Warning: mysqli_fetch_row() expects parameter 1 to be mysqli_result, null given

Question

What is the problem, It is printing the same whatever I do? A help would be appreciated.

<?php 

if(isset($_POST['submit'])) {



$username = $_POST['username'];
$password = $_POST['password'];


$connection = mysqli_connect('localhost', 'root', '', 'ayaneshapp');


    if($connection) {

    echo "Logged in!";

    } else {

    die("Sorry! Something went wrong.");

    }



    $query = "SELECT * FROM ayaneshtable ";

    $result = mysqli_query($connection, "SELECT * FROM ayaneshtable");

    if(!$result) {

    die('404 ERROR');

    }

}


?>

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Document</title>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
</head>
<body>

<div class="container">

    <div class="col-sm-6">

        <?php 

            while($row = mysqli_fetch_row($result)) {

                echo $row['username'];

            }

        ?>

    </div>


</div>  

</body>
</html>

score 0 · Accepted Answer · answered Sep 07 '18 at 08:35

$result is null because your code didn't go into the if statement. Every time you visit your page when $_POST['submit'] is not set, it will throw this error.

So to solve it you have to either also wrap your while loop in the same if statement, or make sure that $result gets filled regardless of there being something in $_POST['submit'].

Warning: mysqli_fetch_row() expects parameter 1 to be mysqli_result, null given

1 Answers1