mikkiharu
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Fatal error: Uncaught TypeError: mysqli_num_rows(): Argument #1 ($result) must be of type mysqli_result...on line 28
in PHP Coding Help
Posted
<!DOCTYPE html>
<html>
<head>
<title>Registered Users</title>
<link rel="stylesheet" href="includes/style.css" type="text/css" media="screen"/>
<link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.25/css/jquery.dataTables.min.css">
</head>
<body>
<?php
include('includes/header.html');
?>
<center>
<table id="records">
<thead>
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Email</th>
</tr>>
</thead>
<tbody>
<?php
require_once('mysqli_connect.php');
$query = "SELECT * FROM users";
$result = mysqli_query($dbc,$query);
if(mysqli_num_rows($result) > 0) -------------------------------------------------Line 28
{
while($row = mysqli_fetch_assoc($result))
{?>
<tr>
<td><?php echo $row["id"];?></td>
<td><?php echo $row["First Name"];?></td>
<td><?php echo $row["Last Name"];?></td>
<td><?php echo $row["Email"];?></td>
</tr>
<?php
}
}
mysqli_close($dbc);
?>
</tbody>
</table>
</center>
</body>
<?php include('includes/footer.html');?>
</html>