正则表达式在第一个空格后排除

Question

<?php

$url = '/#123_abc" data-text="something-something"';
preg_match_all ('(/#.*)', $url, $result);


var_dump($result);

How can I get only /#123_abc ? I want to exclude data-text="something-something"

Answer 1

You can use a negated character class as

/(^\/#[^"]+)/

• ^ Anchors the regex at the start

• [^"]+ Negated character class, matches anything other that a ".

  + quantifies the pattern, matches one or more occurence.

Regex Demo

---

Test

$url = '/#123_abc" data-text="something-something"';
preg_match_all ('/(^\/#[^"]+)/', $url, $result);
var_dump($result[0]);
// => Outputs
// /#123_abc

---

P.S : You can prevent the escaping of \ by using different pair of delimiters as

$url = '/#123_abc" data-text="something-something"';
preg_match_all ('~(^/#[^"]+)~', $url, $result);
var_dump($result[0]);
// => Outputs
// /#123_abc