正则表达式匹配一个数字后跟一个特定的字符串

Question

I need to find a number followed by a specific string, within another string.

The initial string could be: some text 0.25mcg some more text some text 25mcg some more text

so the number could be a decimal. I need to be able to return the number (so 0.25 or 25) where ever the number is followed by 'mcg'

Can anyone help me out. This doesn't work:

if(preg_match('(\d+mcg)', $item, $match))

Answer 1

Another option is to capture a digit with an optional decimal part \\d+(?:\\.\\d+)? and use a word boundary \\b to prevent the match being part of a larger word.

\b(\d+(?:\.\d+)?)mcg\b

Regex demo | Php demo

Code example

$re = '/\b(\d+(?:\.\d+)?)mcg\b/';
$str = 'some text 0.25mcg some more text some text 25mcg some more text';

preg_match_all($re, $str, $matches);
print_r($matches[1]);

Output

Array
(
    [0] => 0.25
    [1] => 25
)

If you want a match only instead of a capturing group you might also opt for a positive lookahead (?= instead.

\b\d+(?:\.\d+)?(?=mcg\b)

Regex demo | Php demo

Answer 2

It's a job for preg_match_all

preg_match_all('/([\d.]+)mcg/', $item, $matches);

[\\d.]+ matches 1 or more digits or dot.

Answer 3

here is simple version:

<?php
$item1 = 'some text 0.25mcg some more text';
$item2 = 'some text 25mcg some more text';

if (preg_match('/([0-9\\.]+)\\s*mcg/', $item1, $match)) echo $match[1] . '<br>';
if (preg_match('/([0-9\\.]+)\\s*mcg/', $item2, $match)) echo $match[1] . '<br>';