 = {array} [2]
     0 = ""
     1 = "TAG:Link2"

I'm expecting only the first array without the curly braces...what am i missing with the regex? thx.

Just use $pattern = "/{(.*?Link.*?)}/"; and print_r($matches[1]); — Thamilhan
– Thamilhan, Commented May 17, 2016 at 17:52
yeap! spot on...if you want to submit it as an answer so i can accept it :) — Danny Valariola
– Danny Valariola, Commented May 17, 2016 at 17:56

chris85 · Accepted Answer · 2016-05-17 17:56:20Z

1

preg_match_all is global and finds all matches. If you want to find it just once use preg_match.

Demo: https://eval.in/572825

The 0 index in your current example is all matches. The 1 is the first capture group Link.*?, and the 2 is your second capture group .*Link.*?.

answered May 17, 2016 at 17:56

chris85

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Comments

Thamilhan · Accepted Answer · 2016-05-17 17:59:30Z

0

Just use /{(.*?Link.*?)}/ and your matches will be in index 1

<?php

$str = "asdasd{Link1111:TAG}\r\n {TAG:Link2sds}";

$pattern = "/{(.*?Link.*?)}/";
preg_match_all($pattern, $str, $matches);

print_r($matches[1]);

Your eval

answered May 17, 2016 at 17:59

Thamilhan

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1 Comment

Casimir et Hippolyte Over a year ago

Wrong! You can't use .*? before Link. Consider the string {toto123:tag} {Link111:tag}. Remember that the regex engine walks from left to right and try to succeed at each position.

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