如何在使用Php方法时接收和处理来自jQuery脚本的响应

Question

如何在使用jQuery Ajax方法时接收和处理来自PHP的响应？

javascript

$.ajax({
    type: 'post',
    url: 'show.php',
    data: {name: name},
    dataType: 'json',
    success: function(response) {
       //I want my json response back here
    }
});

show.php

$hostelName = $_POST['name'];
$sql = //here is the actual sql containing the $hostelname
$query = mysql_query($sql);
$obj = mysql_fetch_object($query); 
$sum = $obj->sum;
$tour = $obj->tour;
echo json_encode(
    array(
        "sum" => $sum,
        "tour" => $tour
    )
);

Answer 1

试试这个：

    $.ajax({
    type: 'post',
    url: 'show.php',
    data: "name="+ name,
    dataType: 'json',
    success: function(response) {
    //here I'd like back the php query
    }

您的PHP代码如下：

    $hostelName =mysql_escape_string($_POST['name']);

    $sql = //here is the actual sql containing the $hostelname

    $query = mysql_query($sql);

    $reusult = mysql_fetch_assoc($query);
    echo json_encode($reusult);

Answer 2

在PHP代码中，您将以JSON的形式返回响应。所以只需要在下面的代码片段中解析

 $.ajax({
    type: 'post',
    url: 'show.php',
    data: {name: name},
    dataType: 'json',
    success: function(response) {
       //response is json you need to parse it
         var json = response,
         obj = JSON.parse(json);
         alert(obj.sum);
         alert(obj.tour);

    }