如何在MySQL的无行日中每天对行数进行分组？

Question

So i have this table. It has some hundreds of rows. Every row has a datetime field in it. And what i need to acomplish is to get how much rows there are in given period of time but not for the whole period but for every day in this period. Till this point i know what to do. But moreover i need to have also rows for the days that does not have any row in the table with value 0.

So for example:

2012-01-01 12:13
2012-01-01 43:32
2012-01-03 23:32

Should give me the result like this:

2012-01-01 2
2012-01-02 0
2012-01-03 1

Anyone can help??

Answer 1

To deal with dates with 0 corresponding records, my normal practice is to use a calendar table to join on.

For example, create a table with one field called calendar_date and populate it with every date from 1st Jan 2000 to 31st Dec 2070 , or some other range that suits your reporting purposes.

Then use something like...

SELECT
  calendar.calendar_date,
  COUNT(*)
FROM
  calendar
LEFT JOIN
  yourData
    ON  yourData.timeStamp >= calendar.calendar_date
    AND yourData.timeStamp <  calendar.calendar_date + 1
WHERE
      calendar.calendar_date >= '01 Jan 2012'
  AND calendar.calendar_date <  '04 Jan 2012'
GROUP BY
  calendar.calendar_date

This table can have many extra uses, such as flagging bank holidays, starts of weeks and months. With prudent uses of flags and indexes, you can get a lot out of it.

Answer 2

If you don't have a table with a list of dates in it, you can simulate one with something like the following query:

select * from 
(select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) calendar_date from
 (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
 (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
 (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
 (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
 (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
where calendar_date between ? /*start of date range*/ and ? /*end of date range*/

- so Dems' query could become:

SELECT
  calendar.calendar_date,
  COUNT(*)
FROM
  (select * from 
    (select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) calendar_date from
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
   WHERE calendar_date >= '01 Jan 2012'
     AND calendar_date <  '04 Jan 2012'
  ) calendar
LEFT JOIN
  yourData
    ON  yourData.timeStamp >= calendar.calendar_date
    AND yourData.timeStamp <  calendar.calendar_date + 1

Answer 3

try

SELECT DATE_FORMAT(YourDate, '%Y-%m-%d') as YourDate, COUNT(*)
FROM TableName where YourDate= 'date' 
GROUP BY YourDate