Question

  define('DB_HOST', 'localhost');
  define('DB_USER', '******');
  define('DB_PASSWORD', '************');
  define('DB_NAME', '***********');

 $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);


 class User {

 function checkUser($uid, $oauth_provider,$username,$email,$twitter_otoken,$twitter_otoken_secret) 
{
    $query = mysqli_query($dbc,"SELECT * FROM si_table WHERE oauth_uid = '$uid' and oauth_provider = '$oauth_provider'");
    $result = mysqli_fetch_array($query);
    if (!empty($result)) {
        # User is already present
    } else {
        #user not present. Insert a new Record
        $query = mysqli_query($dbc,"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')") ;
        $query = mysqli_query($dbc,"SELECT * FROM si_table WHERE oauth_uid = '$uid' and oauth_provider = '$oauth_provider'");
        $result = mysqli_fetch_array($query);
        return $result;
    }
    return $result;
}



}

我收到此错误！我不知道为什么？ mysqli_query（）期望参数1为mysqli，在行14,15,20,21和22中给出null！

Answer 1

将global $dbc;添加到checkUser的开头。变量未在函数范围中定义。

Mysqli_query（）期望参数1是mysqli，null给定？混乱？

1 个答案: