MySQL group by month and day for current year - mysql

Question

I was wondering if this is possible:
I have some data where i have an datetime field. Now i want to make an sql query where i can make groups by month and in each month by day.
Something like this:
Month day COUNT(*)
1 1 200
1 2 300
1 3 500
2 1 600
2 2 0
Why i need this? I need to make an sql query to make an chart XY and show fill this requeriments:
SELECT series,value1,value2 FROM...WHERE...GROUP BY...ORDER BY.
So i want to make each month to be an SERIE, and then each day is value1, and the count value 2
Hope everyone understand my bot question...
Best Regards and tks in advanced

Answer 1

Is this all you're looking for?
SELECT MONTH(m), DAY(d), COUNT(*)
FROM sparkles
WHERE YEAR(y) = 2013
GROUP BY MONTH(m), DAY(d)
If your dates have gaps, you will need to use a date lookup table.

Answer 2

Use the MONTH() and DAYOFMONTH() functions.
Here is the documentation: dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html

Answer 3

Try this:
SELECT Month, Day, Count(*) FROM yout_table group by Month, Day

Answer 4

First you are going to need a table that holds every day this year:
CREATE TABLE DaysThisYear
(
dt datetime not null,
mm int, dd int,
primary key (dt)
);
INSERT INTO DaysThisYear (dt,mm,dd)
SELECT ymd,MONTH(ymd),DAY(ymd) FROM
(SELECT IFNULL(ymd + INTERVAL 0 SECOND,'1980-01-01 00:00:00') ymd
FROM (SELECT CONCAT(yy,'-',SUBSTR(mm+100,2),'-',SUBSTR(dd+100,2)) ymd,yy
FROM (SELECT 1 dd UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8
UNION SELECT 9 UNION SELECT 10 UNION SELECT 11 UNION SELECT 12
UNION SELECT 13 UNION SELECT 14 UNION SELECT 15 UNION SELECT 16
UNION SELECT 17 UNION SELECT 18 UNION SELECT 19 UNION SELECT 20
UNION SELECT 21 UNION SELECT 22 UNION SELECT 23 UNION SELECT 24
UNION SELECT 25 UNION SELECT 26 UNION SELECT 27 UNION SELECT 28
UNION SELECT 29 UNION SELECT 30 UNION SELECT 31 UNION SELECT 32) AAA,
(SELECT YEAR(NOW()) yy,mm FROM
(SELECT 1 mm UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5
UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION
SELECT 10 UNION SELECT 11 UNION SELECT 12) M) BBB) AA) A
WHERE YEAR(ymd) <> 1980
ORDER BY ymd;
To see that every day for this year was loaded, run this:
SELECT * FROM DaysThisYear;
Now, if you have a table with a DATETIME column, you can join the DaysThisYear table to it
For example, lets say your table looks like this:
CREATE TABLE mydata
(
id int not null auto_increment,
dt DATETIME,
.
.
.
PRIMARY KEY (id),
KEY dt (dt)
);
You could perform something like this:
SELECT A.mm,A.dd,SUM(IF(ISNULL(B.mm),0,1)) mmdd_count
FROM DaysThisYear A LEFT JOIN
(SELECT MONTH(dt) mm,DAY(dy) dd FROM mydata) B
ON A.mm=B.mm AND A.dd=B.dd;
Give it a Try !!!