week number definition for Linux ?

Hi all,

I have a question about how week number should be defined/counted (in Linux) in terms of week number of a year and week number of a month.

For instance, 2011-01-07:

Code:

January 2011

Su	Mo	Tu	We	Th	Fr	Sa
 	 	 	 	 	 	1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31

It is in the week 1 of the year 2011 which obtained by using

Code:

 size_t strftime ( char * ptr, size_t maxsize, const char * format, const struct tm * timeptr );

and printf the value of %U of the given struct tm for 2011-01-07 .

However, I am not sure 2011-01-07 is considered to be in the week #1 or week #2 of the month. What would be the right value?

I need to know the correct week number of the month because it will be used to set TZ env as CST+6CDT,M4.1.0/2,M10.5.0/2 for daylight saving time. and the number in Bold above is the week number in month needed to be specified.

Could you kindly offer your comments on this.
Thanks in advance. :-)

Read the man page for strftime(), and see which format matches your idea of week numbers.

%u The day of the week as a decimal, range 1 to 7, Monday being 1. See also %w. (SU)

%U The week number of the current year as a decimal number, range 00 to 53, starting with the first Sunday as the first day of week 01. See also
%V and %W.

%V The ISO 8601 week number (see NOTES) of the current year as a decimal number, range 01 to 53, where week 1 is the first week that has at least
4 days in the new year. See also %U and %W. (SU)

%w The day of the week as a decimal, range 0 to 6, Sunday being 0. See also %u.

%W The week number of the current year as a decimal number, range 00 to 53, starting with the first Monday as the first day of week 01.

Who are the people who are defining your date format? They are the ones you should be asking.

Never mind. :-)

%U: Week number with the first Sunday as the first day of week one (00-53), so the return value 1 actually means week #2.
I think that is the expected week number to be used.

Got it. Thanks!