多重查询的复杂SQL问题

我有一个复杂的sql需要解决，这个想法是确定一个特定的CUSTOMER_ID的ADMIN_ID。确定ADMIN_ID的规则如下：多重查询的复杂SQL问题

要确定ADMIN_ID特定帐户，系统应查询从底部帐户到父，和第一帐户TXT01不为空
它是父母。
的ADMIN_ID仅适用于客户群E.
在外壳顶部的帐户有没有TXT01的，它应该给一个警告信息
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

客户组

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code..still做什么，但停留在一些点。在这里欣赏大师可以GV我的手

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

来源

2012-01-20 user871695

请检查你的老问题。正如有人说，你有一堆已经回答的问题，显然对你有帮助，但你还没有接受过这样的问题。请别做，别人可能不会倾向于帮助你。 –

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 
;

这里是我的结果：

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.

来源

2012-01-20 10:36:54

查询对我来说是相当复杂的..任何地方我会试着去理解它..谢谢哥们儿 – user871695

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

来源

2012-01-20 09:35:07

多重查询的复杂SQL问题

回答

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