php 如何使用 jquery ajax 提供和取回数据？
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How to give and get data back with jquery ajax?
提问by Boldizsar
I created a jquery ajax request and trying to pass a variable to the php to do a query and give back data to the actual context.
我创建了一个 jquery ajax 请求，并尝试将一个变量传递给 php 以进行查询并将数据返回给实际上下文。
The actual context, handling the ajax:
处理 ajax 的实际上下文：
$.ajax({
type: 'post',
url: 'show.php',
data: {name: name},
dataType: 'json',
success: function(response) {
//here I'd like back the php query
}
The php:
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
)
);
回答by Karthick Kumar
Try this:
尝试这个：
$.ajax({
type: 'post',
url: 'show.php',
data: "name="+ name,
dataType: 'json',
success: function(response) {
//here I'd like back the php query
}
and your PHP code to this:
和你的 PHP 代码：
$hostelName =mysql_escape_string($_POST['name']);
$sql = //here is the actual sql containing the $hostelname
$query = mysql_query($sql);
$reusult = mysql_fetch_assoc($query);
echo json_encode($reusult);
回答by Jason W
In your PHP code you are returning response as JSON. so just need to parse as below in code snippet
在您的 PHP 代码中，您将响应作为 JSON 返回。所以只需要在代码片段中解析如下
$.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);
}
