Question

在stackoverflow上有很多关于SQL Top N的问题，但我似乎找不到与我所拥有的情况相匹配的问题。我想在前n个查询中执行一些分组。我的数据看起来像这样（显然是假值）。

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

该表的日期范围涵盖数月，每月有数千行。我想做的是有一个查询告诉我每个月最常发生的10个IP地址。我可以使用以下内容执行此操作一个月：

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

但我真正想要的是能够在数据集中看到每个月的前n个。这基本上禁止我使用我指定的where子句。当然，当我这样做的时候，我只是在所有月份都达到10。我正在寻找的结果应该是这样的：

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.

可以在MySQL中完成吗？似乎我遇到的障碍是MySQL不允许在UNION中包含的查询语句中使用ORDER BY。谢谢你的帮助！

Answer 1

我刚刚尝试了一个与@Charles Bretana的given非常类似的查询，它确实有效。我使用了VIEW来帮助澄清事情。

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

插入一堆日期/ IP地址对（未显示）......

为每月所有计数和IP地址创建一个视图：

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 | 
+--------+-----------------+------+

现在每月显示前三个IP地址：

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 | 
+--------+-----------------+------+

Answer 2

这是第一次粗略猜测，但试试这个

Select Month, Address
From  (Select DATE_FORMAT(MY_DATE, '%b-%y') Month, 
       IP_Address Address, Count(*) AddressCount
         From MY_DATA
       Group By DATE_FORMAT(MY_DATE, '%b-%y'), IP_Adress) Z
  Join(Select DATE_FORMAT(MY_DATE, '%b-%y') Month, 
          IP_Address Address, Count(*) AddressCount
       From MY_DATA
       Group By DATE_FORMAT(MY_DATE, '%b-%y'), IP_Adress) ZZ
    On ZZ.Month = Z.Month 
       And ZZ.AddressCount >= Z.AddressCount 
Group By Z.Month, Z.Address
Where Count(ZZ.AddressCount) >= 10

在MySQL中按前N分组

2 个答案: