简单的PHP语句，我无法让它工作！

如何获得$ schedule = true`和$ schedule2 = true的工作？简单的PHP语句，我无法让它工作！

我知道这很简单，我可以忽略一些简单的事情！

这里是全码：（！它看起来在PHP日期，并告诉它什么时候在我们的新闻网站到期）

我基本上要安排和了schedule2工作

$where = array(); 
$where = run_filters('also-allow', $where); 

if ($allow_full_story or $allow_add_comment) { 
    $post = 'full'; 

    if ($title){ 
     $where[] = "url = $title"; 
    } elseif ($time){ 
     $where[] = "date = $time"; 
    } elseif ($id){ 
     $where[] = "id = $id"; 
    } 
} else { 
    $post = 'short'; 

    if (!$is_logged_in or $is_logged_in and $member['level'] == 4) { 
     $where[] = 'hidden = 0'; 
     $where[] = 'and'; 
    } 

    if ($user or $author) { 
     $where[] = 'author = '.($author ? $author : $user); 
     $where[] = 'and'; 
    } 

    if ($year and !$month) { 
     $where[] = 'date > '[email protected](0, 0, 0, 1, 1, $year); 
     $where[] = 'and'; 
     $where[] = 'date < '[email protected](23, 59, 59, ($year == date("Y") ? date("n") : 12), ($year == date("Y") ? date("d") : 31), $year); 
    } elseif ($year and $month and !$day) { 
     $where[] = 'date > '[email protected](0, 0, 0, $month, 1, $year); 
     $where[] = 'and'; 
     $where[] = 'date < '[email protected](23, 59, 59, $month, (($year == date("Y") and $month >= date("n")) ? date("d") : 31), $year); 
    } elseif ($year and $month and $day) { 
     if($year == date("Y") and $month >= date("n") and $day >= date("d")) { 
      $where[] = 'hidden = 2'; 
     } 
     else { 
      $where[] = 'date > '[email protected](0, 0, 0, $month, $day, $year); 
      $where[] = 'and'; 
      $where[] = 'date < '[email protected](23, 59, 59, $month, $day, $year); 
     } 
    } 
    else { 
    if ($schedule) { 
     $where[] = 'date > '.(time() + $config_date_adjust * 60 - 432000); 
    } 
    else { 
     $where[] = 'date < '.(time() + $config_date_adjust * 60); 
    } 
    $schedule = false; 
} 
else { 
    if ($schedule2) { 
     $where[] = 'date > '.(time() + $config_date_adjust * 60 - 86400); 
    } 
    else { 
     $where[] = 'date < '.(time() + $config_date_adjust * 60); 
    } 
    $schedule2 = false; 
}

来源

2010-04-24 eberswine

什么？你能解释一下你期望这段代码做什么，它实际上在做什么？ – 2010-04-24 00:53:28

@Tyler：看起来他正在创建一个sql语句 – SeanJA 2010-04-24 00:56:27

@SeanJA是的，但我的意思是关于他想用'$ schedule'来做什么，以及他无法完成的功能。 – 2010-04-24 01:00:53