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Home/ Questions/Q 7731465
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T06:29:12+00:00 2026-06-01T06:29:12+00:00

I am getting this warning even though everything functions perfectly. Here is the code

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I am getting this warning even though everything functions perfectly.

Here is the code snippet:

$selected_Id = isset($_GET["url"])?$_GET["url"]:""; 
// if no services selected, go to first service
$mysqliContent = new CustomMySqli();
if ($selected_Id=="") {
    $queryContent = "SELECT servicesID, ServiceUrl, title, content FROM services order by orderField asc LIMIT 1";
    $stmtContent = $mysqliContent->prepare($queryContent);
} else { 
$queryContent = "SELECT servicesID, ServiceUrl, title, content FROM services WHERE ServiceUrl = '$url'";
$stmtContent = $mysqliContent->prepare($queryContent);  
$stmtContent->bind_param('d',$selected_Id);
}
/* execute query */
$resultContent = $stmtContent->execute(); 

// bind results
$stmtContent->bind_result($idContent, $urlContent, $titleContent, $contentContent);
$contentContent = stripslashes($contentContent);

/* fetch values */ 
$stmtContent->fetch();
$selected_Id = $idContent;
/* close statement */
$stmtContent->close();
?>

Any help would be greatly appreciated.

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  1. Editorial Team
    Editorial Team
    2026-06-01T06:29:14+00:00Added an answer on June 1, 2026 at 6:29 am

    It should be:

    $queryContent = "SELECT servicesID, ServiceUrl, title, content FROM services WHERE ServiceUrl = ?";
    

    Prepared statements replace all question marks with the variable(s) supplied to bind_param().

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