Split text into an array of URL and non-URL strings

Question

I need to split a PHP array into an array containing text and URLs. For instance, assuming

$string = "Hello, my name is http://www.audio.com/1234.mp3/. Today is https://radio.org/weather.wav";

The expected output should be something like:

$a[0] = "Hello, my name is";
$a[1] = "http://www.audio.com/1234.mp3/";
$a[2] = ". Today is";
$a[3] = "https://radio.org/weather.wav";

Why do you need it like this? If you want to convert the urls to links, it's easier to use preg_replace — Czechnology
– Czechnology, Commented Feb 26, 2011 at 18:31
My goal is to produce a function that reads text out loud, replacing text by already existing audio files whenever appropriate. For instance, Say("hello $name!"); might produce different results whether $name is text or the url of a sound file... — Leo Burd
– Leo Burd, Commented Feb 27, 2011 at 14:28

mario · Accepted Answer · 2011-02-26 19:18:08Z

1

You cannot split it easily. But a workaround would be to match it in pairs using something like:

preg_match_all('#(.*?)(https?://\S+(?<![,.]))\K#s', $str, $m,
               PREG_SET_ORDER);
$list = call_user_func_array("array_merge", $m);

The call_user_func_array is another workaround to avoid flatteing the array manually. This method will lead to empty entries in between however:

Array
(
    [0] => 
    [1] => Hello, my name is 
    [2] => http://www.audio.com/1234.mp3/
    [3] => 
    [4] =>  Today is 
    [5] => https://radio.org/weather.wav
)

Also note that the simplistic URL regex ate up the period. (Use exact character groups instead of lookbehind.)

answered Feb 26, 2011 at 19:18

mario

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5 Comments

Marc B Over a year ago

There is preg_split(), which has a few options to capture delimiters and whatnot as well.

Leo Burd Over a year ago

Thanks so much for your answer and comments, mario and Mark B! How difficult would it be to get the punctuation (!.,?;) after the url? Is that possible at all? As mentioned above, my goal is to produce a function that reads text out loud, replacing text by already existing audio files whenever appropriate. For instance, Say("hello $name!"); might produce different results whether $name is text or the url of a sound file. If that's not possible, one alternative would be to require urls to always have a space after them... Would that make sense?

mario Over a year ago

@user362794: You could indeed use preg_split as Marc said. It might return the desired result without workarounds: $list = preg_split('#(https?://\S+(?<![,.]))#', $str, -1, PREG_SPLIT_DELIM_CAPTURE);

Leo Burd Over a year ago

@mario and @MarcB, Sorry for the additional questions, but what do I need to do in order to prevent punctuation such as ! ? ; : from being included in the parsed URL? Any ideas? That's the sort of thing that might happen in situations such as Say("Hi $audio_name_url!"); or Say("Weather for $location: $weather."); ... Thanks once again!

mario Over a year ago

@elbee: Extend the negative lookbehind assertion (?<![,?...]). See regular-expressions.info/lookaround.html -- Btw, only one @notification per comment works.

mickmackusa · Accepted Answer · 2024-08-18 22:18:43Z

0

Directly populate the desired result by calling preg_split() and including the delimiting URL substrings in the flat output array. Write \s* to optionally match zero or more whitespaces before or after the URLs -- this will exclude unwanted/trimmable whitespaces from the array values.

Code: (Demo)

$string = "Hello, my name is http://www.audio.com/1234.mp3/. Today is https://radio.org/weather.wav";
var_export(
    preg_split(
        '#\s*(https?://\S+(?<![,.]))\s*#',
        $string,
        flags: PREG_SPLIT_NO_EMPTY | PREG_SPLIT_DELIM_CAPTURE
    )
);

Output:

array (
  0 => 'Hello, my name is',
  1 => 'http://www.audio.com/1234.mp3/',
  2 => '. Today is',
  3 => 'https://radio.org/weather.wav',
)

answered Aug 18, 2024 at 22:18

mickmackusa♦

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