Question

我有一个表格，用户必须输入他们的预订ID和姓氏。如果数据库中这两个值匹配，那么我需要从数据库中返回相应的值。

看起来即使输入了错误的预订ID和名称，也将其视为成功。页面将重新加载并显示foobar。

这是我的代码

test06.php

<?php

$conn = mysqli_connect("","","","");

$reservation_id=$_POST['reservation_id'];
$guest_last_name=$_POST['guest_last_name'];

$stmt = $conn->prepare("SELECT reservation_id, guest_last_name, guest_full_name, unit_number, floor, key_sa 
                        FROM reservations2 
                            INNER JOIN guest ON (reservations2.reservation_id=guest.reservation_idg) 
                            INNER JOIN unit USING (unit_id) 
                        WHERE reservation_id=? 
                        AND guest_last_name=?");
$stmt->bind_param("ss", $reservation_id, $guest_last_name);
$stmt->execute();
$stmt->bind_result($reservation_id, $guest_last_name, 
                   $guest_full_name, $unit_number,  
                   $floor, $key_sa);
if ($stmt->errno) {
    die("Query failed to execute: " . $stmt->error);
}
if ($stmt->fetch()) {
    echo json_encode(array("reservation_id" => $reservation_id, 
                           "guest_last_name" => $guest_last_name,  
                           "guest_full_name" => $guest_full_name,  
                           "unit_number" => $unit_number,  
                           "floor" => $floor,  
                           "key_sa" => $key_sa));
} else {
    $error="Not matching record";
    echo json_encode($error);
}
$stmt->close();

?>

内部HTML页面

<p id='guest_full_name'></p>
<p id='unit_number'></p>
<p id='floor'></p>
<p id='error'></p>

<script>
function validateReservation(){
  var reservation_id = document.getElementById("reservation_id").value;
  var guest_last_name = document.getElementById("guest_last_name").value;

    $.ajax({
        type: 'POST',
        url: 'test06.php',
        // dataType: 'json',
        data: {
                'reservation_id': reservation_id,
                'guest_last_name' : guest_last_name
            }, 
        success: function(json) {
            var json = JSON.parse(json);
            console.log(json);
            $('#guest_full_name').html(json.guest_full_name);
            $('#unit_number').html(json.unit_number);   
            $('#floor').html(json.floor);
            $('#key_sa').html(json.key_sa);
            }, 
        error: function(err) {
            console.log(err);
            $("#error").html("Error!");
            } 
    }); 
}

Answer 1

您必须发送状态码。回显JSON只会给您200响应。

header('Content-Type: application/json');
http_response_code(500); 
echo $json;
exit;

http://php.net/manual/en/function.http-response-code.php

Ajax的响应总是成功的

1 个答案: