How to get first letter of every word using regex in PHP

Question

I have a string variable and I want to get the first letter of every word of it. I want the end result to be an array of first letters. ``` $language = 'Sample Language'; preg_match('/(\b[a-z])+/i', $language, $match); print_r($match); ``` What I am getting is only the first letter of the first word. The above prints Array([0]=>S [1]=>S) If I change it to ``` preg_match('/(\bL)+/i', $language, $match); ``` I can get a match for the L of the second word. So obviously after the first match it stops looking for the rest. I am not very good with regular expressions. Can anyone point out what I have done wrong? words in input string are not always separated by space. It is quite unpredictable. Some formats I have come across: "MainLanguage: Language" "MainLanguage, Language" "MainLanguage: Language-SubLanguage" here I want to get M, L and S

Accepted Answer

First of all use `preg_match_all` for this, and secondly you don't need the `+` quantifier:

```
$language = 'Sample Language';
preg_match_all('/\b\w/', $language, $match);
print_r($match);
```

- `\b`: Matches a word boundary, a word boundary is a position that separates a word character from a non-word character. Word characters are usually `[a-zA-Z0-9_]`.

- `\w`: Matches a word character.

- `\b\w` Matches a word character that is at a word boundary position, in this case right after the space or the start-of-string that separates the words.

In case you want camel case situations then you can combine the previous expression with another one like this:

```
\b\w|(?<=\p{Ll})\p{Lu}
```

The second part of the expression, namely `(?<=\w)\p{Lu}` should match any word character if it is an uppercase character `\p{Lu}` following a lowercase one `\p{Ll}` which should cover the camel case situation, the original expressions covers the situation when a hyphen `-` is used to separate the two words.

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How to get first letter of every word using regex in PHP

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