首页 > 解决方案 > 警告:mysqli_fetch_array() 期望参数 1 为 mysqli_result,布尔值在 C:\xampp\htdocs\kurinchi\kkmalar.php 第 41 行给出

问题描述

我无法找出这段代码有什么问题。由于查询执行和插入也有效,但错误消息显示:

警告:mysqli_fetch_array() 期望参数 1 为 mysqli_result,布尔值在 C:\xampp\htdocs\kurinchi\kkmalar.php 第 41 行给出

$q = intval($_GET['q']);
$availability = isset($_GET['availability']);
$con = mysqli_connect('localhost','root','','kurinchi');

if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"ajax_demo");
$sql="UPDATE rooms SET availability=(availability-1) WHERE ID = (SELECT ID from (SELECT * from rooms WHERE ID = '".$q."')AS innerResult)" ;
$result = mysqli_query($con,$sql);

if($result === FALSE) { 
    die(mysql_error()); 
}

echo "<table>
<tr>
    <th>ROOM TYPE</th>
    <th>AC TYPE</th>
    <th>PRICE</th>
    <th>AVAILABILITY</th>
    <th>BOOK </th>
</tr>";

while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['room_type'] . "</td>";
    echo "<td>" . $row['ac_type'] . "</td>";
    echo "<td>" . $row['price'] . "</td>";
    echo "<td>" . $row['availability'] . "</td>";
    echo "<td>" . $row['booking_status'] . "</td>";
    echo "</tr>";
}

echo "</table>";
mysqli_close($con);

标签: phpmysqli

解决方案


之后试试这个mysqli_query()

if (!$result) {
    printf("Error: %s\n", mysqli_error($con));
    exit();
}

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