php - 函数返回空数据

Question

我使用这个函数来检查记录是否存在于我的数据库中并采取相应的行动。第二部分数据不匹配时出现错误:本应插入一条新记录，但没有插入新数据，返回空数据。我哪里错了？

 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;
}



}

Answer 1

您的插入查询有一个错误:您只为 6 个字段提供了 4 个值。

对 mysqli_query($dbc, $query2); 的返回值做一些错误检查