sql - Grouping by a Top N in MySQL

Question

There are a lot of SQL Top N questions on stackoverflow but I can't seem to find one that matches the situation I'm having. I would like to perform some grouping within a top n query. My data looks like this (obviously with fake values).

MY_DATE    IP_ADDRESS
1/1/09     999.999.999.999
1/1/09     999.999.999.999
1/1/09     999.999.999.998
... a lot more rows

The date range for the table covers several months and has many thousands of rows per month. What I would like to do is have a single query tell me which 10 IP Addresses occurred the most frequently for each month. I can do this for a single month using the following:

SELECT DATE_FORMAT(MY_DATE, '%b-%y') AS "MONTH", IP_ADDRESS, COUNT(*) AS HITS
FROM MY_DATA
WHERE DATE_FORMAT(MY_DATE, '%b-%y') = 'JAN-09'
GROUP BY DATE_FORMAT(MY_DATE, '%b-%y'), IP_ADDRESS
ORDER BY HITS DESC
LIMIT 10

But what I really want is to be able to see the top n for every month in the data set. That essentially prohibits me from using the where clause I specified. Of course, when I do that, then I just get the to 10 for all months. The result I'm looking for should look like this:

MONTH    IP_ADDRESS        COUNT(*)
JAN-09   999.999.999.999   200
JAN-09   999.999.999.998   150
... ( 8 more rows of January )
FEB-09   999.999.999.999   320
FEB-09   999.999.999.998   234
... ( 8 more rows of February)
MAR-09   999.999.999.999   440
... ETC.

Can this be done in MySQL? It seems the barrier I'm hitting is that MySQL doesn't allow an ORDER BY within a query statement included in a UNION. Thanks for the help!

Answer 1

I just tried a query very similar to the one given by @Charles Bretana and it does work. I used a VIEW to help clarify things.

CREATE TABLE my_data (
 my_date DATE,
 ip_address CHAR(15)
);

Insert a bunch of date/IPaddress pairs (not shown)...

Create a view for all counts per month and IP address:

CREATE VIEW my_data_per_month as
 SELECT EXTRACT(YEAR_MONTH FROM my_date) AS month,
   ip_address, COUNT(*) AS hits
 FROM my_data
 GROUP BY month, ip_address;

SELECT * FROM my_data_per_month
ORDER BY month ASC, hits DESC;

+--------+-----------------+------+
| month  | ip_address      | hits |
+--------+-----------------+------+
| 200901 | 999.999.999.999 |    8 | 
| 200901 | 999.999.999.998 |    6 | 
| 200901 | 999.999.999.997 |    5 | 
| 200901 | 999.999.999.996 |    4 | 
| 200901 | 999.999.999.995 |    3 | 
| 200901 | 999.999.999.994 |    2 | 
| 200902 | 999.999.999.998 |    8 | 
| 200902 | 999.999.999.997 |    6 | 
| 200902 | 999.999.999.996 |    5 | 
| 200902 | 999.999.999.995 |    4 | 
| 200902 | 999.999.999.994 |    3 | 
| 200902 | 999.999.999.993 |    2 | 
| 200903 | 999.999.999.997 |    8 | 
| 200903 | 999.999.999.996 |    6 | 
| 200903 | 999.999.999.995 |    5 | 
| 200903 | 999.999.999.994 |    4 | 
| 200903 | 999.999.999.993 |    3 | 
| 200903 | 999.999.999.992 |    2 | 
+--------+-----------------+------+

Now show the top three IP addresses per month:

SELECT m1.month, m1.ip_address, m1.hits
FROM my_data_per_month m1
LEFT OUTER JOIN my_data_per_month m2
  ON (m1.month = m2.month AND m1.hits < m2.hits)
GROUP BY m1.month, m1.ip_address
HAVING COUNT(*) < 3
ORDER BY m1.month ASC, m1.hits DESC;

+--------+-----------------+------+
| month  | ip_address      | hits |
+--------+-----------------+------+
| 200901 | 999.999.999.999 |    8 | 
| 200901 | 999.999.999.998 |    6 | 
| 200901 | 999.999.999.997 |    5 | 
| 200902 | 999.999.999.998 |    8 | 
| 200902 | 999.999.999.997 |    6 | 
| 200902 | 999.999.999.996 |    5 | 
| 200903 | 999.999.999.997 |    8 | 
| 200903 | 999.999.999.996 |    6 | 
| 200903 | 999.999.999.995 |    5 | 
+--------+-----------------+------+