[英]Sql oracle group by and condition
我有这些数据:
year code points
-------------
2017 M 1
2018 L 3
2019 L 5
我需要2019年的总分,取2019年之前与2019年代码匹配的所有代码
所以结果应该是:
y c p
----------
2019 L 8
我有一个 oracle 数据库,不知道该怎么做(使用“连接方式”?)
轻松测试请求:
WITH test AS (
SELECT 2017 y, 'M' code, 1 point FROM DUAL
UNION
SELECT 2018 y, 'L' code, 3 point FROM DUAL
UNION
SELECT 2019 y, 'L' code, 5 point FROM DUAL
)
SELECT * FROM test
您可以按如下方式进行self join
以达到所需的结果:
SQL> WITH test AS (
2 SELECT 2017 y, 'M' code, 1 point FROM DUAL
3 UNION
4 SELECT 2018 y, 'L' code, 3 point FROM DUAL
5 UNION
6 SELECT 2019 y, 'L' code, 5 point FROM DUAL
7 )
8 SELECT
9 T1.Y,
10 T1.POINT + COALESCE(T2.POINT,0) AS POINTS
11 FROM
12 TEST T1
13 LEFT JOIN TEST T2 ON T1.CODE = T2.CODE
14 AND T1.Y > T2.Y
15 WHERE T1.Y = '2019'
16 ;
Y POINTS
---------- ----------
2019 8
SQL>
或者您可以使用如下analytical function
:
SQL> WITH test AS (
2 SELECT 2017 y, 'M' code, 1 point FROM DUAL
3 UNION
4 SELECT 2018 y, 'L' code, 3 point FROM DUAL
5 UNION
6 SELECT 2019 y, 'L' code, 5 point FROM DUAL
7 )
8 SELECT Y, POINTS FROM
9 (SELECT
10 T1.Y,
11 SUM(T1.POINT) OVER (PARTITION BY CODE ORDER BY Y) AS POINTS
12 FROM
13 TEST T1)
14 WHERE Y = '2019'
15 ;
Y POINTS
---------- ----------
2019 8
SQL>
我想我会去:
select sum(points)
from (select code, sum(points) as points
from t
where y <= 2019
group by code
having max(y) = 2019
) x;
这处理两种边缘情况:
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