Question

Possible Duplicate:
Get next/previous ISO week and year in PHP

I am trying to write a script that will display the days of a week in a table and that will advance a week if a button is clicked. I have managed to get it working up until the point where it reaches the end of the year and then the dates all go wrong. He is what I have so far…

 <?
     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>

Some of the values have been echo’d so I can check the values that are being passed are correct and I know I have probably taken extra steps I didn’t need to but I am fairly new to this and wanted to make the code easier to folllow whilst I get it working. As I said, the add button works a treat until it hits new year.

Thanks

Ok, made some advancement, works fine until it gets to 2012 then it just runs through 2012 again rather than starting 2013

 <?
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

In my opinion you are going to be way better served to make all your calculation based on a unix timestamp value and then convert to string only as needed for output. That way you don’t have to deal with week number problems (i.e. week 0), you are not limited to having Monday be the first day of each week (as as is the basis of calculation in date("W")), and you won’t have to make a bunch of hacks to look for edge conditions.

So assuming that $_POST['last_week'] is in your d-m-Y format something like this:

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);

For the part where you are looping through the week to display you can use something like this:

for ($i = 0; $i < 7; $i++) {
    $current_day_ts = $display_week_ts + ($i * 3600 *24);
    echo date('d-m-Y', $current_day_ts); 
}