PHP中GET和POST，ajax问题

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GET和POST，ajax问题遗留

<!DOCTYPE html>

<html lang="en">
<head>
<meta charset="UTF-8">
<title>getPostDemo</title>
<script type="text/javascript" src="jquary/jquery-3.0.0.min.js"></script>
</head>
<body>
<input type="button" onclick="postMessage()" value="点击我"/>
<script type="text/javascript">
function postMessage(){
$.ajax({
type:"post",
url:"getPostDemoTest.php",
data:{
"code":100,
"user":"admin"
},
success:function(data){
var result = eval("("+data+")");
console.log(result.code);
console.log(data);
}
});
}
</script>
</body>

</html>

<?php
/**

* Created by PhpStorm.

* User: liugaoyang

* Date: 2016/7/15
* Time: 11:39
*/

//当客户端发送数据时。服务器返回数据
if($_POST["code"]==100){
$arr = array("code"=>100,"user"=>"admin");
echo json_encode($arr);
}else{
echo json_encode(array("code"=>"youcuowu"));

}

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>getPostDemo</title>
<script type="text/javascript" src="jquary/jquery-3.0.0.min.js"></script>
</head>
<body>
<input type="button" onclick="postMessage()" value="点击我"/>
<script type="text/javascript">
// post只可以在data里面传参，如果在url里面传参默认为是get.
function postMessage(){
$.ajax({
type:"post",
url:"getPostDemo.php",
data:{
"code":11
},
success:function(data){
alert(data);
}
});
}
</script>
</body>

</html>

echo "get";
//print_r($_GET);
echo json_encode($_GET);
echo "post";
//print_r($_POST);
echo json_encode($_POST);

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