Function returns empty data

Question

I use this function to check whether the record exists in my database and act accordingly. There is some error in the second part when the data is not matched: it is supposed to insert a new record, but it does not insert new data, and it returns empty data. Where am I going wrong?

 class User {

function checkUser($uid, $oauth_provider, $username,$email,$twitter_otoken,$twitter_otoken_secret) 
{

     // Define database connection constants
define('DB_HOST', 'localhost');
define('DB_USER', '*********');
define('DB_PASSWORD', '********');
define('DB_NAME', '******');
    $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

    $query ="SELECT * FROM si_table WHERE oauth_uid = '$uid' and oauth_provider = '$oauth_provider'";
    $data=mysqli_query($dbc,$query);
    $result = mysqli_fetch_array($data);
    if (!empty($result)) {

        # User is already present
    } else {

        #user not present. Insert a new Record

     $query2 ="INSERT INTO si_table (oauth_provider, oauth_uid, user_name,email_id,twitter_oauth_token,twitter_oauth_token_secret) VALUES ('$oauth_provider',$uid,'$username','$email')";
        mysqli_query($dbc,$query2);
        $query1 ="SELECT * FROM si_table WHERE oauth_uid = '$uid' and oauth_provider = '$oauth_provider'";
        $data1=mysqli_query($dbc,$query1);
        $row = mysqli_fetch_array($data1);

        return $row;

    }
    return $result;
}



}

Why are you using mysqli and concatenated query strings? Please strongly consider using parameterized queries - it is a small investment that will save you the horror of SQL injections. — Polynomial
– Polynomial, Commented Dec 5, 2011 at 14:05
More importantly, what's going on with the INSERT statement? That one looks wrong. Better check for errors: de3.php.net/manual/en/mysqli.error.php — Carsten
– Carsten, Commented Dec 5, 2011 at 14:08
I am sorry am little ignorant about this..can you please guide me? — Robbie Dc
– Robbie Dc, Commented Dec 5, 2011 at 14:09

Carlos Campderrós · Accepted Answer · 2011-12-05 14:06:49Z

2

You have an error in your insert query: you are providing just 4 values for 6 fields.

Do some error checking on the return value of mysqli_query($dbc, $query2);

answered Dec 5, 2011 at 14:06

Carlos Campderrós

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