本文实例讲述了php基于jquery的ajax技术传递json数据简单实现方法。分享给大家供大家参考，具体如下：
html页面：
<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">一会看显示结果</div><form id="my" action="" method="post"> <p><span>姓名：</span> <input type="text" name="username" /></p> <p><span>年龄：</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页面：
<?phpheader("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 + ")");//转为Object对象 var str = res.username + res.age + res.job; $("#result").html(str); }); }); });</script> 
