jQuery实现form表单基于ajax无刷新提交方法

表单基于ajax无刷新提交方法"/>

jQuery实现form表单基于ajax无刷新提交方法

一级标题

jQuery实现form表单基于ajax无刷新提交方法:

<script type="text/javascript" src=".3.1/jquery.js"></script><form id="form" onsubmit="return false" action="##" method="post">     <p class="title">user name</p><input type="text" autofocus="autofocus" name="username" class="input" id="Username" placeholder="Enter Username"><br><p class="title">Password</p><input type="password" class="input" id="Password" name="password" placeholder="Enter Password"><br><input type="button" class="submit" value="login" onclick="login()">//错误提示<div style="margin-top: 20px;margin-left: 30px;color: red;height: 5px" id="msg_test"></div></form>        

<script>function login(){$.ajax({type: "POST", dataType: "json", url: "./login.php" , data: $('#form').serialize(),success: function (data) {if (data.status == 200) {//根据返回值进行跳转location.href = data.url;}else{//提示层document.getElementById('msg_tests').innerHTML =　data.msg;}},});}
</script>

后端 PHP语言 Laravel框架演示返回信息：

     $tab = Orders::find($order_id);$tab->demo = $request->input('demo','');if($tab->save()){ $data = ['msg' => '修改成功', 'status' => '200'];return $data;}else{$data = ['msg' => '修改失败', 'status' => '400'];return $data;}