CODEWITHSUNDEEP
phpjqueryajax
user1375823
user1375823

Reputation: 105

Trying to send data through ajax to 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.

Upvotes: 0

Views: 41

Answers (2)

Charlie
Charlie

Reputation: 524

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;
};

Upvotes: 1

Rakesh Sharma
Rakesh Sharma

Reputation: 13728

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;
}

Upvotes: 1

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