<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>JSON.parse()</title>
<script type="text/javascript">
//演示样例1：此演示样例使用 JSON.parse 将 JSON 字符串转换为对象
var jsontext = '{"firstname":"Jesper","surname":"Aaberg","phone":["555-0100","555-0120"]}';//JSON 字符串
var contact = JSON.parse(jsontext);
document.write(contact.surname + ", " + contact.firstname + ", "+ contact.phone);
//演示样例2：和实例1是一样的效果
var jsontext2 = {"firstname":"Jesper","surname":"Aaberg","phone":["555-0100","555-0120"]};//JSON 对象
//var contact2 = JSON.parse(jsontext2);
document.write("<br /><br />"+jsontext2.surname + ", " + jsontext2.firstname + ", "+ jsontext2.phone);
</script>
</head>
<body>
</body>
</html>
输出：
Aaberg, Jesper, 555-0100,555-0120
Aaberg, Jesper, 555-0100,555-0120
************************************************************************************
前端页面接收JSON对象的实例：
<script>
sendRecord('1');
function sendRecord(record){
var req = {
user_id:<?php echo $userId;?>,
record:record,
}
$.ajax({
url: "./a.php",
type:"post",
data:req,
dataType:"JSON", //返回数据格式为JSON对象
success: function(res){
alert(Object.prototype.toString.apply(res));//alert输出[object Object]，不用JSON.parse()解析
if(res.result==1){
alert('11');
}else if(res.result==2){
alert('22');
}else if(res.result==3){
alert('33');
}
},
error: function(){
alert('error000');
console.log(this);
}
});
}
</script>a.php
<?php
$record = $_POST['record'];
if ($record==1) {
$json['result'] = 1;
}elseif($record==2){
$json['result'] = 2;
}elseif($record==3){
$json['result'] = 3;
}
$json = json_encode($json);
echo $json; //{"result":1}
?>
***************************************************************************
<script type="text/javascript" src="./jquery.min.js"></script>
<script>
sendRecord('2');
function sendRecord(record){
var req = {
record:record,
}
$.ajax({
url: "./b.php",
type:"post",
data:req,
dataType:"JSON", //返回数据格式为JSON对象
success: function(res){
//加上dataType:"JSON"，alert输出[object Array]，不须要JSON.parse()解析
//不加dataType:"JSON"，alert输出[object String]，须要JSON.parse()解析
alert(Object.prototype.toString.apply(res));
//var res2 = JSON.parse(res);
alert(res);
},
error: function(){
alert('error000');
console.log(this);
}
});
}
</script>b.php
<?php $record = $_POST['record']; if ($record==1) { $json = array('a','aa'); }elseif($record==2){ $json = array('b','bb'); }elseif($record==3){ $json = array('c','cc'); } $json = json_encode($json); echo $json; //["b","bb"] ?
>
总结：假设后台是通过echo json_encode();输出json字符串的话，并且前台使用了dataType:"JSON", 那么则不须要JSON.parse();把json字符串转换成对象或数组。