php+jquery+ajax+json的一個最簡單實例
html頁面：
<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<script type="text/javascript" src="jquery-1.8.2.min.js"></script>
<script type="text/javascript">
$(function(){
$("#send").click(function(){
var cont = $("input").serialize();
$.ajax({
url:'ab.php',
type:'post',
dataType:'json',
data:cont,
success:function(data){
var str = data.username + data.age + data.job;
$("#result").html(str);
}
});
});
});
</script>
</head>
<body>
<div id="result">一會看顯示結果</div>
<form id="my" action="" method="post">
<p><span>姓名：</span> <input type="text" name="username" /></p>
<p><span>年齡：</span><input type="text" name="age" /></p>
<p><span>工作：</span><input type="text" name="job" /></p>
</form>
<button id="send">提交</button>
</body>
</html>
/*******************************************/
php頁面：
<?php
header("Content-type:text/html;charset=utf-8");
$username = $_POST['username'];
$age = $_POST['age'];
$job = $_POST['job'];
$json_arr = array("username"=>$username,"age"=>$age,"job"=>$job);
$json_obj = json_encode($json_arr);
echo $json_obj;
?>
使用post方式
<script type="text/javascript">
$(function(){
$("#send").click(function(){
var cont = {username:$("input")[0].value,age:$("input")[1].value,job:$("input")[2].value};
var url = 'ab.php';
$.post(url,cont,function(data){
var res = eval("(" + data + ")");//轉為Object對象
var str = res.username + res.age + res.job;
$("#result").html(str);
});
});
});
</script>