$newstr = preg_replace('/[^a-zA-Z0-9\']/', '_', "There wouldn't be any");
$newstr = str_replace("'", '', $newstr);

I put them on two separate lines to make the code a little more clear.

Note: If you're looking for Unicode support, see Filip's answer below. It will match all characters that register as letters in addition to A-z.

Python's re.sub doesn't support array-style inputs the way PHP's preg_replace does. You would need to iterate over the replacements instead e.g.

def clean(text):
    utf8 = {
        '[áàâãªä]'   :   'a',
        '[ÁÀÂÃÄ]'    :   'A',
        '[ÍÌÎÏ]'     :   'I',
        '[íìîï]'     :   'i',
        '[éèêë]'     :   'e',
        '[ÉÈÊË]'     :   'E',
        '[óòôõºö]'   :   'o',
        '[ÓÒÔÕÖ]'    :   'O',
        '[úùûü]'     :   'u',
        '[ÚÙÛÜ]'     :   'U',
        'ç'          :   'c',
        'Ç'          :   'C',
        'ñ'          :   'n',
        'Ñ'          :   'N',
        '–'          :   '-', # UTF-8 hyphen to "normal" hyphen
        '[’‘‹›‚]'    :   ' ', # Literally a single quote
        '[“”«»„]'    :   ' ', # Double quote
        ' '          :   ' ', # nonbreaking space (equiv. to 0x160)
    }
    text = text.strip()
    for pat, repl in utf8.items():
        text = re.sub(pat, repl, text, 0, re.U)
    return text

Note also that python does not use delimiters around regexes, and you pass the u flag to re.sub directly. I've adjusted your code to deal with those issues.

Sample usage:

print(clean('ÂôÑ‹Î'))

Output:

AoN I

You just need to escape special characters used in regexes. There is this great function called preg_quote that will do it automatically.

$pattern_array = array();
$replace_array = array();
$pattern_array[] = '#' . preg_quote('[vc_row][vc_column width="1/2" css=".vc_custom_1503896034143{margin-bottom: 40px !important;}"][shortcode_testimonials_carousel', '#') . '#i';
$replace_array[] = '[vc_row fixed_width="true"][vc_column width="1/2" css=".vc_custom_1503896034143{margin-bottom: 40px !important;}"][shortcode_testimonials_carousel';
echo preg_replace($pattern_array, $replace_array, $content);

try this : $show=preg_replace(str_replace('$','\$',$ch),' ',$show, 1); If you can't use preg_quote and have multiple escape ( in this case $ is your problem), you could put your special character in an array and filter from it.

You do not need to escape & only the ?

$reg = preg_replace("#&&#", "&", $reg);

$reg = preg_replace("#\?&#", "?", $reg);

$str=preg_replace("/[^0-9a-zA-Z ]/u", "_", $str_test);

Notice 'u' modifier! Explanation: http://www.php.net/manual/en/reference.pcre.pattern.modifiers.php#107498

If the _subject_ contains utf-8 sequences the 'u' modifier should be set, otherwise a pattern such as /./ could match a utf-8 *sequence as two to four individual ASCII characters*.

There is actually no need to even loop the array.
Str_replace accepts arrays.
Add the %% at start and end of each key in the array and this code will replace them all.

$vars = array(
  '%%name%%' => $name,
  '%%email%%' => $email,
  '%%your_name%%' => $directorname,
  '%%logininfo%%' => $loginInfo,
  '%%directorname%%'  =>  $directorname,
  '%%campname%%'  =>  $campname,
  '%%numvideos%%' =>  NUM_VIDEOS);


$message = str_replace(array_keys($vars) , $vars , $message);

See here for a working example https://3v4l.org/T8YGs

Try this :

 <a href="<?php echo $row->id;?>/
 <?=trim(preg_replace('#[^\w\-]+#','_', strtolower($row->showTitle())), '_')?>.html"> 

It will remove any trailing underscore.

try to replace the regular expectation change

preg_replace('/[^A-Za-z0-9\-]/', '', $string);

with

preg_replace("/[^A-Za-z0-9\-\']/", '', $string);  // escape apostraphe

or

you can str_replace It is quicker and easier than preg_replace() Because it does not use regular expressions.

$text = str_replace("'", '', $string);

You have to escape those characters that have a special meaning in regular expression in that case.

preg_replace ('/[^\.,?!\'":;\-]/', '' ,$value);

preg_quote can also be used:

$toKeep = preg_quote ('.,?!\'":;', '/');
preg_replace ('/[^' . $toKeep . ']/', '', $value); 

You can use:

$str = preg_replace('/^=+\h*\K.+?(?=\h*=)/m', 'DONE', $str);

RegEx Demo

RegEx Breakup:

^          # Line start
=+         # Match 1 or more =
\h*        # Match or more horizontal spaces 
\K         # resets the starting point of the reported match
.+?        # match 1 or more of any character (non-greedy)
(?=\h*=)   # Lookahead to make sure 0 or more space followed by 1 = is there

If I understand you correctly, use:

/[^\p{Latin}0-9()]/u

That will match anything that is not parentheses, letters or numbers.

So the full code:

$string = preg_replace('/[^\p{Latin}0-9()]/u', '', $string);

That regex just seems to have a lot of unnecessary problems. As Otala said, the hypen isn't escaped, the pipe isn't escaped and it's not looking for periods. Also it had two checks for hyphens.

You should simplify it by simply looking for all non-pipe characters up to the pipe:

preg_replace("/([^|\r\n]+)\|/", '<h3 id="${1}">${1}</h3>', $content);