dongtuo7364
2016-05-31 03:06
浏览 49

将值从ajax传递给php类函数

Q. Is there a way to pass values from ajax to a certain php class having functions? Let's say validating a username on the registration form whether the user exist or not.

This is a simple form that will accept username and has a span tag to display the message.

<form action="" method="POST">
  <input type="text" name="username"><span class="check"></span>
  <input type="submit" name="signup">
</form>

And for the php class:

<?php 
  class User {
    function isUserExist($username) {
      $query = mysql_query("SELECT username FROM users WHERE username='$username'");
      $result = mysql_num_rows($query);

      return ($result !== 0 ? true : false);
    }
  }
?>

It is initialized on the php class that established connection to the database. So calling to the php page will become like this: $user->isUserExist($_POST['username']);.

So is it possible to pass values from the form to ajax and send it to the php class function?

</div>

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Q. 有没有办法将值从ajax传递给具有函数的某个php类? 假设用户是否存在,请在注册表上验证用户名。

这是一个简单的表单,它将接受用户名并有一个span标记来显示该消息。

 &lt; form action =”“method =”POST“&gt; 
 
&lt;输入 type =“text”name =“username”&gt;&lt; span class =“check”&gt;&lt; / span&gt; 
 
&lt; input type =“submit”name =“signup”&gt; 
 
&lt;  ; / form&gt;   
 
  
 
  
 
 
 
 

对于php类: \ n

 &lt;?php 
 
 class User {
 
 function isUserExist($ username){  
 
 $ query = mysql_query(“SELECT username FROM users WHERE username ='$ username'”); 
 
 $ result = my  sql_num_rows($ query); 
 
 
 
返回($ result!== 0?  true:false); 
 
} 
 
} 
 
?&gt;   
 
  
 
  
 
  
 
 

它在建立与数据库连接的php类上初始化。 调用php页面将变成这样:$ user-&gt; isUserExist($ _ POST ['username']);.

那么可以将表格中的值传递给ajax并将其发送到php类函数吗?

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2条回答 默认 最新

  • dongyashun2559 2016-05-31 03:32
    已采纳

    From Html to ajax

    var username = $("input[name='username']").value;
    

    Fetch in ajax & Send it to php(server)

      var xhttp = new XMLHttpRequest();
      xhttp.onreadystatechange = function() {
        if (xhttp.readyState == 4 && xhttp.status == 200) {
          //set your span to this -> xhttp.responseText;
        }
      };
      xhttp.open("POST", "your php script url", true);
      xhttp.send("username="+username);
    

    Receive it on the server(php)

    $mUsername = $_POST['username'];
    echo $mUsername;
    

    Read this tutorial for more help Tutorial on PHP + AJAX

    已采纳该答案
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  • dongying7667 2016-05-31 06:15
    Try this,
    
    <script type="text/javascript">
    $(document).ready(function(){
        $("input[name = 'signup']").click(function(e) {
            var username = $("input[name = 'username']").val();
            $.ajax ({
                url: "isUserExist_function_existing_file.php", 
                data: { username : username },
                success: function( result ) {
                    if(result)
                        alert("Name allready Exist");
                    else
                        alert("Name available");
                }
            });
        });
    });
    </script>
    
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