Question

可能重复：
Get next/previous ISO week and year in PHP

我正在尝试编写一个脚本，该脚本将在一个表格中显示一周的日期，如果单击一个按钮，则会提前一周。我已经设法让它一直工作，直到它到达年底，然后日期都出错了。他就是我到目前为止......

 <?
     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>

有些值已被回复，所以我可以检查传递的值是否正确，我知道我可能已经采取了我不需要的额外步骤，但我对此很新，并希望制作当我开始工作时，代码更容易被删除。正如我所说的那样，添加按钮会有效，直到它迎来新的一年。

由于

好的，取得了一些进步，一直运行到2012年然后它再次运行到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>

Answer 1

在我看来，您将更好地根据unix时间戳值进行所有计算，然后根据输出的需要转换为字符串。这样你不必处理周数问题（即第0周），你不仅限于星期一是每周的第一天（这是date("W")中计算的基础），并且你不需要做一堆黑客来寻找边缘条件。

因此假设$_POST['last_week']符合您的d-m-Y格式：

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); 
}

Php.Advance每周日历一周

1 个答案: