Regex replace in case beetwen determinated chars (PHP)

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i have a mysql database that store up 1m ads. This ads have field "Ad_Content" and have random examples (Ad_Content):

1 CASE: \r\n{text...text...text...}\r\n\r\n\r\n <- This row is good

2 CASE: {text...text...}\r\n \r\n\r\n <- Wrong, have 1 space beetwen string "\r\n\r\n"

3 CASE: \r\n \r\n{text...}\r\n\r\n\r\n \r\n <- Wrong, have 2 spaces beetwen string "\r\n\r\n"

4 CASE: \r\n \r\n\r\n\r\n{text...text...}\r\n \r\n\r\n \r\n <- Wrong, have 3 spaces beetwen string "\r\n*"

I try to remove all spaces found in the string "\r\n group" with REGEX but i can't get its.

Expected examples:

1 CASE: \r\n{text...text...}\r\n\r\n <- This row is good

2 CASE: {text...text...}\r\n\r\n\r\n (good)

3 CASE: \r\n\r\n{text...}\r\n\r\n\r\n\r\n (good)

4 CASE: \r\n\r\n\r\n\r\n{text...text...}\r\n\r\n\r\n\r\n (good)

My code is:

$adContent  = $getDataAd->Ad_Content;
// This REGEX its ok, replace all \r\n\r\n\r\n\r\n... for 2 <br> but broken if detect a space :(
$pattern    = "/(?:\r?\n[ ]*)+/"; 
$adContent  = preg_replace($pattern, "<br><br>", $adContent);

Thanks :)

laravel php regex
1 Answers

You may do a regex replacement on the following pattern:

\r\n\s+(?=\r\n)

Sample script, using your third example as input:

$input = "\r\n \r\n\r\n\r\n{text...text...}\r\n \r\n\r\n \r\n";
$output = preg_replace("/\r\n\s+(?=\r\n)/", "\r\n", $input);

The trick here is to use the lookahead (?=\r\n) to assert, but not consume, a required following CRLF. This way, we allow for the possibility of that following CRLF to also have unwanted spaces in front of it.

user_18632290
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