Function not working (using LIKE and PDO)

[oz11]

Been working on this part of my site,...

<?php
function getLinkTree($pdo, $url) {
    $res = str_replace(array('https://www.', 'http://www.', 'https://','http://'), '', $url);
    $stmt = $pdo->prepare("SELECT url FROM links WHERE url LIKE ?");
    $stmt->execute(['%$res%']); 
    while ($row = $stmt->fetch()) {
        echo $row['url']."<br />\n";
    }
}

getLinkTree($pdo, "https://bbc.co.uk");

?>

Instead of it coming back with all the other urls which contain "bbc.co.uk" it doesnt do anything. However in phpmyadmin, when I run the query ..

Quote

SELECT * FROM `links` WHERE url LIKE '%bbc.co.uk%';

.. it brings back the correct results. What am I doing wrong guys and gals?

Edited November 12, 2022 by oz11

[oz11]

Changed 

Quote

$stmt->execute(['%$res%']);

to

Quote

$stmt->execute(["%$res%"]);

and seems to work now. Thanks guys for being here.

[maxxd]

PHP won't interpolate variables inside single quotes, so your search string was literally %$str% in the first example. Inside double quotes, however, PHP will replace the variable with the value it contains.