Using date() inside preg_replace causes dates to be replace with 1970-CodePudding

I'm trying to add 24 hours to every date using preg_replace.

I'm having trouble passing the date to strtotime(). All dates are replaced with 1970.

I tried using strtotime() with time() and preg_replace() $data = "Hello world Jan 19, 2023 8:06:47am testing Feb 09, 2021 2:06:47pm zooph foobar"; echo preg_replace("/[a-zA-Z]{3} \d{2}, \d{4} [0-9]{1,2}\:\d{2}\:\d{2}[ap]m/", date("F j, Y, g:i a", (strtotime("$0")) 86400),$data);

CodePudding user response：

Your regex pattern is correct, but in order to use a function of the match as the replacement, you need to use preg_replace_callback() here:

$data = "Hello world Jan 19, 2023 8:06:47am testing Feb 09, 2021 2:06:47pm zooph foobar";
$out = preg_replace_callback(
    "/[a-zA-Z]{3} \d{2}, \d{4} [0-9]{1,2}\:\d{2}\:\d{2}[ap]m/",
    function($m) {
        return date("F j, Y, g:i a", (strtotime($m[0])) 86400);
        },
    $data
);

echo $out;

This prints:

Hello world January 20, 2023, 8:06 am testing February 10, 2021, 2:06 pm zooph foobar

CodePudding user response：

I think you need to add preg_replace_callback function:

$data = "Hello world Jan 19, 2023 8:06:47am testing Feb 09, 2021 2:06:47pm zooph foobar";

echo preg_replace_callback("/[a-zA-Z]{3} \d{2}, \d{4} [0-9]{1,2}\:\d{2}\:\d{2}[ap]m/", function ($matches) {
    return date("F j, Y, g:i a", strtotime($matches[0])   86400);
}, $data);

Result :

Hello world January 20, 2023, 8:06 am testing February 10, 2021, 2:06 pm zooph foobar