1

可能重复：
在 PHP 中获取下一个/上一个 ISO 周和年

我正在尝试编写一个脚本，该脚本将在表格中显示一周中的几天，如果单击按钮，它将提前一周。我已经设法让它工作到年底，然后日期都出错了。他是我目前所拥有的...

 <?
     if(isset($_POST['add_week'])){
     $week = date('d-m-Y', strtotime($_POST['last_week']));
     $new_week =  strtotime ( '+1 week' , strtotime ( $week ) ) ;
     $new_week = date('d-m-Y', $new_week);

     $week_number = date("W", strtotime( $new_week));
     $year = date("Y", strtotime( $new_week));
  }else{

         $week_number = date("W");
         $year = date("Y");
  }

  if($week_number < 10){
      $week_number = "0".$week_number;
   }

  $week_start = date('d-m-Y', strtotime($year."W".$week_number,0));
 echo $week.' '.$new_week.' '.$week_number;
?>

<table name="week">
<tr>
        <?
            for($day=1; $day<=7; $day++)
  {
echo '<td>';
    echo date('d-m-Y', strtotime($year."W".$week_number.$day))." | \n";
echo '</td>';
  }
?>
</tr>
<tr>
 <form name="move_weeks" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
 <input type="hidden" name="last_week" value="<? echo $week_start; ?>" />
  <td colspan="7"><input type="submit" name="back_week" value="back_week" />
 <input ype="submit" name="add_week" value="add_week" />
</td>
</form>
</tr>
</table>

一些值已被回显，因此我可以检查正在传递的值是否正确，我知道我可能已经采取了我不需要的额外步骤，但我对此相当陌生，并且想让代码更容易在我让它工作的时候跟随它。正如我所说，添加按钮在新年到来之前一直有效。

谢谢

好的，取得了一些进步，直到 2012 年都可以正常工作，然后它再次运行到 2012 年，而不是从 2013 年开始

 <?
if(isset($_POST['add_week'])){
    $week = date('d-m-Y', strtotime($_POST['last_week']));
    $new_week =  strtotime ( '+1 week' , strtotime ( $week ) ) ;
    $new_week = date('d-m-Y', $new_week);


    $week_number = date("W", strtotime( $new_week));
    $year = date("Y", strtotime( $new_week));
}else if(isset($_POST['back_week'])){
    $week = date('d-m-Y', strtotime($_POST['last_week']));
    $new_week =  strtotime ( '-1 week' , strtotime ( $week ) ) ;
    $new_week = date('d-m-Y', $new_week);


    $week_number = date("W", strtotime( $new_week));
    $year = date("Y", strtotime( $new_week));
}else{

$week_number = date("W");
$year = date("Y");
}
/*if($week_number < 10){
   $week_number = "0".$week_number;
}*/
$week_start = date('d-m-Y', strtotime($year."W".$week_number,0));
echo $week.' '.$new_week.' '.$week_number;
?>

<table name="week">
    <tr>
<?
for($day=1; $day<=7; $day++)
{
    echo '<td>';
    echo date('d-m-Y', strtotime($year."W".$week_number.$day))." | \n";
    echo '</td>';
}
?>
</tr>
<tr>
<form name="move_weeks" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="last_week" value="<? echo $week_start; ?>" />
<td colspan="7"><input type="submit" name="back_week" value="back_week" /><input type="submit" name="add_week" value="add_week" />
</td>
</form>
</tr>
</table>
4

1 回答 1

3

在我看来，您将更好地根据 unix 时间戳值进行所有计算，然后仅在输出需要时转换为字符串。这样您就不必处理周数问题（即第 0 周），您不限于将星期一作为每周的第一天（作为计算的基础date("W")），并且您不会有做一堆黑客来寻找边缘条件。

所以假设$_POST['last_week']你的 dmY 格式是这样的：

if(isset($_POST['add_week'])){
     $last_week_ts = strtotime($_POST['last_week']);
     $display_week_ts = $last_week_ts + (3600 * 24 * 7);
} else if (isset($_POST['back_week'])) {
     $last_week_ts = strtotime($_POST['last_week']);
     $display_week_ts = $last_week_ts - (3600 * 24 * 7);
} else {
    $display_week_ts = floor(time() / (3600 * 24)) * 3600 * 24;
}

$week_start = date('d-m-Y', $display_week_ts);

对于您循环显示一周的部分，您可以使用以下内容：

for ($i = 0; $i < 7; $i++) {
    $current_day_ts = $display_week_ts + ($i * 3600 *24);
    echo date('d-m-Y', $current_day_ts); 
}
于 2012-09-04T15:27:55.847 回答