CODEWITHSUNDEEP
mysqldate
user1043036
user1043036

Reputation: 83

Week of the year for weeks starting with Saturday

We have customers that currently have defined weeks starting either on Sat, Sun or Monday. Came across these DATE_FORMAT options which nicely handle the week starting on Sunday and Monday but can't find a way to do same for week starting on Saturday. Any suggestions? 

%U    Week (00..53), where Sunday is the first day of the week
%u    Week (00..53), where Monday is the first day of the week

Upvotes: 2

Views: 2564

Answers (3)

deciomsousa
deciomsousa

Reputation: 31

I had a similar issue: I needed to calculate week numbers based on the following rules:

  • Week starts on Friday
  • The remainder days of an year (all the days after the last Friday of the year that do not complete a week) should be counted in the first week of the next year.

For example:

  • 27/12/2012 (Thursday) should be Week 52 of 2012
  • 28/12/2012 (Friday) should be Week 1 of 2013
  • Week 1 2013 goes from 28/12/2012 to 3/1/2013

I made this statement that calculates both the YEAR and WEEKNUMBER based on these rules that you can easily adapt to your circunstance:

SELECT IF(ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7) > 52, YEAR(current_date)+1, YEAR(current_date)), 
        IF(ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7) > 52, 1, ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7));

The tricky part is just this expression:

ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7)

The rest (If clauses) are just for adapting the result of the expression to make year+1 and week = 1 on week 53.

I'll try to explain the expression as best as I can. The following expression gives you the week number pure simple (the day of the year divided by 7 days in a week rounded up):

ceil(( dayofyear(current_date))/7)

But now you want to make it start on Friday (or any other day). To do this you need to add to the current day, the days of the first week that were part of the previous year (it's like your current actually started a few days before, because your first week contains days from the previous year). This expression calculates that offset based on the weekday on Jan/1:

dayofweek(date_format(current_date, '%Y-01-01'))+OFFSET

The offset is the difference between 7 and the weekdaynumber you want the week to start:

  • 0 for Saturday
  • 1 for Friday
  • 2 for Thursday
  • 3 for Wednesday ...

So now you just have to add it to the previous one resulting in the above mentioned expression that calculates the week numbers starting on any weekday and assuming week 1 to start on the previous year:

ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+OFFSET )/7)

Then I just added an IF that turns week 53 into week 1 and another to add 1 to the year if it's week 53.

Upvotes: 3

vyegorov
vyegorov

Reputation: 22855

It took me a while to think on this question.

ISO standard defines the first week to start on Monday and contain 4th day of the year. MySQL's functions provide much more choices.

date_format() flags %U and %u are using notation where first week is the one where Sunday or Monday is first met. As this is not according to the ISO, I will provide both variants.

If you want to count week numbers starting from Saturday and first year's week is the one containing Saturday, you can use one of the following expressions:

SELECT sign(dayofweek(current_date) - 7) + ceil(dayofyear(current_date)/7);
SELECT ceil((dayofyear(current_date)+
            (dayofweek(date_format(current_date, '%Y-01-01'))%7-7))/7);

If first year's week is the one where 4th day of the year falls into, use:

SELECT ceil((dayofyear(current_date)+
            (dayofweek(date_format(current_date, '%Y-01-04'))%7-4+1))/7);

The very first expression is quite straightforward.

I will elaborate on the 2nd and 3rd ones. I calculate week number by taking current day of the year, dividing by 7 and ceiling up, quite simple. Week number needs to be adjusted based on the situation at the beginning of the year though.

  • for the first case (first week starts with the first Saturday), I take day-of-week for the Jan/1 of the year in subject, make Saturday as the day 0 and then adjust day-of-year by the difference. This this makes all days before first saturday yielding negative adjustment number and it ceils up to zero;
  • for the second case (first week is the one where 4 day of the year falls in), I take day-of-week for the Jan/4 of the year in subject, make Saturday as the day 0. The -4+1 formula gives adjustment to the first Saturday before Jan/4, +1 is used as days of the year starts from 1, not from 0. Negative adjustment means 1st day of the year is not in the first week of the year.

Here're some test dates on the SQL Fiddle.

If you want to count weeks from any other day, you just have to change the formula, making that day being 0 in the sequence. Say, to count weeks starting from Wednesday, use:

SELECT ceil((dayofyear(current_date)+
    ((dayofweek(date_format(current_date, '%Y-01-04'))+3)%7-4+1))/7);

+3 is used as it complements dayofweek() value for Wednesday to the 7.

Upvotes: 2

eggyal
eggyal

Reputation: 125835

Make an adjustment based on DAYOFWEEK().

Upvotes: 0

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