我有一个复杂的sql需要解决，想法是确定特定CUSTOMER_ID的ADMIN_ID。确定ADMIN_ID的规则如下：
要确定特定帐户的ADMIN_ID，系统应查询 从下到上的帐户，以及TXT01的第一个帐户不为空 它是父母。
ADMIN_ID仅适用于CUSTOMER E组。
如果顶级帐户缺少TXT01，则应提供警告消息
只有CUSTOMER_ID = PAYING_ACCOUNT_ID需要分析
帐户表
CUSTOMER_ID PAYING_ACCOUNT_ID PARENT_ACCOUNT_ID
3271516 3271516 719216
1819276 1819276 810546
719216 719216 719216
810546 810546 810547
810547 810547 810547
999999 111111 111111
111111 111111 111111
123456 123456 231
231 231 231
客户表
CUSTOMER_ID TXT01
719216 TOM
810546 NULL
810547 JIM
3271516 NULL
1819276 NULL
999999 NULL
111111 BEN
123456 NULL
231 NULL
客户群
CUSTOMER_ID GROUP
719216 E
810546 E
810547 E
3271516 E
1819276 E
999999 E
111111 E
123456 E
231 E
888 A
输出
CUSTOMER_ID PAYING_ACCOUNT_ID PARENT_ACCOUNT_ID ADMIN_ID
3271516 3271516 719216 TOM
1819276 1819276 810546 JIM
719216 719216 719216 TOM
810546 810546 810547 JIM
810547 810547 810547 JIM
111111 111111 111111 BEN
123456 123456 231 Warning!!
231 231 532 Warning!!
DDL
CREATE TABLE ACCOUNT (CUSTOMER_ID NUMBER(20) NOT NULL,
PAYING_ACCOUNT_ID NUMBER(20),
PARENT_ACCOUNT_ID NUMBER(20));
CREATE TABLE CUSTOMER (CUSTOMER_ID NUMBER(20) NOT NULL,
TXT01 VARCHAR2(20));
CREATE TABLE CUSTOMER_GROUP (CUSTOMER_ID NUMBER(20) NOT NULL,
GROUP VARCHAR2(20));
INSERT INTO ACCOUNT VALUES (3271516,3271516,719216);
INSERT INTO ACCOUNT VALUES (1819276,1819276,810546);
INSERT INTO ACCOUNT VALUES (719216,719216,719216);
INSERT INTO ACCOUNT VALUES (810546,810546,810547);
INSERT INTO ACCOUNT VALUES (810547,810547,810547);
INSERT INTO ACCOUNT VALUES (999999,111111,111111);
INSERT INTO ACCOUNT VALUES (111111,111111,111111);
INSERT INTO ACCOUNT VALUES (123456,123456,231);
INSERT INTO ACCOUNT VALUES (231,231,231);
INSERT INTO CUSTOMER VALUES (719216,'TOM');
INSERT INTO CUSTOMER VALUES (810546,NULL);
INSERT INTO CUSTOMER VALUES (810547,'JIM');
INSERT INTO CUSTOMER VALUES (3271516,NULL);
INSERT INTO CUSTOMER VALUES (1819276,NULL);
INSERT INTO CUSTOMER VALUES (999999,NULL);
INSERT INTO CUSTOMER VALUES (111111,'BEN');
INSERT INTO CUSTOMER VALUES (123456,NULL);
INSERT INTO CUSTOMER VALUES (231,NULL);
INSERT INTO CUSTOMER_GROUP VALUES (719216,'E');
INSERT INTO CUSTOMER_GROUP VALUES (810546,E);
INSERT INTO CUSTOMER_GROUP VALUES (810547,'E');
INSERT INTO CUSTOMER_GROUP VALUES (3271516,'E');
INSERT INTO CUSTOMER_GROUP VALUES (1819276,'E');
INSERT INTO CUSTOMER_GROUP VALUES (999999,'E');
INSERT INTO CUSTOMER_GROUP VALUES (111111,'E');
INSERT INTO CUSTOMER_GROUP VALUES (123456,'E');
INSERT INTO CUSTOMER_GROUP VALUES (231,'E');
INSERT INTO CUSTOMER_GROUP VALUES (888,'A');
这是我的代码......仍然在做，但在某些方面陷入困境。欣赏大师在这里可以给我一只手
WITH myData AS (
SELECT CUSTOMER_ID, TXT01 FROM CUSTOMER WHERE CUSTOMER_ID IN
(SELECT CUSTOMER_ID from CUSTOMER_GROUP WHERE GROUP = 'E')
)
SELECT v.*
FROM
(SELECT m.* ,
CASE WHEN TXT01 IS NOT NULL THEN TXT01
ELSE ( *STUCK HERE*
END ADMIN_ID
FROM myData m) v
答案 0 :(得分：0)
SELECT
C.CUSTOMER_ID, nvl(a.TXT01, 'warning!')
FROM CUSTOMER c
JOIN CUSTOMER_GROUP cg ON (cg.customer_id = c.customer_id and cg.group = 'E')
JOIN ACCOUNT a ON (a.customer_id = c.customer_id)
WHERE c.CUSTOMER_ID = a.PAYING_ACCOUNT_ID
答案 1 :(得分：0)
试试这个
with acc_hierarchy as
(select level lev, ac.*, connect_by_root(ac.paying_account_id) account_id
from account ac
connect by prior ac.parent_account_id = ac.paying_account_id
and prior ac.paying_account_id <> prior ac.parent_account_id),
acc_customers as
(select ah.account_id, min(c.txt01) keep(dense_rank first order by ah.lev) admin_id
from acc_hierarchy ah
left join (select cus.customer_id, cus.txt01
from customer cus
join customer_group cg
on cg.customer_id = cus.customer_id
and cg."GROUP" = 'E') c
on c.customer_id = ah.customer_id
where c.txt01 is not null
group by ah.account_id)
select a.customer_id, a.paying_account_id, a.parent_account_id, NVL(ac.admin_id, 'Warning!!') admin_id
from account a
left join acc_customers ac
on ac.account_id = a.paying_account_id
where a.customer_id = a.paying_account_id
;
以下是我的结果：
SQL> with acc_hierarchy as
2 (select level lev, ac.*, connect_by_root(ac.paying_account_id) account_id
3 from account ac
4 connect by prior ac.parent_account_id = ac.paying_account_id
5 and prior ac.paying_account_id <> prior ac.parent_account_id),
6 acc_customers as
7 (select ah.account_id, min(c.txt01) keep(dense_rank first order by ah.lev) admin_id
8 from acc_hierarchy ah
9 left join (select cus.customer_id, cus.txt01
10 from customer cus
11 join customer_group cg
12 on cg.customer_id = cus.customer_id
13 and cg."GROUP" = 'E') c
14 on c.customer_id = ah.customer_id
15 where c.txt01 is not null
16 group by ah.account_id)
17 select a.customer_id, a.paying_account_id, a.parent_account_id, NVL(ac.admin_id, 'Warning!!') admin_id
18 from account a
19 left join acc_customers ac
20 on ac.account_id = a.paying_account_id
21 where a.customer_id = a.paying_account_id;
CUSTOMER_ID PAYING_ACCOUNT_ID PARENT_ACCOUNT_ID ADMIN_ID
----------- ----------------- ----------------- --------------------
3271516 3271516 719216 TOM
1819276 1819276 810546 JIM
719216 719216 719216 TOM
810546 810546 810547 JIM
810547 810547 810547 JIM
111111 111111 111111 BEN
123456 123456 231 Warning!!
231 231 231 Warning!!
8 rows selected.