Reputation: 389
How can I calculate the day of month in PHP with giving month, year, day of week and number of week.
Like, if I have September 2013 and day of week is Friday and number of week is 2, I should get 6. (9/6/2013 is Friday on the 2nd week.)
Upvotes: 4
Views: 514
Reputation: 11302
This way its a little longer and obvious but it works.
/* INPUT */
$month = "September";
$year = "2013";
$dayWeek= "Friday";
$week = 2;
$start = strtotime("{$year}/{$month}/1"); //get first day of that month
$result = false;
while(true) { //loop all days of month to find expected day
if(date("w", $start) == $week && date("l", $start) == $dayWeek) {
$result = date("d", $start);
break;
}
$start += 60 * 60 * 24;
}
var_dump($result); // string(2) "06"
Upvotes: 0
Reputation: 72981
One way to achieve this is using relative formats for strtotime()
.
Unfortunately, it's not as straightforward as:
strtotime('Friday of second week of September 2013');
In order for the weeks to work as you mentioned, you need to call strtotime()
again with a relative timestamp.
$first_of_month_timestamp = strtotime('first day of September 2013');
$second_week_friday = strtotime('+1 week, Friday', $first_of_month_timestamp);
echo date('Y-m-d', $second_week_friday); // 2013-09-13
Note: Since the first day of the month starts on week one, I've decremented the week accordingly.
Upvotes: 4
Reputation: 173562
I was going to suggest to just use strtotime()
in this fashion:
$ts = strtotime('2nd friday of september 2013');
echo date('Y-m-d', $ts), PHP_EOL;
// outputs: 2013-09-13
It seems that this is not how you want the calendar to behave? But it is following a (proper) standard :)
Upvotes: 3