PHP / MySQL - display list of dates for all Mondays in year

Question

I need to populate a drop-down list on a form with dates for all Mondays in the next year?

What is the best way to do this? With a MySQL table populated with dates or is there a smarter method of generating these dates without the need for MySQL?

Desired output would be something like:

enter image description here

The value of the selected option would just be the date commencing, so for the first option in the image, 11 Mar 13 would be the value.

Thanks for the help and apologies if this is a silly question.

Whilst people are help to help it's polite to have attempted something yourself first. What solutions to your problem have you tried and what problems did you have. This also stops people from repeating the same mistakes you might have made, or attempts you discarded, when answering your question. — Ben, Mar 12 '13 at 11:16
Possible duplicate of http://stackoverflow.com/questions/2045736/getting-all-dates-for-mondays-and-tuesdays-for-the-next-year ? — Jacob Tomlinson, Mar 12 '13 at 11:16
the best way would be to do it in php and not use a database. I don't know what would make you think of putting the dates into a database first. — Popnoodles, Mar 12 '13 at 11:17

Mark Baker · Accepted Answer · 2013-03-12T12:19:18.310

Edit: Modified to support passing in a start date rather than simply a year

function getDays($year, $startMonth=1, $startDay=1, $dayOfWeek='monday') {
    $start = new DateTime(
        sprintf('%04d-%02d-%02d', $year, $startMonth, $startDay)
    );
    $start->modify($dayOfWeek);
    $end   = new DateTime(
        sprintf('%04d-12-31', $year)
    );
    $end->modify( '+1 day' );
    $interval = new DateInterval('P1W');
    $period   = new DatePeriod($start, $interval, $end);

    foreach ($period as $dt) {
        echo $dt->format("d/m/Y") . '<br />';
    }
}

getDays(2013, 3, 12, 'monday');

Formatting of the output as a dropdown should be a straightforward exercise

score 1 · Answer 2 · answered Mar 12 '13 at 11:19

1

You could simply use PHP and then the DateTime::add method. Then you can take the first monday of the year and use the method to add 7 days 52 times. That will give you all mondays. If you add 6 days you get the till date.

answered Mar 12 '13 at 11:19

Hugo Delsing

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Jacob Tomlinson · Answer 3 · 2013-03-12T11:37:10.780

This code will output a list of the dates you want

for($i = 0; $i <= 365; $i ++){
  $startdate = strtotime("today + $i day");
  $enddate = strtotime("today + " . ($i + 6) . " day");
  if(date('D', $startdate) == 'Mon'){
      echo date('d M y', $startdate) . " to " . date('d M y', $enddate) . "\n";
  }
}

To have this display as a dropdown list you would need to add the HTML to it like this

<select id="date" name="date"> 
  <?
    for($i = 0; $i <= 365; $i ++){
      $startdate = strtotime("today + $i day");
      $enddate = strtotime("today + " . ($i + 6) . " day");
      if(date('D', $startdate) == 'Mon'){
          echo '<option value="' . date('d-m-Y', $startdate) . '">' .date('d M y', $startdate) . " to " . date('d M y', $enddate) . "</option>";
      }
    } 
  ?> 
</select>

I have given the value of the select option the date of the Monday in dd-mm-yyyy format so when dealing with the form once submitted you will have the Monday which was selected. So if you were to select the row 27 May 13 to 02 Jun 13 and submit the form the php file on the other end of the form would have the variable $_POST['date'] and it's value would be 27-05-2013.

PHP / MySQL - display list of dates for all Mondays in year

3 Answers3