if(isset($ _ GET []的问题)

Question

我有以下脚本的问题.
我建立了我的网站,以便链接看起来像这个website.com/contact
现在我想有一个子网页,如website.com/contact/inschrijven或website.com/contact/bouw_aanmelding
但是如果我有多个GETS出错了.1 GET一切顺利.我究竟做错了什么？

代码:

.htaccess中:

RewriteEngine on
RewriteBase /

RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME}\.php -f
RewriteRule ^(.*)$ $1.php

RewriteRule ^contact/([^/]*) contact.php?tab=$1

PHP代码(我已经尝试过了):

<?php    
   if(isset($_GET['tab']) == 'bouw_aanmelding') {
      $pagetitle = "Contacteer ons: Bouwer worden!";
   }elseif(isset($_GET['tab']) == 'inschrijven') {
      $pagetitle = "Contacteer ons: Inwoner worden!";
   }elseif(isset($_GET['tab']) == '') {
      $pagetitle = "Contacteer ons";
   }
?>

PHP代码(等等):

<?php    
   if(isset($_GET['tab']) == 'bouw_aanmelding') {
       $pagetitle = "Contacteer ons: Bouwer worden!";
   }
   if(isset($_GET['tab']) == 'inschrijven') {
      $pagetitle = "Contacteer ons: Inwoner worden!";
   }
   if(isset($_GET['tab']) == '') {
      $pagetitle = "Contacteer ons";
   }
?>

这两种方法都没有帮助.希望有人可以帮助我.谢谢.

Ps,对不起,我的英文不好,我是荷兰人.

Answer 1

isset返回一个boolean.你应该使用:

 if (isset($_GET['tab']) && $_GET['tab'] == 'inschrijven') {

 }

更好的是:

$pagetitle = 'default title';

if (isset($_GET['tab'])) {
  switch ($_GET['tab']) {
    case 'bouw_aanmelding':
       $pagetitle = "Contacteer ons: Bouwner worden!";
       break;
    case 'inschrijven':
       $pagetitle = "Contacteer ons: Inwoner worden!";
       break;
  }
}