SQL：更新GROUP BY以包括基于另一列的最大值的值

Question

题

在查询中使用GROUP BY语句和聚合函数时，如何从列中添加特定值？

概观

这是我表的一个示例：

id  | year | quarter | wage | comp_id | comp_industry |
123 | 2012 | 1       | 1000 | 456     | abc           |
123 | 2012 | 1       | 2000 | 789     | def           |
123 | 2012 | 2       | 1500 | 789     | def           |
456 | 2012 | 1       | 2000 | 321     | ghi           |
456 | 2012 | 2       | 2000 | 321     | ghi           |

为了按quarter和wage计算每个人的wage值之和，我运行了以下查询：

SELECT SUM(wage) AS sum_wage
FROM t1
GROUP BY id, year, quarter, sum_wage;

结果是

id  | year | quarter | sum_wage | 
123 | 2012 | 1       | 3000     |
123 | 2012 | 2       | 1500     |
456 | 2012 | 1       | 2000     |
456 | 2012 | 2       | 2000     |

期望的输出

我想更新查询以包括comp_industry列，其中每个quarter和year个人的wage最高。 我不确定从哪里开始，所以我只返回人们每个quarter和year赚钱最多的行业。

id  | year | quarter | sum_wage | comp_industry
123 | 2012 | 1       | 3000     | def
123 | 2012 | 2       | 1500     | def
456 | 2012 | 1       | 2000     | ghi
456 | 2012 | 2       | 2000     | ghi

我看过基于另一列分组的另一列的最大值获取值，并获取具有列的最大值的行，但是不确定从那里开始。

任何帮助或建议，将不胜感激！

Answer 1

您可以尝试将窗口函数与SUM和ROW_NUMBER一起使用。

使行号按id ， year ， quarter列按wage desc排序，然后得到rn = 1 。

模式（PostgreSQL v9.6）

CREATE TABLE T (
   id INT, 
   year INT,
   quarter INT,
   wage INT,
   comp_id INT,
  comp_industry VARCHAR(50)
);


INSERT INTO T VALUES (123 , 2012 , 1 , 1000 , 456    ,'abc');
INSERT INTO T VALUES (123 , 2012 , 1 , 2000 , 789    ,'def');
INSERT INTO T VALUES (123 , 2012 , 2 , 1500 , 789    ,'def');
INSERT INTO T VALUES (456 , 2012 , 1 , 2000 , 321    ,'ghi');
INSERT INTO T VALUES (456 , 2012 , 2 , 2000 , 321    ,'ghi');

---

查询＃1

SELECT id, year,quarter ,sum_wage, comp_industry FROM (
  SELECT *,
           SUM(wage)  OVER (PARTITION BY  id, year, quarter  order by year ) sum_wage,
           ROW_NUMBER() OVER (PARTITION BY  id, year, quarter order by wage desc) rn
    FROM T
) t1
where rn = 1;

| id  | year | quarter | sum_wage | comp_industry |
| --- | ---- | ------- | -------- | ------------- |
| 123 | 2012 | 1       | 3000     | def           |
| 123 | 2012 | 2       | 1500     | def           |
| 456 | 2012 | 1       | 2000     | ghi           |
| 456 | 2012 | 2       | 2000     | ghi           |

---

在数据库小提琴上查看

Answer 2

我不是100％肯定我理解这个问题，这对您有用吗？

SELECT id, 
       year, 
       quarter, 
       comp_industry, 
       SUM(wage)
  FROM (SELECT id, 
               year, 
               quarter,
               comp_industry, 
               wage
          FROM (SELECT TMP.*,
                       RANK() OVER
                         ( PARTITION BY id, 
                                        year, 
                                        quarter
                               ORDER BY wage_sum DESC         
                         ) wage_rnk
                  FROM (SELECT t1.*,
                               SUM(wage) OVER
                                 ( PARTITION BY id, 
                                                year, 
                                                quarter 
                                 ) wage_sum
                        FROM t1
                        GROUP BY id, 
                                 year, 
                                 quarter
                       ) TMP
               ) TMP2
         WHERE wage_rnk = 1
       ) TMP3
 GROUP  
    BY id, 
       year, 
       quarter, 
       comp_industry;