complicated date

[Anon]

i'm not one to post algorithm requests on here, but i figure there has got to be an easy way to do this...

i am making a status report for that deals with weeks that end on wednesdays.

the first week of the year before the first wednesday is week 1. (if the first day of the year is a wednesday it is also week 1)
up to the next wednesday is week 2 and so on...

for any date i need to retrieve then current week with the rules above, and also the maximum week. the maximum week will be the number of wednesdays in that year (i'm SURE you can do that somehow with date())

could anyone help? thanks!

[Anon]

if its any help,

strftime("%U", time());

will return the current week number, starting on sundays.

my probably would be if jan 1st was a tuesday, i would like that to return as the max week of the previous year.

i just need a function that takes 0-6 (weekday) and a timestamp as a parameter, and returns the current week starting at the weekday specified, or 0 if the timestamp is before that day at the beginning of the year.

[Anon]

figured it out..... if anyone cares, here is the code:

/****** CUSTOMIZE ****/
$week_ends_on = 4; // 1 is sunday
/***************************/

$time = time();
$current_week = strftime("%U", $time);
$max_week = strftime("%U", mktime(0,0,0,12,31,date("Y")));
$current_day = strftime("%u", $time);
$current_day++;
$current_day %= 7;
for($i=1;$i<$week_ends_on;$i++) {
if($current_day==$i) {
$current_week--;
}
}
$jan_1 = strftime("%u", mktime(0,0,0,1,1,date("Y")));
$jan_1++;
$jan_1 %= 7;
for($i=2;$i<=$week_ends_on;$i++) {
if($jan_1==$i) {
$current_week++;
$max_week++;
}
}