Mysql 如何在SQL中应用`GROUPBY`查找同一列中的差异？

我需要找到按

Id

列分组的点之间的差异

Id | Year | Points
---+------+-------
1  | 2017 | 10
1  | 2018 | 20
2  | 2017 | 13
2  | 2018 | 16
3  | 2017 | 25
3  | 2018 | 20

预期结果：

Id | Points
---+-------
1  | 10
2  | 3
3  | -5

---

如果您想要年份之间的差异，您不需要

分组依据

：

select t2017.id, t2017.points as points_2017,
       t2018.points as points_2018,
       (t2018.points - t2017.points) as diff
from t t2017 join
     t t2018
     on t2017.id = t2018.id and
        t2017.year = 2017 and
        t2018.year = 2018;

您可以使用条件聚合执行非常类似的操作：

select id,
       sum(case when year = 2017 then points end) as points_2017,
       sum(case when year = 2018 then points end) as points_2018,
       (sum(case when year = 2018 then points end) -
        sum(case when year = 2017 then points end)
       ) as diff
from t
group by id;

PS.需要MySQL 8+。

进行聚合

select
id,sum(case when year = 2018 then points end) -sum(case when year = 2017 then points end) as diff
 from tablename group by id

---

分享您的预期结果@Zaynulabadinthin

select
id,sum(case when year = 2018 then points end) -sum(case when year = 2017 then points end) as diff
 from tablename group by id