I'm working on weekly schedule report. I'm generating week on the basis of current date. Refer following code. <?php $date = mktime(0, 0, 0, date('m'), date('d'), date('Y')); $week = (int)date('W', $date); $year = date('Y'); for($day=1; $day<=7; $day++) { echo date('M dS, Y', strtotime($year."W".$week.$day)); } ?> This is working fine, but when I try to navigate between weeks, I'm facing issues. For week 9 I'm getting date 01-03-1970. It's not generating week dates for February. Also tried with static date 03-02-2012 but getting same random date. $week =9; //(int)date('W', $date); In another Issue echo date('N', strtotime('03-09-2012')); above satement returning 1 . I'm not convinced with the result here. Please let me know what should I do to over come with this or help me to find out way to get no of day of the week for given date.

Issue with date function in PHP

Answers (8)

Reputation: 533

Try this. It works fine for me:

<?php 
if(isset($_GET['diffweek'])){
  $diffweek=$_GET['diffweek'];
}else{
  $diffweek=0;}

$date = mktime(0, 0, 0, date('m'), date('d'), date('Y')); 
$week = date('W', $date+$diffweek*604800); 
$year = date('Y');

for($day=1; $day<=7; $day++)
  {
  echo date('M dS, Y', strtotime($year."W".$week.$day));
  echo '<hr>';
  }

echo '<a href="?diffweek='.($diffweek-4).'">4 weeks back</a><br>';
echo '<a href="?diffweek='.($diffweek-1).'">prew. week</a><br>';
echo '<a href="?diffweek=0">this week</a><br>';
echo '<a href="?diffweek='.($diffweek+1).'">next week</a><br>';
echo '<a href="?diffweek='.($diffweek+4).'">4 week later</a><br>';
?>

Upvotes: 1

Bryan Plasters

Reputation: 69

using the DateTime class (PHP 5 >= 5.2.0):

    $date = date_create();

    for($d=1; $d<=7; $d++){

        echo $date->format('M dS, Y')."<br />";         
        $date->modify('+1 day');

    }

Upvotes: 1

kdvy

Reputation: 404

Trying to keep the solution as close to your current code as possible.

For your 1970 issue:

It appears that strtotime requires zero-leading week numbers to work correctly for those weeks under 10, otherwise it outputs to 1970, ignoring the year (this seems like a bug to me?). Though, support for weeks and day of week isn't in the official documentation for strtotime.

So I would replace

    $week = (int)date('W', $date);

with:

    $week = str_pad(date('W', $date),2,'0',STR_PAD_LEFT);

Regarding the following code:

    echo date('N', strtotime('03-09-2012'));

This is returning 1 (Monday) as strtotime is interpreting 03-09-2012 in the UK format of d-m-Y as opposed to the US format of m-d-Y. To avoid confusion you should use the ISO date format of Y-m-d. See PHP manual on Date Formats for more information.

So if you wanted 9th March 2012 you would write:

    echo date('N', strtotime('2012-03-09'));

I hope this helps!

Upvotes: 1

Michael

Reputation: 12806

This might help you. It's a function to return the calendar dates of the given year:

<?php

  /**
   * Get all the dates in the given year's calendar
   *
   * @param NULL|integer $year
   * @param string $format
   * @return array
   */
  function getCalendar($year = NULL, $format = 'Y-m-d')
  {

    // If the year is not set then use the current
    if (!isset($year))
    {
      $year = date('Y');
    }

    // Get the first day of the year
    $start = $year . '-01-01';

    // Iterate over an incremental day increase
    for ($d = 0; ; ++$d)
    {

      // Get the date
      $date = strtotime($start . " + $d day");

      // Return the current list of dates if the current date is for next year
      if ($date >= strtotime($year + 1 . '-01-01'))
      {
        return $dates;
      }

      // Get the week number
      $w = ceil(($d + 1) / 7);

      // Add the date to the array
      $dates[isset($dates[$w][7]) ? ++$w : $w][date('N', $date)] = date($format, $date);

    }

    // Return the dates array
    return $dates;

  }

  // Get the calendar for 2012
  echo '<pre>' . print_r(getCalendar(2012, 'M dS, Y'), true) . '</pre>';

This will output:

As you can see, it's in the form $array[$week_number][$day_number] (where a $day_number of 1 is Monday and 7 is Sunday) and the $week_number will be up to 53 or 54 (depending on the year).

You can then get a specific week's dates using:

$calendar = getCalendar();

$calendar[9];

// Or if you have PHP 5.4.0

getCalendar()[9];

I've intentionally not included days from the previous or next years at the beginning and the end.

Upvotes: 1

David Stockton

Reputation: 2251

This should work for you. The 'today' in the first line can be changed to whatever date you want to test it for.

<?php 
$today = strtotime('today');
$startWeek = strtotime(date('o-\\WW', $today));
$date = mktime(0, 0, 0, date('m', $startWeek), date('d', $startWeek), date('Y', $startWeek)); 
$week = (int)date('W', $date); 
$year = date('Y', $date);

for($day=0; $day < 7; $day++)
{
    echo date('M dS, Y', $today + ($day * 24 * 60 * 60)), "\n";
}
?>

Hope this helps.

Upvotes: 0

Alexei Tenitski

Reputation: 9360

Following code

$ts = mktime(0, 0, 0, 4, 5, 2012);

$offset = date('N', $ts);

for ($i = 1 - $offset, $j = 0; $j++ < 7; $i++) {
    echo date('r', $ts + $i * 3600 * 24) . "\n";
}

outputs days of the week the specified day belongs to:

Apr 02nd, 2012
Apr 03rd, 2012
Apr 04th, 2012
Apr 05th, 2012
Apr 06th, 2012
Apr 07th, 2012
Apr 08th, 2012

I hope this is what you were after.

As for issue with strtotime check if the date returned is really the date your were after by doing

echo date('r', strtotime('03-09-2012'));

It is safer to pass date to strtotime as YYYY-MM-DD.

Upvotes: 1

tim

Reputation: 2724

I think weeks in strtotime is supposed to be formatted like yeer-week-weekday i.e. 2012-W14-3.

for ($day=1; $day<=7; $day++) {
    echo date('M dS, Y', strtotime(date('Y') ."-W". date('W') .'-'. $day));
}

Upvotes: 0

Cyril N.

Reputation: 39859

I don't know which version of PHP you are running, but you should use the DateTime object, with DateInterval

It's way better and more easier to use.

Upvotes: 2

Issue with date function in PHP

Answers (8)

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