How to group by week in MySQL?

Question

Oracle's table server offers a built-in function, TRUNC(timestamp,'DY'). This function converts any timestamp to midnight on the previous Sunday. What's the best way to do this in MySQL?

Oracle also offers TRUNC(timestamp,'MM') to convert a timestamp to midnight on the first day of the month in which it occurs. In MySQL, this one is straightforward:

TIMESTAMP(DATE_FORMAT(timestamp, '%Y-%m-01'))

But this DATE_FORMAT trick won't work for weeks. I'm aware of the WEEK(timestamp) function, but I really don't want week number within the year; this stuff is for multiyear work.

Answer 1

Figured it out... it's a little cumbersome, but here it is.

FROM_DAYS(TO_DAYS(TIMESTAMP) -MOD(TO_DAYS(TIMESTAMP) -1, 7))

And, if your business rules say your weeks start on Mondays, change the -1 to -2.

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Edit

Years have gone by and I've finally gotten around to writing this up. http://www.plumislandmedia.net/mysql/sql-reporting-time-intervals/

Answer 2

You can use both YEAR(timestamp) and WEEK(timestamp), and use both of the these expressions in the SELECT and the GROUP BY clause.

Not overly elegant, but functional...

And of course you can combine these two date parts in a single expression as well, i.e. something like

SELECT CONCAT(YEAR(timestamp), '/', WEEK(timestamp)), etc...
FROM ...
WHERE ..
GROUP BY CONCAT(YEAR(timestamp), '/', WEEK(timestamp))

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Edit: As Martin points out you can also use the YEARWEEK(mysqldatefield) function, although its output is not as eye friendly as the longer formula above.

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Edit 2 [3 1/2 years later!]:
YEARWEEK(mysqldatefield) with the optional second argument (mode) set to either 0 or 2 is probably the best way to aggregate by complete weeks (i.e. including for weeks which straddle over January 1st), if that is what is desired. The YEAR() / WEEK() approach initially proposed in this answer has the effect of splitting the aggregated data for such "straddling" weeks in two: one with the former year, one with the new year.
A clean-cut every year, at the cost of having up to two partial weeks, one at either end, is often desired in accounting etc. and for that the YEAR() / WEEK() approach is better.

Answer 3

The accepted answer above did not work for me, because it ordered the weeks by alphabetical order, not chronological order:

2012/1
2012/10
2012/11
...
2012/19
2012/2

Here's my solution to count and group by week:

SELECT CONCAT(YEAR(date), '/', WEEK(date)) AS week_name, 
       YEAR(date), WEEK(date), COUNT(*)
FROM column_name
GROUP BY week_name
ORDER BY YEAR(DATE) ASC, WEEK(date) ASC

Generates:

YEAR/WEEK   YEAR   WEEK   COUNT
2011/51     2011    51      15
2011/52     2011    52      14
2012/1      2012    1       20
2012/2      2012    2       14
2012/3      2012    3       19
2012/4      2012    4       19

Answer 4

You can get the concatenated year and week number (200945) using the YEARWEEK() function. If I understand your goal correctly, that should enable you to group your multi-year data.

If you need the actual timestamp for the start of the week, it's less nice:

DATE_SUB( field, INTERVAL DAYOFWEEK( field ) - 1 DAY )

For monthly ordering, you might consider the LAST_DAY() function - sort would be by last day of the month, but that should be equivalent to sorting by first day of the month ... shouldn't it?

Answer 5

Just ad this in the select :

DATE_FORMAT($yourDate, \'%X %V\') as week

And

group_by(week);

Answer 6

If you need the "week ending" date this will work as well. This will count the number of records for each week. Example: If three work orders were created between (inclusive) 1/2/2010 and 1/8/2010 and 5 were created between (inclusive) 1/9/2010 and 1/16/2010 this would return:

3 1/8/2010
5 1/16/2010

I had to use the extra DATE() function to truncate my datetime field.

SELECT COUNT(*), DATE_ADD( DATE(wo.date_created), INTERVAL (7 - DAYOFWEEK( wo.date_created )) DAY) week_ending FROM work_order wo GROUP BY week_ending;