我有这样一张桌子:
+-----+------+
| acc | CASE |
+-----+------+
| 001 | a |
| 001 | b |
| 001 | c |
| 002 | a |
| 002 | b |
| 003 | b |
| 003 | c |
| 004 | a |
| 005 | b |
| 006 | b |
| 007 | a |
| 007 | b |
| 007 | c |
| 008 | a |
| 008 | b |
| n | x |
+-----+------+
我不知道如何分组和统计数据
+-----------+-----------+
| case | count_acc |
+-----------+-----------+
| a | 1 |
| b | 2 |
| c | 0 |
| a+b | 2 |
| b+c | 1 |
| a+b+c | 2 |
| a+b+c+…+x | n |
+-----------+-----------+
如果a + b,b + c ... a + b + c + ... + x我不能将案例分组并计算acc。你有什么想法分组和数数?
select b.case,count(distinct(a.acc)) as account from
test a , (select acc , rtrim(case,'+') case
from ( select acc , case , rn from test
model
partition by (acc)
dimension by (row_number() over (partition by acc order by case) rn )
measures (cast(case as varchar2(10)) case)
rules
(case[any] order by rn desc = case[cv()]||'+'||case[cv()+1])
)
where rn = 1) b
where a.acc = b.acc
group by b.case
您可以在oracle中使用LISTAGG实现相同的功能
with test as ( select 001 acc , 'a' case FROM DUAL UNION
SELECT 001 , 'b' FROM DUAL UNION
SELECT 001 , 'c' FROM DUAL UNION
SELECT 002 , 'a' FROM DUAL UNION
SELECT 002 , 'b' FROM DUAL UNION
SELECT 003 , 'b' FROM DUAL UNION
SELECT 003 , 'c' FROM DUAL UNION
SELECT 004 , 'a' FROM DUAL UNION
SELECT 005 , 'b' FROM DUAL UNION
SELECT 006 , 'b' FROM DUAL UNION
SELECT 007 , 'a' FROM DUAL UNION
SELECT 007 , 'b' FROM DUAL UNION
SELECT 007 , 'c' FROM DUAL UNION
SELECT 008 , 'a' FROM DUAL UNION
SELECT 008 , 'b' FROM DUAL )
select case,count(1) from
(
SELECT count(1),acc, LISTAGG(case, '+') WITHIN GROUP (ORDER BY acc) AS case
FROM test
GROUP BY acc) GROUP BY case order by 1;
首先,您必须组织数据,对ACC进行分组并对CASE进行聚合
SELECT
LISTAGG (case,'+') WITHIN GROUP (ORDER BY case) case,
ACC
FROM TEST
GROUP BY ACC
你可以数数:
SELECT case, count(*) FROM (
SELECT
LISTAGG (case,'+') WITHIN GROUP (ORDER BY case) case,
ACC
FROM TEST
GROUP BY ACC
) GROUP BY case
ORDER BY CASE;
如果你没有功能LISTAGG,例如11g以下,请参考这个网站:
http://oracle-base.com/articles/misc/string-aggregation-techniques