试图通过ajax将数据发送到php

问题描述：

So I'm trying to create some code that checks if a username is taken. I'm kind of half there and having trouble. I'm new to it and trying to learn how to do it & the code will be messy.

the jquery:

$('#signusername').keyup(function()
        {       
        var username=$('#signusername').val();         
        if(username != ''){

            $.post('username_check.php', {signusername :username}, function(result)

            {
                    if(result==''){
                        $('.error').text('Avaliable');
                        } else{
                        $('.error').text('Taken');
            }
            }
            );
        }else{
        $('.error').text('???');//this is the the only thing that outputs correctly

        }

the php:

function checkUsername($signusername, $conn) {
            $stmt = $conn->prepare("SELECT * FROM user_info where username= '".$signusername."'");
                $stmt->bindParam(1, $signusername);
                $stmt->execute();
                if($stmt->rowCount() == 1) {
                   return TRUE;
                }
                };


if(isset($_POST['signusername']) && !empty($_POST['signusername'])){
    $signusername= $_POST['signusername'];
    checkUsername($signusername, $conn);
    $result='';
    if(checkUsername($signusername, $conn) == TRUE){
    $result='';
    }else{
    $result='';
}
echo $result;
};

I use the same code to check if the username is taken when the form is submitted so I don't think that is the problem. I assume I'm doing something wrong with moving the username variable across? Hope you can help.

所以我正在尝试创建一些代码来检查用户名是否被占用。我有一半在那里遇到麻烦。我是新手，并试图学习如何做到这一点＆amp; 代码将是混乱的。 p>

jquery： p>

  $（'＃signusername'）。keyup（function（）
 {  
 var username = $（'＃signusername'）。val（）; 
 if（username！=''）{
 
 $ .post（'username_check.php'，{signusername：username}，function（ 结果）
 
 {
 if（result ==''）{
 $（'。error'）。text（'Avaliable'）; 
} else {
 $（'。error'）。  text（'Taken'）; 
} 
} 
）; 
} else {
 $（'。error'）。text（'???'）; //这是唯一的东西 输出正确
 
} 
  code>  pre> 
 
  php： p> 
 
 
 函数checkUsername（$ signusername，$ conn）  {
 $ stmt = $ conn-＆gt; prepare（“SELECT * FROM user_info where username ='”。$ signusername。“'”）; 
 $ stmt-＆gt; bindParam（1，$ signusername）; 
 $  stmt-＆gt; execute（）; 
 if（$ stmt-＆gt; rowCount（）== 1）{
 返回TRUE; 
} 
}; 
 
 
if（isset（$ _ POST ['signusername']）＆amp;＆amp;  ！empty（$ _ POST ['signusername']））{
 $ signusername = $ _POST ['signusername']; 
 checkUsername（$ signusername，$ conn）; 
 $ result =''; 
 if（checkUsername  （$ signusername，$ conn）== TRUE）{
 $ result =''; 
} else {
 $ result =''; 
} 
echo $ result; 
}; 
  code  >  pre> 
 
 我使用相同的代码来检查在提交表单时是否采用了用户名，所以我认为这不是问题所在。 我假设我在移动用户名变量时遇到了什么问题？ 希望你能帮忙。 p> 
  div>

答

Check your console for syntex error }; remove semicolon after }

Also remove function calling two time and you sending result blank in both condition so send some response back to ajax in also fail condition

if(isset($_POST['signusername']) && !empty($_POST['signusername'])){
    $signusername= $_POST['signusername'];
    $result='';
    if(checkUsername($signusername, $conn) == TRUE){
    $result='user found';  
    }else{
    $result='user not found';
    }
   echo $result;
}

答

Try this code

function checkUsername($signusername, $conn) {
            $stmt = $conn->prepare("SELECT * FROM user_info where username= '".$signusername."'");
               $stmt->bindParam(1, $signusername);
                $stmt->execute();
                if($stmt->rowCount() == 1) {
                   return TRUE;
                }
return false;
                };


if(isset($_POST['signusername']) && !empty($_POST['signusername'])){
    $signusername= $_POST['signusername'];
    $result = checkUsername($signusername, $conn);
    if($result != TRUE){
       $result='';
    }else{
}
echo $result;
};

试图通过ajax将数据发送到php

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