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Home/ Questions/Q 8617167
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T05:47:32+00:00 2026-06-12T05:47:32+00:00

I’m trying to get strings inside a quote. I’m using regex but i have

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I’m trying to get strings inside a quote.
I’m using regex but i have problems with escaped quotes.

For example, i have this:

$var = "SELECT * FROM TABLE WHERE USERNAME='Carasuman'";
preg_match_all('~([\'"])(.*?)\1~s', $var, $result);
$new = preg_replace('~([\'"])(.*?)\1~s',"<#################>",$var);

The code Works perfect. I got a replaced value in $new and quoted value in $result[1]

$new = "SELECT * FROM TABLE WHERE USERNAME=<#################>";
$result[1] = "Carasuman";

My problem is when i add a scaped quote inside quotes:

$var = "SELECT * FROM TABLE WHERE USERNAME='Carasuman\'s'";

I got this:

$new = "SELECT * FROM TABLE WHERE USERNAME=<#################>'s";
$result[1] = "Carasuman\" //must be "Carasuman\'s";

How I can avoid this error and get $new and $result[1] like first example?:

$new = "SELECT * FROM TABLE WHERE USERNAME=<#################>";
$result[1] = "Carasuman\'s";

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T05:47:33+00:00Added an answer on June 12, 2026 at 5:47 am

    for the match, you’re never going to get Carasuman’s without the \ as a single matched element since you can have match skip over chars within a single match. its either going to grab the Carasuman or Carasuman\’s

    just use str_replace to get rid of the backslash

    preg_match_all('~([\'"])(.*)\1~s', $var, $result);
    $result[2] = str_replace('\\','',$result[2]);
    

    for the replace, the ? in the (.*?) group makes it ungreedy, meaning it will stop at the first match. Remove the ? in (.*?) to make it greedy, meaning it will keep going until the last match

    preg_replace('~([\'"])(.*)\1~s',"<#################>",$var);
    

    Edit

    Rather than doing the str_replace after the match on $result[2], it would probably be better to just do beforehand on the initial string like:

    $var = str_replace("\\'","'",$var); 
    preg_match_all('~([\'"])(.*)\1~s', $var, $result);
    $new = preg_replace('~([\'"])(.*)\1~s',"<#################>",$var);
    

    You still need to make your wildcard match greedy like (.*?) to (.*) in order to have the apostrophe in the name included in the match/replace instead of being counted as the terminating single quote

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