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Home/ Questions/Q 7958169
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T04:22:46+00:00 2026-06-04T04:22:46+00:00

Would someone be able to enlighten me as to why when the following code

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Would someone be able to enlighten me as to why when the following code block is run in my script, the error is triggered, but when the sql query is attempted directly via php myadmin, there are no problems.

There is a mysql connection running in the script.

// Generate a random key and add details to specials table
$email = $_POST['specials'];
$randomKey = substr(str_shuffle(str_repeat('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789',12)),12);
$sql = "
    INSERT INTO specials SET
    id='',
    email='$email',
    verified='0',
    `key`='$randomKey',
    timestamp=NOW()
";
if (!mysqli_query($link, $sql))
{
    $error = 'Error adding email address to specials table.';
    include "$docRoot/html/main/error.html.php";
    exit();
}

I am quite sure that I have coded this correctly and before i go through a trail and error testing session, i thought thatone could have a brief look over this and tell me if they spot any obvious errors that may have eluded me.

Thanks for taking the time to read through this!

EDIT:

With the following code :

if (!mysqli_query($link, $sql12))
{
    echo mysqli_error();
    $error = 'Error adding email address to specials table.';
    include "$docRoot/html/main/error.html.php";
    exit();
}

I get the following output:

Warning: mysqli_error() expects exactly 1 parameter, 0 given in /serveraddress/includes/inc/assistors/controllers.inc.php on line 161

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T04:22:47+00:00Added an answer on June 4, 2026 at 4:22 am

    Try this to see what exactly its returning

    $result = mysqli_query($link, $sql);
    var_dump($result);
    
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