我有一个表单,用户必须输入他们的预订id和姓氏。如果这两个值在数据库中匹配,那么我需要从数据库返回相应的值。
我有两个文件,一个是使用ajax的html文件,另一个是php文件。当点击按钮时,没有返回任何内容,我没有看到任何具体的错误,我确信我输入的值是正确的。
<script>
var ajax = getHTTPObject();
function getHTTPObject()
{
var xmlhttp;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else if (window.ActiveXObject) {
// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
} else {
//alert("Your browser does not support XMLHTTP!");
}
return xmlhttp;
}
function updateCityState()
{
if (ajax)
{
var reservation_id = document.getElementById("reservation_id").value;
var guest_last_name = document.getElementById("guest_last_name").value;
if(reservation_id)
{
var param = "?reservation_id=" + reservation_id + "&guest_last_name=" + guest_last_name;
var url = "test04.php";
ajax.open("GET", url + param, true);
ajax.onreadystatechange = handleAjax;
ajax.send(null);
}
}
}
function handleAjax()
{
if (ajax.readyState == 4)
{
var guest_full_name = document.getElementById('guest_full_name');
var unit_number = document.getElementById('unit_number');
var floor = document.getElementById('floor');
var key_sa = document.getElementById('key_sa');
if(!!ajax.responseText) {
var result = JSON.parse(ajax.responseText);
if(!!result){
guest_full_name.innerHTML = (!!result.guest_full_name) ? result.guest_full_name : '';
unit_number.innerHTML = (!!result.unit_number) ? result.unit_number : '';
floor.innerHTML = (!!result.floor) ? result.floor : '';
key_sa.innerHTML = (!!result.key_sa) ? result.key_sa : '';
}
}
}
}
</script>
<p id='employee_name'></p>
<p id='employee_age'></p>
<p id='safe_code'></p>
我的test04.php
<?php
$conn = mysqli_connect("","","","");
$reservation_id = mysqli_real_escape_string($conn, $_GET['reservation_id']);
$guest_last_name = mysqli_real_escape_string($conn, $_GET['guest_last_name']);
$query = "SELECT reservation_id, guest_full_name, guest_last_name unit_number, floor, key_sa FROM reservations2 INNER JOIN guest ON (reservations2.reservation_id=guest.reservation_idg) INNER JOIN unit USING (unit_id) where reservation_id ='".$reservation_id."'AND guest_last_name ='".$guest_last_name."";
$result = mysqli_query($conn, $query) or die(mysql_error());
$response = array();
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$response['guest_full_name'] = ($row['guest_full_name'] != '') ? $row['guest_full_name'] : '';
$response['unit_number'] = ($row['unit_number'] != '') ? $row['unit_number'] : '';
$response['floor'] = ($row['floor'] != '') ? $row['floor'] : '';
$response['key_sa'] = ($row['key_sa'] != '') ? $row['key_sa'] : '';
}
}
echo json_encode($response, true);
?>
1条答案
按热度按时间zqdjd7g91#
我没有看到任何具体的错误
你在找什么?
您是否检查了php脚本的原始响应,或者只是查看了浏览器中呈现的内容?
您是否验证了错误日志记录是否正常工作,是否检查了日志?
php的逻辑还不清楚—json数据和php数组不能处理多个记录,但您可以处理多个记录。正确地实现rest是很好的。这也应该应用身份验证,并使用csrf来实现安全性——但我假设您出于说明的目的忽略了这些内容。
编写代码不是为了处理失败或丢失的数据。考虑(注意与您发布的内容的所有差异):