PHP - Get date from specified weekday in a given week - php

Question

I wan't to get a PHP date() from specified weekday in a given week.
For example:
Weekday: Thursday - in week:8 - year:13 (means 2013).
I would like to return a date from these specified values. The phpdate will in this case return: "21 Feb 2013", which is a Thursday in week 8 of 2013.
Please fill in this php-method:
function getDateWithSpecifiedValues($weekDayStr,$week,$year) {
//return date();
}
Where the example:
getDateWithSpecifiedValues("Tuesday",8,13);
will return a phpdate of "19 Feb 2013"

Answer 1

First, you have to define what you mean by "week of the year". There are several different definitions. Does the first week start on Jan 1? On the first Sunday or Monday? Is it number 1 or 0?
There is a standard definition, codified in ISO 8601, which says that weeks run from Monday through Sunday, the first one of the year is the one with at least 4 days of the new year in it, and that week is number 1. Your example expected output is consistent with that definition.
So you can convert the values by putting them into a string and passing that string to strptime, along with a custom format string telling it what the fields in the string are. For example, the the week number itself should be indicated in the format string by %V.
For the weekday, the format depends on how you want to provide it as input to your function. If you have the full name (e.g. "Thursday"), that's %A. If you have the abbreviated name (e.g. "Thu"), that's %a. If you have a number (e.g. 4), that's either %w (if Sundays are 0) or %u (if Sundays are 7). (If you're not sure, you can always just use %w and pass the number % 7.)
Now, the year should be %G (full year) or %g (just the last two digits). It's different from the normal calendar year fields (%Y for 2014 and %y for 13) because, for example, week 1 of 2014 actually started on December 30, 2013, which obviously has a '%Y' of 2013 where we want 2014. However, the G fields don't work properly with strptime, so you'll have to use the Y's.
For example:
$date_array = strptime("$weekDayStr $week $year", '%A %V %y');
That's a good start, but the return value of strptime is an array:
array('tm_sec' => seconds, 'tm_min' => minutes, tm_hour => hour,
tm_mday => day of month, tm_mon => month number (0..11), tm_year => year - 1900)
And that array is not the input expected by any of the other common date or time functions, as far as I can tell. You have to pull the values out yourself and modify them in some cases and pass the result to something to get what you want. For instance:
$time_t = mktime($date_array['tm_hour'], $date_array['tm_min'],
$date_array['tm_sec'], $date_array['tm_mon']+1,
$date_array['tm_mday'], $date_array['tm_year']+1900);
And then you can return that in whatever form you need. Here I'm returning it as a string:
function getDateWithSpecifiedValues($weekDayStr,$week,$year) {
$date_array = strptime("$weekDayStr $week $year", '%A %V %y');
$time_t = mktime($date_array['tm_hour'], $date_array['tm_min'],
$date_array['tm_sec'], $date_array['tm_mon']+1,
$date_array['tm_mday'], $date_array['tm_year']+1900);
return strftime('%d %b %Y', $time_t);
}
For example,
php > print(getDateWithSpecifiedValues('Thursday',8,13)."\n");
21 Feb 2013

Answer 2

Try this function:
function getDateWithSpecifiedValues($weekDayStr, $week, $year) {
$dt = DateTime::createFromFormat('y, l', "$year, $weekDayStr");
return $dt->setISODate($dt->format('o'), $week, $dt->format('N'))->format('j M Y');
}
demo

Answer 3

I'm fairly certain neither 21st of January or the 19th of January is in week 8.
You can however use strptime to parse custom formats:
var_dump(strptime("Thursday 8 13", "%A %V %y"));
array(9) {
["tm_sec"]=>
int(0)
["tm_min"]=>
int(0)
["tm_hour"]=>
int(0)
["tm_mday"]=>
int(21)
["tm_mon"]=>
int(1)
["tm_year"]=>
int(113)
["tm_wday"]=>
int(4)
["tm_yday"]=>
int(58)
["unparsed"]=>
string(0) ""
}
See the documentation for strptime for the meaning of each value in the returned array.