Question

例如我有一张桌子＆＃34;信息＆＃34;如下。

+--------+------------+------+------------+
| Entity | Department | Code | Code_count |
+--------+------------+------+------------+
| E1     | D1         |  123 |          5 |
| E1     | D1         |  234 |         10 |
| E1     | D1         |  345 |         20 |
| E1     | D2         |  456 |          2 |
| E1     | D2         |  567 |          5 |
| E1     | D2         |  678 |         10 |
+--------+------------+------+------------+

我的查询应该是这样的：对于每个Entity和Department对，请选择code的{{1}}。

非常感谢任何帮助。感谢。

Answer 1

select t1.*
from your_table t1
join
(
  select entity, department, max(Code_count) as Code_count 
  from your_table
  group by entity, department
) t2 on t1.entity = t2.entity 
    and t1.department = t2.department 
    and t1.Code_count = t2.Code_count

Answer 2

这应该适合你：

SELECT ENTITY, DEPARTMENT, CODE
FROM TABLE
WHERE CODE_COUNT IN (
  SELECT MAX(CODE_COUNT) FROM TABLE GROUP BY ENTITY, DEPARTMENT)

SQL小提琴：http://sqlfiddle.com/#!2/482397/12

Answer 3

使用Window Function获取结果。

SELECT Row_number()OVER(
                       partition BY Entity, Department
                       ORDER BY Code_count DESC) RN,
                   *
            FROM   table_name

partition BY将查询结果集划分为多个分区。 window函数分别应用于每个分区，并为每个分区重新开始计算。

SELECT Entity,
       Department,
       Code,
       Code_count
FROM   (SELECT Row_number()OVER(
                   partition BY Entity, Department
                   ORDER BY Code_count DESC) RN,
               *
        FROM   table_name) A
WHERE  rn = 1

基于条件的选择SQL查询

3 个答案: