在MySQL中按年和月分组

Question

I have a table with 25 date columns. Now i am building an application in PHP and have to show a month and year based summary. I have mentioned the Source table layout and expected Summary table below.

Here ID represents each row. Source table ID will be used in the summary table. The summary table will be created per year. I need some help to prepare this summary table. Here I would like to mention that i will have around 30K rows on the source table.

Source Table:

ID  |  DATE 1      |  DATE 2        |   DATE 3
--------------------------------------------------
1   |  2017-01-14  |   2017-01-19   |   2017-01-25
2   |              |   2017-03-19   |   2017-03-25
3   |  2017-03-15  |                |   2017-05-25
4   |  2017-04-24  |   2017-05-19   |   
5   |  2017-04-10  |   2017-06-19   |   2017-07-25
6   |  2017-05-11  |   2017-06-19   |   2017-08-25

Summary Table

ID  |   YEAR | JAN | FEB | MAR | APR | MAY | JUN | JUL | AUG | SEP | OCT | NOV | DEC
------------------------------------------------------------------------------------
1   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0 
2   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0 
3   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0 
4   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0 
5   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0 
6   |   2017 |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0  |  0

Answer 1

You need to unpivot the data and re-aggregate it. The structure of the query is:

select id, year(date) as yyyy,
       sum( month(date) = 1 ) as Jan,
       sum( month(date) = 2 ) as Feb,
       . . .
       sum( month(date) = 12 ) as Dec
from ((select id, date1 as date from source) union all
      (select id, date2 as date from source) union all
      . . .
      (select id, date25 as date from source) 
     ) d
where date is not null
group by id, year(date)
order by id, year(date);

The . . . means fill in the blanks.