PHP regex how to replace with the same string

the following statement works:

$display_html =
preg_replace(
   "/".$search_string."/i",
   "<b class='highlight'>".$search_string."</b>",
    $row['title'] );

where search_string is what is being searched for, and 'title'
is the  LIKE-matching result from a mysql database.

currently the regex substitutes the CASE that was used as typed in by the searcher.
I would like to have the CASE retained to that of what the field was in the database
(could be lower, initial cap, all caps, etc.)

how can the regex substitution (which surrounds the title search phrase with HTML tags for highlighting)
be altered to retain the the capitalization in the returned field result.
willsherwoodAsked:
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käµfm³d 👽Commented:
Add in capturing parens around the thing being sought, then use the appropriate backreference (most likely 1) to refer to what was captured by the parens.

e.g.

$display_html = (
   "/(".$search_string.")/i", 
   "<b class='highlight'>$1</b>",
    $row['title'] );

Open in new window


Because I have only used one set of capturing parens, the backreference I use is for group 1 (that's the $1 bit in the replacement).
0
käµfm³d 👽Commented:
Also, since you are searching for stuff based on what the user types in, you'll want to use the escape function (preg_quote) so that the user doesn't inadvertently (or maliciously) type in special regex characters into your input.

e.g.

$display_html = (
   "/(".preg_quote($search_string).")/i", 
   "<b class='highlight'>$1</b>",
    $row['title'] );

Open in new window

0

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willsherwoodAuthor Commented:
thanks for the detailed help!
0
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