SAP HANA grouping sets
原文链接：https://my.oschina.net/corleone/blog/131576
2019独角兽企业重金招聘Python工程师标准>>> hot3.png
本文只是对SAP HANA SQL Reference Manual 中关于group by的部分做一些个人解释。
数据准备：
create column table t1 ( id int primary key, customer varchar(5), year int, product
varchar(5), sales int );
insert into t1 values(1, ‘C1’, 2009, ‘P1’, 100);
insert into t1 values(2, ‘C1’, 2009, ‘P2’, 200);
insert into t1 values(3, ‘C1’, 2010, ‘P1’, 50);
insert into t1 values(4, ‘C1’, 2010, ‘P2’, 150);
insert into t1 values(5, ‘C2’, 2009, ‘P1’, 200);
insert into t1 values(6, ‘C2’, 2009, ‘P2’, 300);
insert into t1 values(7, ‘C2’, 2010, ‘P1’, 100);
insert into t1 values(8, ‘C2’, 2010, ‘P2’, 150);
GROUPING SETS
select customer, year, product, sum(sales) from t1
group by GROUPING SETS ((customer, year),(customer, product));
select customer, year, NULL, sum(sales) from t1 group by customer, year
union all
select customer, NULL, product, sum(sales) from t1 group by customer, product;
两结果集是一致的。
ROLLUP
select customer, year, sum(sales) from t1 group by ROLLUP(customer, year)
类似于：select customer, year, sum(sales)
from t1 group by grouping sets ( (customer, year), (customer))
union all
select NULL, NULL, sum(sales) from t1;
CUBE
select customer, year, sum(sales) from t1 group by CUBE(customer, year);
按顺序组合group by并union结果集，非group by 字段且不被聚合的列，默认为null.
类似于 select customer, year, sum(sales)
from t1
group by grouping sets
(
(customer, year),
(customer),
(year)
)
union all
select NULL, NULL, sum(sales)
from t1;
BEST
文档中给出的例子不是特别容易说明问题，换一个例子
select customer, year, sum(sales) from t1 group by CUBE best 1 (customer, year);
select customer, year, sum(sales) from t1 group by CUBE best 2 (customer, year);
由上面的图，对照前面的 cube 结果集，可知，best N是取 前N个 group by 组合的结果并union。
同时也支持N 为负数，如果负数的话，即为逆序的前N个group by 组合的结果并union。
limit
select customer, year, product, sum(sales) from t1 group by grouping sets LIMIT 2
( (customer, year), (product) );
只是简单的，每种分组取N条结果集并合并成一个大结果集。
至此，还是比较简单的。下面开始稍微有点意思了。
with subtotal
select customer, year, product, sum(sales)
from t1 group by grouping sets LIMIT 2 WITH SUBTOTAL
( (customer, year), (product) );
由此结果集可以看出，在每种group by 下都有一行记录，且记录是此group by 返回结果集的total.在分组中，但是因为limit 而没有返回的结果集，不算在total中，因此顾名思义，为subtotal。那有朋友就能想象出来，with total就是包含该分组的所有结果集的total。下面会提到。
with balance
文档给出的例子也不好，再改：
select customer, year, sum(sales) from t1 group by CUBE limit 1 with balance (customer, year)
和上面的很类似，也在每中分组下多了一条对该分组总结性的记录。不过这条记录是平衡的，或者是补差值的记录。
由前面结果集可见，sum(sales)=1250.由于每种分组的只取一条。因此每条都会有一个非null的差值记录来平衡。
官方的解释是这条记录来表示没有返回的结果集的subtotal.说的更好。
with total
select customer, year, product, sum(sales)
from t1 group by grouping sets LIMIT 2 WITH TOTAL
( (customer, year), (product) )
对照前面的with subtotal ， 顾名思义，这会就不是subtotal。不多解释。
text_filter
select customer, year, product, sum(sales), text_filter(customer),
text_filter(product) from t1 group by grouping sets TEXT_FILTER ‘*2’
( (customer, year), (product) )
官方定义为按每中分组的第一个分组列来过滤，返回符合的结果集
fill up
select customer, year, product, sum(sales),
text_filter(customer), text_filter(product)
from t1 group by grouping sets TEXT_FILTER ‘*2’ FILL UP
( (customer, year), (product) )
与text_filter对照看，这里返回的结果集里没有过滤掉不符合条件的。
sort matchs to top
结合fill up 看 ，顾名思义，就是按分组把匹配的记录置顶