Question

我有一個問題，將其轉換爲mysqli從mysql，我試圖什麼PHP文檔說，但我仍然無法得到它。任何幫助，將不勝感激。

return (mysql_result(mysql_query("SELECT COUNT(id) FROM users WHERE (username = '$username' OR email = '$username') AND password = '$password'"), 0) == 1) ? $user_id : false; 

我嘗試這樣做：

return (mysqli_data_seek(mysqli_query($con,"SELECT COUNT(id) FROM users WHERE (username = '$username' OR email = '$username') AND password = '$password'"), 0) == 1) ? $user_id : false; 

整體功能：

function login($username, $password){ 
$con=mysqli_connect("127.0.0.1","root","","frostbase"); 
    $user_id = user_id_from_username($username); 


    $username = sanitize($username); 
    $password = md5($password); 

    return (mysqli_data_seek(mysqli_query($con,"SELECT COUNT(id) FROM users WHERE (username = '$username' OR email = '$username') AND password = '$password'"), 0) == 1) ? $user_id : false; 
    mysqli_close($con); 
} 

Answer 1

你可以做到這一點的方式更好實際。

safeMysql是一個工具，可以幫助您將醜陋的舊的基於mysql的API代碼轉換爲新的醜陋的基於mysqli API的代碼，但使用方式更好的方法。

的一點是，你不能在應用程序代碼中使用原始API調用，無論是mysql，mysqli或PDO。但是將它們包裝在一個庫中，可以完成所有必需的操作，如數據綁定，抓取，錯誤處理，分析等等。

坦率地說，你只寫你的查詢 - 其餘的是由LIB完成：

$sql = "SELECT 1 FROM users WHERE (username = ?s OR email = ?s) AND password = ?s"; 
return (bool)$db->getOne($sql, $username, $username, $password); 

，或者在功能

function login($username, $password){ 
    global $con; // you have to connect ONCE per application 
    $sql = "SELECT 1 FROM users WHERE (username = ?s OR email = ?s) AND password = ?s"; 
    return (bool)$con->getOne($sql, $username, $username, $password); 
}