douli1306 2014-08-04 12:54
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试图通过ajax将数据发送到php

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.

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  • douhe1002 2014-08-04 12:58
    关注

    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;
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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