php+jQuery+Ajax implements page asynchronous refresh function
This article mainly introduces php jQuery Ajax to implement asynchronous page refresh, which has certain reference value. Interested friends can refer to it
The details are as follows:
## The code in #JQueryAjax.html is as follows (using the simpler $.post)<html> <head> <meta charset="UTF-8"> <title>JQueryAjax+PHP</title> <script type="text/javascript" src="https://code.jquery.com/jquery-3.0.0.min.js"></script> </head> <body> 用户名:<input type="text" id="username" name="username" /><br> 密码:<input type="password" id="password" name="password" /><br> <button type="button" class="butn">ajax提交</button><br> <span class="con"></span> <script type="text/javascript"> $(document).ready(function(){ $(".butn").click(function(){ var username = $("#username").val(); var password = $("#password").val(); $.post('ajax.php',{name:username,pwd:password},function(data) { alert(data); $(".con").html(data); }) }) }) </script> </body> </html>
<?php echo '用户名:',$_POST['name'],',密码:',$_POST['pwd']."<br>"; //这里可以进行一些操作,比如数据库交互 echo "操作完毕"; ?>
<html> <head> <meta charset="UTF-8"> <title>JQueryAjax+PHP</title> <script type="text/javascript" src="https://code.jquery.com/jquery-3.0.0.min.js"></script> </head> <body> 用户名:<input type="text" id="username" name="username" /><br> 密码:<input type="password" id="password" name="password" /><br> <button type="button" class="butn">ajax提交</button><br> <span class="con"></span> <script type="text/javascript"> $(document).ready(function(){ $(".butn").click(function(){ var username = $("#username").val(); var password = $("#password").val(); $.ajax({ url: "ajax.php", type: "POST", data:{name:username,pwd:password}, dataType: "json", error: function(){ alert('Error loading XML document'); }, success: function(data,status){//如果调用php成功 alert(status); alert(data); $('.con').html("用户名:"+data[0]+"密码:"+data[1]); } }); }) }) </script> </body> </html>
<?php $name = $_POST['name']; $pwd = $_POST['pwd']; $array = array("$name","$pwd"); //这里进行一个些操作,比如数据库交互 echo json_encode($array);//json_encode方式是必须的 ?>
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