我在代码下面运行hte时显示上述错误,以显示从数据库进行的预订.
<?php
$servername = "localhost";
$username = "*********";
$password = "********";
$dbname = "thelibr1_fyp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, tablename, numseats, person FROM confirms";
$result = $conn->query($sql);
?>
<table id="Confirms" border ="2" style="length:900px;width:350px;">
<thead>
<tr style= "background-color: #A4A4A4;">
<td>Booking ID:</td>
<td>Table No.:</td>
<td>No. of Seats:</td>
<td>Person:</td>
</tr>
</thead>
<tbody>
<?php
while(($row = $result->fetch_assoc()) !== null){
echo
"<tr>
<td>{$row['id']}</td>
<td>{$row['tablename']}</td> …Run Code Online (Sandbox Code Playgroud)