Question

我有一个包含交易的表格，其中的列为id，created_at和company_id。我想对每家公司的四个第一笔交易进行分组，并在每一行上返回每笔交易的created_at值。

换句话说，我希望我的输出的每一行对应于每个公司的四个第一笔交易（因此按company_id分组），其中的列显示company_id和每个的created_at这四笔交易。

我该怎么做？

示例数据：

  id  | company_id | created_at
---------------------------------
 1123 |    abcd    | 10/12/2015
 8291 |    abcd    | 10/14/2015
 9012 |    abcd    | 10/15/2015
 9540 |    abcd    | 10/16/2015
10342 |    abcd    | 10/21/2015
10456 |    abcd    | 10/22/2015
 2301 |    efgh    | 10/13/2015
 4000 |    efgh    | 11/01/2015
 4023 |    efgh    | 11/03/2015
 6239 |    efgh    | 11/08/2015
 7500 |    efgh    | 11/14/2015

示例输出：

  company_id | created_at_1 | created_at_2 | created_at_3 | created_at_4
--------------------------------------------------------------------------
     abcd    |  10/12/2015  |  10/14/2015  |  10/15/2015  |  10/16/2015
     efgh    |  10/13/2015  |  11/01/2015  |  11/03/2015  |  11/08/2015

Answer 1

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,company_id VARCHAR(12) NOT NULL
,created_at DATE NOT NULL
);

INSERT INTO my_table VALUES
( 1123,'abcd','2015/10/12'),
( 8291,'abcd','2015/10/14'),
( 9012,'abcd','2015/10/15'),
( 9540,'abcd','2015/10/16'),
(10342,'abcd','2015/10/21'),
(10456,'abcd','2015/10/22'),
( 2301,'efgh','2015/10/13'),
( 4000,'efgh','2015/11/01'),
( 4023,'efgh','2015/11/03'),
( 6239,'efgh','2015/11/08'),
( 7500,'efgh','2015/11/14');

SELECT x.* 
  FROM my_table x 
  JOIN my_table y 
    ON y.company_id = x.company_id 
   AND y.created_at <= x.created_at 
 GROUP 
    BY x.id 
HAVING COUNT(*) <= 4 
 ORDER 
    BY company_id
     , created_at;
+------+------------+------------+
| id   | company_id | created_at |
+------+------------+------------+
| 1123 | abcd       | 2015-10-12 |
| 8291 | abcd       | 2015-10-14 |
| 9012 | abcd       | 2015-10-15 |
| 9540 | abcd       | 2015-10-16 |
| 2301 | efgh       | 2015-10-13 |
| 4000 | efgh       | 2015-11-01 |
| 4023 | efgh       | 2015-11-03 |
| 6239 | efgh       | 2015-11-08 |
+------+------------+------------+

带变量的解决方案会快几个数量级，例如......

SELECT a.id
     , a.company_id
     , a.created_at
  FROM 
     ( SELECT x.*
            , CASE WHEN @prev = x.company_id THEN @i:=@i+1 ELSE @i:=1 END i, @prev:=x.company_id prev 
         FROM my_table x
            , (SELECT @i:=1,@prev:=null) vars 
        ORDER 
           BY x.company_id
            , x.created_at
     ) a
 WHERE i <= 4;

Answer 2

一种可能的方法如下：

<?xml version="1.0" encoding="utf-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://java.sun.com/xml/ns/javaee"
         xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
         http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" version="2.5">

  <!-- Servlets -->
  <servlet>
    <servlet-name>greetServlet</servlet-name>
    <servlet-class>ch.cern.atlas.emergency.status.server.GreetingServiceImpl</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>greetServlet</servlet-name>
    <url-pattern>/atlas_emergency_status_page/greet</url-pattern>
  </servlet-mapping>

  <!-- Default page to serve -->
  <welcome-file-list>
    <welcome-file>ATLAS_Emergency_Status_Page.html</welcome-file>
  </welcome-file-list>

  <!-- Servlets -->
  <servlet>
    <servlet-name>LoginService</servlet-name>
    <servlet-class>ch.cern.atlas.emergency.status.server.LoginServiceImpl</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>LoginService</servlet-name>
    <url-pattern>/atlas_emergency_status_page/LoginService</url-pattern>
  </servlet-mapping>

  <servlet>
    <servlet-name>DBService</servlet-name>
    <servlet-class>ch.cern.atlas.emergency.status.server.DBServiceImpl</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>DBService</servlet-name>
    <url-pattern>/atlas_emergency_status_page/DBService</url-pattern>
  </servlet-mapping>

</web-app>

编辑：

另一种可能性（受this answer启发）是：

select company_id, 
    min(created_at) as created_at_1,
    (select created_at from t where company_id=t1.company_id order by created_at limit 1 offset 1) as created_at_2,
    (select created_at from t where company_id=t1.company_id order by created_at limit 1 offset 2) as created_at_3,
    (select created_at from t where company_id=t1.company_id order by created_at limit 1 offset 3) as created_at_4
from t as t1
group by company_id

Answer 3

可能这样吗？

   SELECT S.company_id,
          A.created_at created_at_1,
          B.created_at created_at_2,
          C.created_at created_at_3,
          D.created_at created_at_4
     FROM sample S
LEFT JOIN sample A on S.company_id = A.company_id AND A.id NOT IN(S.id)
LEFT JOIN sample B on S.company_id = B.company_id AND B.id NOT IN(S.id, A.id)
LEFT JOIN sample C on S.company_id = C.company_id AND C.id NOT IN(S.id, A.id, B.id)
LEFT JOIN sample D on S.company_id = D.company_id AND D.id NOT IN(S.id, A.id, B.id, C.id)
 GROUP BY S.company_id

http://sqlfiddle.com/#!9/c577e/3

但是，它可能效率不高。

它们不合适，因为你的美国日期格式不适合排序。最好切换到TIMESTAMP格式。

组恰好是四行sql

3 个答案: