ajax|jquery筆記

來源:互聯網
上載者:User

1.index.php

<html><head><script language="JavaScript" type="text/javascript" src="js/jquery.js"></script><script language="JavaScript" type="text/javascript" src="js/jquery.timer.js"></script><script type="text/JavaScript">$(document).ready(function(){    $('#submitbt').click(function(){    //var name = $('#name').val();    //var dataString = "name="+name;    var dataPass = {            'name': $("#name").val()        };    $.ajax({        type: "POST",        url: "post.php",                //data: dataString, 字串傳值方式                        data: dataPass,//json傳值方式        success: function (data) {                        alert(data);var re = $.parseJSON(data || "null");        console.log(re);         }    });   });});</script></head><body><form id="form" name="form" id="myform" method="post"><label>Name</label> <input name="name" type="text" id="name" />        <input type="button" class="button" id="submitbt" value="submit" > </form></body></html>


2.data.php

<?php$name = $_POST['name'];echo json_encode(array('name'=>$name));//為什麼return就不行呢。?>

參考:檢測json格式: http://jsonlint.com/

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