php - 将 ID 放在返回函数之前

Question

在下面的代码中，我将功能从 MYSQL 升级到 MYSQLI，我在以前的帖子中得到了一些帮助，他们说应该做什么.. 但现在问题来了怎么做..

事情是在id之前设置，否则无论我登录哪个帐户return，我仍然会收到用户。ID = 1

function user_id_from_username($username) {
    $username = sanitize($username);
    global $db_connect;

    $result = $db_connect->query("SELECT(id) FROM members WHERE username = '$username'");
    if (false === $result) {
        return false;
    }
    return ($result->num_rows == 1) ? id : false;

我尝试按照旧MYSQL代码中的方式进行操作，但随后出现错误。我的旧代码：

    function user_id_from_username($username){
    $username = sanitize ($username);
    return mysql_result(mysql_query("SELECT(id) FROM members WHERE username = '$username'"), 0, 'id');
}

如果有任何帮助，我将不胜感激！

Answer 1

你试过了吗fetch_array？mysqli_fetch_array

$row = $result->fetch_array(MYSQLI_ASSOC);
// or
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);

// and returning
return $row['id'] ? $row['id'] : false;

Answer 2

我没有尝试过，但这应该可以正常工作吗？我还将 $db_connect 作为参数，因为它更安全，但取决于您。

function user_id_from_username($username) {
    $username = sanitize($username);
    global $db_connect;
    $result = $db_connect->query("SELECT id FROM members WHERE username = '".$db_connect->real_escape_string($username)."'");
    if (!$result) {
        return false;
    }
    if($result->num_rows == 1){
        $row = $result->fetch_assoc();
        $output = $row['id'];
    }
    else{
        $output = "Username does not exist in table or is a duplicate.";
    }
    return $output;
}

Answer 3

return ($result->num_rows == 1) ? id : false;

应该

if(mysqli_num_rows($result) ===1 ){
    $row = mysqli_fetch_assoc($result);
    return intval($row['id']);
}
return false;