Date function problem in PHP

Question

Date function problem in PHP

I am working on a weekly schedule.

I am creating a week based on the current date. Refer to the next 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 works fine, but when I try to navigate between weeks I run into problems. For the ninth week, I get the date 01-03-1970. It does not generate weekly dates for February.

Also tried with static date 03-02-2012 but got the same random date.

$week =9; //(int)date('W', $date);

In another issue

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

over return satement 1

. I'm not sure about this.

Please let me know what I should do to finish this or help me find out how to get the day of the week for a given date.

+3

date php

pinaldesai Mar 30 12 at 12:25

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8 answers

I don't know which version of PHP you are using, but you have to use DateTime object , DateInterval

It's better and easier to use.

+3

Cyril N. Mar 30 12 at 12:29

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This might help you. This is a function that returns the calendar dates for a given year:

<?php

  /**
   * Get all the dates in the given year 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, this is in the form $array[$week_number][$day_number]

(where a $day_number

of 1 is Monday and 7 is Sunday) and $week_number

will be up to 53 or 54 (depending on the year).

Then you can get specific weekends using:

$calendar = getCalendar();

$calendar[9];

// Or if you have PHP 5.4.0

getCalendar()[9];

I have deliberately left out days from the previous or next year at the beginning and end.

+1

MichaelRushton 04 Apr '12 at 6:21

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The 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";
}

displays the days of the week of the specified day:

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

Hope this is what you were.

Regarding the strtotime issue, check if the date you were on actually followed

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

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

+1

Alexei Tenitski 04 Apr 12 at 6:35 am

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using 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');

    }

+1

Bryan plasters 04 Apr 12 at 15:17

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Try it. This works great 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>';
?>

+1

Juraj.Lorinc 04 Apr 12 at 17:25

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I think the weeks in strtotime should be formatted as the day of the week of the week, i.e. 2012-W14-3.

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

0

tim 03 Apr 12 at 18:04

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This should work for you. "Today" on the first line can be changed to any date you want to check.

<?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 it helps.

0

David Stockton 04 Apr 12 at 7:12

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kdvy · Accepted Answer · 2012-04-04T11:53:36+0000

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

For your 1970 problem:

It looks like strtotime requires zero week numbers to work correctly for those weeks before 10, otherwise it is output in 1970, ignoring the year (does this seem like a bug to me?). Although support for weeks and days of the week is not included in the official documentation for strtotime.

So, I would replace

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

from:

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

Regarding the following code:

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

This returns 1 (Monday) as strtotime interprets 03-09-2012 in UK dmY format as opposed to US mdY format. To avoid confusion, you should use the ISO Ymd date format . See the Date Formatting Guide for details .

So, if you want March 9, 2012, you have to write:

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

Hope this helps!

Date function problem in PHP

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