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stackoverflow 上有很多 SQL Top N 问题,但我似乎找不到与我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几个月,每月有数千行。我想做的是有一个查询告诉我每个月哪些 10 个 IP 地址最常出现。我可以使用以下方法在一个月内执行此操作:

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看到数据集中每个月的前 n 个。这基本上禁止我使用我指定的 where 子句。当然,当我这样做时,我整个月都会得到 10 分。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.

这可以在 MySQL 中完成吗?我遇到的障碍似乎是 MySQL 不允许在 UNION 中包含的查询语句中使用 ORDER BY。谢谢您的帮助!

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2 回答 2

3

我刚刚尝试了一个与@Charles Bretana给出的非常相似的查询,它确实有效。我使用 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)...

为每月的所有计数和 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 | 
+--------+-----------------+------+

现在显示每月前三个 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 | 
+--------+-----------------+------+
于 2009-08-31T22:45:17.210 回答
0

这是第一个粗略的猜测,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
于 2009-08-31T19:25:37.517 回答