本文實例講述了PHP+JQUERY操作JSON的方法。分享給大家供大家參考，具體如下：
json.html 代碼：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
|
<html><head><meta http-equiv="Content-Type" content="text/html;charset=utf-8"><title>PHP Json傳輸數據</title></head><script type="text/javascript" src="js/jquery.js"></script><script type="text/javascript">$(function(){$("#submit").click(function(){var text = $("input").serialize();$.ajax({'type':"POST",'url':'json_encode.php','dataType':'json','data':text,success:insertData});});});function insertData(data){var str = "姓名="+data.name+"<br/>性別="+data.sex+"<br/>年齡="+data.age;$("#view").html(str);}</script><body>姓名：<input name="name" id="name" type="text" value=""><br/>性別：<input name="sex" id="sex" type="text" value=""><br/>年齡：<input name="age" id="age" type="text" value=""><input type="submit" name="submit" id="submit" value="提交"><div style="font-size:14px;" id="view"></div></body></html> |
json_encode.php 代碼
|
1
2
3
4
5
6
7
8
9
10
11
|
<?phpheader("Content-type:text/html;charset=utf8");include("Json.php");$name = $_POST['name'];$sex = $_POST['sex'];$age = $_POST['age'];$json_arg = array('name'=>$name,'sex'=>$sex,'age'=>$age);$json = new JSON;$json_result = $json->encode($json_arg);echo $json_result;?> |
希望本文所述對大家PHP程序設計有所幫助。
