STEAMID OPEN ID錯誤

Question

我正在爲我的網站創建一個STEAM OPENID，但有些錯誤提示。 我不知道什麼是錯誤..誰能幫我...

錯誤：

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in D:\xampp\htdocs\PTgamersCo\steamid.php on line 69 

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in D:\xampp\htdocs\PTgamersCo\steamid.php on line 73 

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in D:\xampp\htdocs\PTgamersCo\steamid.php on line 76 

這一個是我的DB連接

$username = "root"; 
$password = ""; 
$host = "localhost"; 
$db = "steam_users"; 

$sql_steam = mysqli_connect($host, $username, $password, $db); 

if (!$sql_steam) { 
    echo "Error: Unable to connect to MySQL." . PHP_EOL; 
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL; 
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL; 
    exit; 
} 
mysqli_close($sql_steam); 

此代碼添加到DB。

foreach ($json_decoded->response->players as $player) 
{ 
    $sql_fetch_id = "SELECT * FROM users_steam WHERE steamid = $player->steamid"; 
    $query_id = mysqli_query($db, $sql_fetch_id); 

    //$_SESSION['avatar'] = $player->avatar; 

    if(mysqli_num_rows($query_id) == 0) { 

     $sql_steam = "INSERT INTO users_steam (name, steamid, avatar) VALUES ('$player->personaname', '$player->steamid', '$player->avatar')"; 
     mysqli_query($db, $sql_steam); 
    } 
} 

Answer 1

所有mysqli_*功能需要的第一個參數被變量您mysqli_connect($host, $username, $password, $db);調用返回。

還需要平倉離場，以便能夠使用它的連接，所以在DBConnection的腳本刪除mysqli_close($db);

所以修改你的DbConnection這

$username = "root"; 
$password = ""; 
$host = "localhost"; 
$dbname = "steam_users"; 

$db = mysqli_connect($host, $username, $password, $dbname); 

if (!$db) { 
    echo "Error: Unable to connect to MySQL." . PHP_EOL; 
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL; 
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL; 
    exit; 
} 
// Dont close the connection, or you wont be able to use it later 
//mysqli_close($db); 

你其他的代碼將有在$db可變