[RESOLVED] week question

[scialom]

Hello,

I have these data:
$year = 2007;
$day = "Mon";
$week = 42;

is it possible to get the exact date of Monday on week 42 on year 2007?

Thanks,
Assaf

[bogu]

This should work now:

$year = 2007;
$week = 42;
$day = 1; /* this is the value of Mon in the week day */

/* timestamp of the first day of the year*/
$timestamp_year = mktime(0,0,0,1,1,$year);

/* the timestamp of the 42 week in 2007 */
$timestamp_nr_week = strtotime("+{$week} week",$timestamp_year);
/* determine the day of the week*/
$day_of_the_week = (int) date("w",$timestamp_nr_week);

if ($day_of_the_week>$day) {
	echo date("D, d M Y",strtotime("-".($day_of_the_week-$day)." day",$timestamp_nr_week));
} elseif ($day_of_the_week<$day) {
	echo date("D, d M Y",strtotime("+".($day-$day_of_the_week)." day",$timestamp_nr_week));
} else {
	echo date("D, d M Y",$timestamp_nr_week);
}

[scialom]

Thanks, I took your code and changed it a bit so:
$year = 2007;
$week = 42;
$day = "Tue";

$week = $week -1;
$timestamp_year = mktime(0,0,0,1,1,$year);
echo date("D, d M Y",strtotime("+$week week $day",$timestamp_year));

it works 🙂

[bogu]

I would have never thought about this strtotime format:

strtotime("+42 week Mon");

I hope is working correctly...

This is from GNU manual:

GNU manual wrote:

The explicit mention of a day of the week will forward the date (only if necessary) to reach that day of the week in the future.

so it seems to be OK ...

[Installer]

You shouldn't use the first of January, as that might be in the last week of the previous year. Use the fourth instead. That will always be in week number 1.