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php+JQuery+Ajax简单实现页面异步刷新

2022年07月19日151jirigala

JQueryAjax.html中的代码如下(用的较为简单的$.post)

  1. <html
  2. <head
  3. <meta charset="UTF-8"
  4. <title>JQueryAjax+PHP</title
  5. <script type="text/javascript" src="https://code.jquery.com/jquery-3.0.0.min.js"></script
  6. </head
  7. <body
  8.     用户名:<input type="text" id="username" name="username" /><br
  9.     密码:<input type="password" id="password" name="password" /><br
  10.     <button type="button" class="butn">ajax提交</button><br
  11.     <span class="con"></span
  12. <script type="text/javascript"
  13. $(document).ready(function(){ 
  14.     $(".butn").click(function(){ 
  15.         var username = $("#username").val(); 
  16.         var password = $("#password").val(); 
  17.         $.post('ajax.php',{name:username,pwd:password},function(data) { 
  18.             alert(data); 
  19.             $(".con").html(data); 
  20.         }) 
  21.     }) 
  22. }) 
  23. </script
  24. </body
  25. </html
<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>

ajax.php:

  1. ajax.php 
  2. <?php  
  3. echo '用户名:',$_POST['name'],',密码:',$_POST['pwd']."<br>"; 
  4. //这里可以进行一些操作,比如数据库交互 
  5.  
  6.  
  7. echo "操作完毕"; 
  8. ?> 
ajax.php 
<?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"; 
?> 

在非json格式下,后台只能返回字符串,如果想后台返回数组,可以采用json格式

例如将JQueryAjax中的代码修改为如下形式:

  1. <html
  2. <head
  3. <meta charset="UTF-8"
  4. <title>JQueryAjax+PHP</title
  5. <script type="text/javascript" src="https://code.jquery.com/jquery-3.0.0.min.js"></script
  6. </head
  7. <body
  8.     用户名:<input type="text" id="username" name="username" /><br
  9.     密码:<input type="password" id="password" name="password" /><br
  10.     <button type="button" class="butn">ajax提交</button><br
  11.     <span class="con"></span
  12. <script type="text/javascript"
  13. $(document).ready(function(){ 
  14.     $(".butn").click(function(){ 
  15.         var username = $("#username").val(); 
  16.         var password = $("#password").val(); 
  17.         $.ajax({ 
  18.              url: "ajax.php",   
  19.              type: "POST", 
  20.              data:{name:username,pwd:password}, 
  21.              dataType: "json", 
  22.              error: function(){   
  23.                  alert('Error loading XML document');   
  24.              },   
  25.              success: function(data,status){//如果调用php成功  
  26.                 alert(status); 
  27.                 alert(data); 
  28.                 $('.con').html("用户名:"+data[0]+"密码:"+data[1]); 
  29.              } 
  30.         }); 
  31.     }) 
  32. }) 
  33. </script
  34. </body
  35. </html
<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>

ajax.php

  1. <?php  
  2. $name = $_POST['name']; 
  3. $pwd = $_POST['pwd']; 
  4. $array = array("$name","$pwd"); 
  5. //这里进行一个些操作,比如数据库交互 
  6.  
  7. echo json_encode($array);//json_encode方式是必须的 
  8. ?> 
<?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 
?>

运行效果如下:

http://blog.csdn.net/qq_28602957/article/details/51814437


本文参考链接:https://www.cnblogs.com/xihong2014/p/6029355.html