登錄在PHP帳戶過期顯示消息

-4

Username| Email | Password | Activation | Expirydate| Createddate 
---------------------------------------------------------------------- 
niranjan|[email protected]|12345|NULL| 2015-05-23| 2015-05-08

表字段當用戶登錄到登錄頁面我的查詢如下IWAS書面

if (empty($error)){ 

$date=date('Y-m-d'); 

$query_check_credentials = "SELECT * FROM members WHERE (Email='$Email' AND password='$Password') AND (Activation IS NULL) AND (Expirydate <= '$date')"; 

$result_check_credentials = mysqli_query($dbc, $query_check_credentials); 

if (@mysqli_num_rows($result_check_credentials) == 1) 
{ 

header("Location: course.php"); 

$_SESSION = mysqli_fetch_array($result_check_credentials, MYSQLI_ASSOC); 

}else 
     { 

$msg_error= 'Either Your Account is inactive or Email address /Password is Incorrect'; 

} 

    } 
else { 

echo '<div class="errormsgbox"> <ol>'; 
     foreach ($error as $key => $values) { 

      echo ' <li>'.$values.'</li>'; 

     } 
     echo '</ol></div>'; 

    }}

在這裏，我想，當用戶登錄登錄頁面首先檢查錯誤，然後在表格中查看成員，Expirydate < =當前日期我想顯示錯誤信息，請幫我解決這個問題。

來源

2015-05-08 Niranjan chowdam

我想你忘了粘貼你的代碼/查詢了。 – chris85

請立即檢查 –

現在發生了什麼事？ – chris85

你可能意味着有這樣的：

"SELECT * FROM members WHERE (Email='".$Email."' AND password='".$Password."') AND (Activation IS NULL) AND (Expirydate <= '".$date."')";

假設$電子郵件，$密碼和$日期較早的代碼中定義。此外，你應該確保這些變量正確處理SQL注入。

來源

2015-05-11 06:05:50 Labbed

登錄在PHP帳戶過期顯示消息

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