将登录代码更新为MySQLi

Question

我正在尝试更新这部分旧的MySQL代码，以便与MySQLi一起运行。我知道MySQL_return function被删除了，我尝试过一些东西，但不太确定如何将这段代码转换为使用MySQLi运行？如有任何意见或帮助，将不胜感激。

function user_id_from_username($email_address, $conn) {
    return mysql_return(mysqli_query($conn, "SELECT `userid` FROM `users` WHERE `email_address` = '$email_address'"), 0, 'userid');
}


function login($email_address, $password, $conn) {
    $userid = user_id_from_username($email_address);

    $password = md5($password);

    return  (mysql_return(mysqli_query($conn, "SELECT COUNT(`userid`) FROM `users` WHERE `email_address` = '$email_address' AND `password` = '$password'"), 0) ==1) ? $userid : false;

}

我曾尝试过这样的编辑：

function user_id_from_username($email_address, $conn) {
    return mysqli_fetch_assoc(mysqli_query($conn, "SELECT `userid` FROM `users` WHERE `email_address` = '$email_address'"), 0, 'userid');
}


function login($email_address, $password, $conn) {
    $userid = user_id_from_username($email_address);

    $password = md5($password);

    return  (mysqli_fetch_assoc(mysqli_query($conn, "SELECT COUNT(`userid`) FROM `users` WHERE `email_address` = '$email_address' AND `password` = '$password'"), 0) ==1) ? $userid : false;

}

但没有运气。有什么建议或例子吗？

Answer 1

如果查询成功执行，$mysqli->query();方法将返回一个$mysqli_result，否则将返回FALSE。然后，您可以调用fetch_assoc();从查询中获取第一行。结果数组将以关联数组的形式出现。下面是一些您应该阅读的文档：http://php.net/manual/en/mysqli.query.php和http://php.net/manual/en/mysqli-result.fetch-assoc.php

function getIdFromEmail($email) {
  $emailSanitized=mysqli_real_escape_string($email);
  $mysqli=new mysqli("host","user","password","dbName");
  $res=$mysqli->query("SELECT `userid` FROM `users` WHERE `email_address`='$emailSanitized'");
  $retArray=$res->fetch_assoc();
  return $retArray['userId'];      
}