JQuery Post to PHP Inserts data, but json response is null

Question

I'm building a simple sign up form with AngularJS and sending the data to a PHP page using JQuery's $.post(). When I send the data, it correctly gets inserted into the database. However, the returned json that I am logging is showing my data fields as null.

Console:

{"status":"success","email":null,"id":null,"sessionId":null}

Javascript:

$.post("admin/addUser.php", {
    email: form.email,
    password: form.password
}).done(function(data){
    console.log(data);
});

PHP:

$email = mysql_real_escape_string($_POST["email"]);
$password = md5(mysql_real_escape_string($_POST["password"]));
$sessionId = md5(uniqid(microtime()) . $_SERVER['REMOTE_ADDR'] . $_SERVER['HTTP_USER_AGENT']);

//Add this user to the database
$sql = mysql_query("INSERT INTO users (email, password, sessionId) VALUES ('".$email."', '".$password."', '".$sessionId."')");

if ($sql){
    //Now find the user we just added
    $getUser = mysql_query("SELECT * FROM users WHERE email = '".$email."' AND sessionId = '".$sessionId."'");
    if ($getUser){
        $row = mysql_fetch_array($getUser);
        $user = array(
            'status' => 'success',
            'email' => $row['email'],
            'id' => $row['id'],
            'sessionId' => $row['sessionId']
            );
            echo json_encode($user);
    }else{
        $user = array(
            'error' => mysql_error()
        );
        echo json_encode($user);
    }
}else{
    $user = array(
        'error' => mysql_error()
    );
    echo json_encode($user);
}

iamawebgeek · Accepted Answer · 2014-07-04 12:21:28Z

1

Are you sure that you have only one record in here

$getUser = mysql_query("SELECT * FROM users WHERE email = '".$email."' AND sessionId = '".$sessionId."'");

Try to dump $row and see the response. BTW I would suggest you to add limit

$getUser = mysql_query("SELECT * FROM users WHERE email = '".$email."' AND sessionId = '".$sessionId."' LIMIT 1");

answered Jul 4, 2014 at 12:21

iamawebgeek

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4 Comments

Mark Kamyszek Over a year ago

Thanks for the help. I added if (mysql_num_rows($getUser) == 1) and it doesn't evaluate. Odd, because I don't have any other data in this table. I'm trying to insert the data, and if successful, retrieve it and return it as json.

iamawebgeek Over a year ago

BTW I cant understand why do you select the data after insert?

Mark Kamyszek Over a year ago

@user3796431 Once the user has signed up, I want to get their sessionId and store it locally so they don't have to sign in again. It's also a bit of confirmation for me that this has worked. I don't understand why (mysql_num_rows($getUser) == 1) returns false when there's no one else in the DB. Additionally, the email column requires unique values, as does the sessionId column.

iamawebgeek Over a year ago

Last guess :) $getUser = mysql_query("SELECT COUNT(*) AS num FROM users WHERE email = '".$email."' AND sessionId = '".$sessionId."'"); $num = mysql_fetch_array($getUser); if($num['num'] == 1) echo json_encode(array( 'status' => 'success', 'email' => $email, 'id' => mysql_insert_id(), 'sessionId' => $sessionId ));

Mark Kamyszek · Accepted Answer · 2014-07-08 20:59:57Z

0

Ok, I dug around and found the answer to this. It was a mistake on my end. I was only storing the password and sessionId as varchar(30). When I was generating the sessionId and checking it against the DB, it was being cut off when it was stored, since I was only allowing 30 chars. I update to 255 and works as expected :-P.

answered Jul 8, 2014 at 20:59

Mark Kamyszek

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