ajax|jquery笔记
发布时间:2020-12-16 00:57:14 所属栏目:百科 来源:网络整理
导读:1.index.php htmlheadscript language="JavaScript" type="text/javascript" src="js/jquery.js"/scriptscript language="JavaScript" type="text/javascript" src="js/jquery.timer.js"/scriptscript type="text/JavaScript"$(document).ready(function(){
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,字符串传值方式 data: dataPass,//json传值方式 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));//为什么return就不行呢? ?> 参考:检测json格式: http://jsonlint.com/ (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |