How do I include the split delimiter in results for preg_split()?

Question

I have this simple pattern that splits a text into periods:

$text = preg_split(\"/[\\.:!\\?]+/\", $text);

But I want to include . :

Answer 1

Here you go:

preg_split('/([^.:!?]+[.:!?]+)/', 'good:news.everyone!', -1, PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY);

How it works: The pattern actually turns everything into a delimiter. Then, to include these delimiters in the array, you can use the PREG_SPLIT_DELIM_CAPTURE constant. This will return an array like:

array (
    0 => '',
    1 => 'good:',
    2 => '',
    3 => 'news.',
    4 => '',
    5 => 'everyone!',
    6 => '',
);

To get rid of the empty values, use PREG_SPLIT_NO_EMPTY. To combine two or more of these constants, we use the bitwise | operator. The result:

array (
    0 => 'good:',
    1 => 'news.',
    2 => 'everyone!'
);

Answer 2

No use for PREG_SPLIT_DELIM_CAPTURE if you use a positive lookbehind in your pattern. The function will keep the delimiters.

$text = preg_split('/(?<=[.:!?])/', 'good:news.everyone!', 0, PREG_SPLIT_NO_EMPTY);

If you use lookbehind, it will just look for the character without matching it. So, in the case of preg_split(), the function will not discard the character.

The result without PREG_SPLIT_NO_EMPTY flag:

array (
    0 => 'good:',
    1 => 'news.',
    2 => 'everyone!',
    3 => ''
);

The result with PREG_SPLIT_NO_EMPTY flag:

array (
    0 => 'good:',
    1 => 'news.',
    2 => 'everyone!'
);

You can test it using this PHP Online Function Tester.