使用if語句進行簡單登錄表格PHP

Question

我正在嘗試基於這個序列的會話編寫一個簡單的登錄系統。 在第一個位置登錄用戶正在工作，但第二個用戶沒有登錄。我找不到錯誤在哪裏。 這裏是login.php的代碼形式。

include './user.php'; 

session_start(); 

    if (!empty($_POST['form-username'])) { 

    if($_POST['form-username']==$user && $_POST['form-password']==$pass){ 

     $_SESSION['login'] = $_POST['form-username']; 
     $_SESSION['password'] = $_POST['form-password']; 
     header("Location: ./dashboard.php"); 

    }else if($_POST['form-username']==$user2 && $_POST['form-password']==$pass2){ 

     $_SESSION['login'] = $_POST['form-username']; 
     $_SESSION['password'] = $_POST['form-password']; 
     header("Location: ./dashboard.php"); 
    } 

    else { 

     header("Location: ./index.php?err"); 
     //exit; 
    } 

    } 
    if(!empty($_SESSION['login'])): 
     if($_SESSION['login']==$user AND $_SESSION['password']==$pass || $_SESSION['login']==$user2 AND $_SESSION['password']==$pass2){ 

      header("Location: ./dashboard.php"); 

     } 
    endif; 

這裏是user.php的文件的代碼：

<?php 

    $user = 'panel1'; 
    $pass = 'panel123'; 

    $user2 ='panel2'; 
    $pass2 = 'panel091'; 

Answer 1

我想你應該試着改變過去的if聲明爲：

if (($_SESSION['login'] == $user AND $_SESSION['password'] == $pass) OR ($_SESSION['login'] == $user2 AND $_SESSION['password'] == $pass2)) { 
    // ... 
} 

注意包裝每個條件周圍的paranthesis。