如果包含，PHP将从字符串中删除整个单词

I have code that displays my tweets. In the tweet the pictures are showing up as urls as are the links. I would like to remove the entire 'WORD' if it is a pic or a link. PS I found threads on here that are close to what I am looking for but do not produce my desired effect.

If it contains a "http" or a ".pic" I then want to remove the entire 'word'.

Here is my code:

<?php


$wordlist = array('http','pic');
 $replaceWith  = "";

/* Sample data */
$words = 'This tweet has a pic.twitter.com/00GeQ3zLub and a url http://www.mywebsite.com';

foreach ($wordlist as $v)
  $words = clean($v, $words, $replaceWith);

function clean($word, $value, $replaceWith) {
    return preg_replace("/\w*$word\w*/i", "$replaceWith ",trim($value));
}

echo $words;
?>

ACTUAL OUTPUT: This tweet has a .twitter.com/00GeQ3zLub and a url ://www.mywebsite.com

DESIRED RESULT: This tweet has a and a url

UPDATE for clarification:
I want to remove any "string of characters with no spaces" that contain a ".pic" or a "http". I don't know how to explain it with the right terms... but if a .pic.twitter.com/ia8akd is in my tweet I want the whole thing gone. Same with anything that contains a "http". I want the WHOLE 'string' gone. for example my tweet is "This is my website: http://www.MyWebsite.com. Pretty cool?" I would like this to display as "This is my website: Pretty cool?"

The \w does not match a ., nor :. You should match all continuos non-whitespace characters around your words.

\S*(?:http|pic)\S*

This will remove anything starting with pic though, it is not specific to a URL.

Regex demo: https://regex101.com/r/qZ8tD3/1

PHP Demo: https://eval.in/611103

PHP Usage:

$wordlist = array('http','pic');
 $replaceWith  = "";

/* Sample data */
$words = 'This tweet has a pic.twitter.com/00GeQ3zLub and a url http://www.mywebsite.com';

foreach ($wordlist as $v)
  $words = clean($v, $words, $replaceWith);

function clean($word, $value, $replaceWith) {
    return preg_replace("/\S*$word\S*/i", "$replaceWith ",trim($value));
}

echo $words;

I'd suggest you trim $value first thing, then use a function like this:

function clean($word, $value, $replaceWith) {
    $scan = preg_quote($word);
    return preg_replace("#\\S{$scan}\\S#i", $replaceWith . ' ', $value);
}

This requires $value to contain a whitespace at beginning and end, so you can:

$value = " {$value} ";
foreach ($words as $word) {
    $value = clean($word, $value, $replaceWith);
}
$value = trim($value);

You could also preg_split $value around whitespaces and use array_filter on the resulting array, but this solution could be less performant.

As an optimization, if all words have the same replacement, then you can assemble a single regex out of the words array:

// So [ 'http', '.pic' ] becomes '#\\S(http|\\.pic)\\S#i'
$regex = '#\\S(' 
       . implode('|', array_map('preg_quote', $words))
       . ')\\S#i';

$value = trim(preg_replace($regex, $replaceWith . ' ', " {$value} "));

you can use this...

https://eval.in/611119

$wordlist = array('http','pic');
 $replaceWith  = "";



/* Sample data */
$words = 'This tweet has a pic.twitter.com/00GeQ3zLub and a url http://www.mywebsite.com';

foreach ($wordlist as $v)
  $words = clean($v, $words, $replaceWith);

function clean($word, $value, $replaceWith) {
    $reg_exUrl = "/ (".$word.")(\:\/\/|.)[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(\/\S*)?/ ";
    return preg_replace($reg_exUrl,$replaceWith,trim($value));

}

echo $words;
?>