Question

我写了一个函数＆＃34; attempt_login1（）＆＃34;登录它工作正常。

但当我包含函数＆＃34; login_count（）＆＃34;要计算登录总数（它在代码中注释掉了）它会给出错误

解析错误：语法错误，意外＆＃39; $连接＆＃39;第45行的F：\ XAMP1 \ htdocs \ sandbox \ test.php中的（T_VARIABLE）

我不确定是什么导致了这个错误。谢谢你的帮助。

$username = "user123";
$password = "12345";

attempt_login($username, $password);

    function attempt_login($username, $password){
        $user = find_admin_by_username($username);  //select user by username in database
        $user_set = mysqli_fetch_assoc($user);

            if($user_set){
                $user_password = $user_set["hashed_password"];
                $hashed_password = password_verify($password, $user_password);

                if($hashed_password == true){
            //      login_count($user_set["id"], $user_set["login_count"]);
                    return $user_set;
                }else{
                    return false;
                }
                }else{
                    return false;
                }
            }

    function login_count($user_id, $count){
        $global $connection;

        $count = $count++;
        $query2 = "UPDATE admins SET login_count = $count WHERE id = $user_id";
        $result2 = mysqli_query($connection, $query2);

    }
function find_admin_by_username($username){
    global $connection;

    $query  = "SELECT * ";
    $query .= "FROM admins ";
    $query .= "WHERE username = '{$username}' ";
    $query .= "LIMIT 1";
    $user_set = mysqli_query($connection, $query);
    confirm_query($user_set);

    if($user_set){
        return $user_set;
    }else{
        return NULL;
    }
}

Answer 1

你有拼写错误：

在$global $connection之前，

global $connection应该global没有美元。

Answer 2

使用以下代码为您的函数login_count：

function login_count($user_id, $count){
    global $connection;

    $count = $count++;
    $query2 = "UPDATE admins SET login_count = $count WHERE id = $user_id";
    $result2 = mysqli_query($connection, $query2);

}

希望它会对你有所帮助:)。

在现有函数中添加函数时出错

2 个答案: