jQuery Ajax发布到php没有捕获变量

Question

我究竟做错了什么.PHP似乎没有赶上title并wrapper从$阿贾克斯.代码看起来是否正确.我得到的成功消息表明找不到标题的错误.

jQuery main.html

$.ajax({
   type: "POST",
   url: "process.php",
   data: 'title=test&wrapper=testing',
   success: function(msg){
     alert( "Data Saved: " + msg );
   } 
});

PHP process.php

<?php 
$title = $_REQUEST['title'];
$wrapper = $_REQUEST['wrapper'];
...
?>

Answer 1

看一下:jQuery.ajax()

数据参数最好是一个键/值对对象,它更干净,更容易调试:)

$.ajax({
   type: "POST",
   url: "process.php",
   data: {
     title: 'test',
     wrapper: 'testing'
     },
   success: function(msg){
     alert( "Data Saved: " + msg );
   } 
});