我对编码完全不熟悉所以我的问题可能是首先对此感到愚蠢。
我有一个CUST_REFERRED代表CUST_NUMBER的数据库谁推荐了某人
CUST_NUM NAME_S NAME_F ADDRESS Z_CODE CUST_REFERRED
1001 MORALES BONITA P.O. BOX 651 32328
1002 THOMPSON RYAN P.O. BOX 9835 90404
1003 SMITH LEILA P.O. BOX 66 32306
1004 PIERSON THOMAS 69821 SOUTH AVENUE 83707
1005 GIRARD CINDY P.O. BOX 851 98115
1006 CRUZ MESHIA 82 DIRT ROAD 12211
1007 GIANA TAMMY 9153 MAIN STREET 78710 1003
1008 JONES KENNETH P.O. BOX 137 82003
1009 PEREZ JORGE P.O. BOX 8564 91510 1003
1010 LUCAS JAKE 114 EAST SAVANNAH 30314
1011 MCGOVERN REESE P.O. BOX 18 60606
1012 MCKENZIE WILLIAM P.O. BOX 971 02110
1013 NGUYEN NICHOLAS 357 WHITE EAGLE AVE 34711 1006
1014 LEE JASMINE P.O. BOX 2947 82414
1015 SCHELL STEVE P.O. BOX 677 33111
1016 DAUM MICHELL 9851231 LONG ROAD 91508 1010
1017 NELSON BECCA P.O. BOX 563 49006
1018 MONTIASA GREG 1008 GRAND AVENUE 31206
1019 SMITH JENNIFER P.O. BOX 1151 07962 1003
1020 FALAH KENNETH P.O. BOX 335 08607
我的想法是找到推荐max book的客户。因此，您可以看到3次1003号参考书名为LEILA SMITH
我尝试了一个代码;
SELECT
CUST_REFERRED,
COUNT(*)
FROM
CUSTOMER
GROUP BY
CUST_REFERRED
ORDER BY CUST_REFERRED ASC;
这段代码告诉我：
1003 3
1006 1
1010 1
首先，我的问题是我无法使用LIMIT函数来查找最大数量 第二个问题是如何添加客户的更多信息？
答案 0 :(得分：1)
试试这个：
Select CUST_REFERRED, z.cnt from
(SELECT CUST_REFERRED, COUNT(*) cnt
FROM CUSTOMER where CUST_REFERRED is Not null
GROUP BY CUST_REFERRED) Z
where z.cnt =
(select Max(cnt) from
(SELECT COUNT(*) cnt
FROM CUSTOMER where CUST_REFERRED is Not null
GROUP BY CUST_REFERRED) ZZ)
答案 1 :(得分：1)
SELECT NAME_F,
NAME_S,
ADDRESS,
CUST_REFERRED
FROM CUSTOMER
WHERE CUST_NUM = (SELECT MOST_CUS_REF
FROM (SELECT CUST_REFERRED MOST_CUS_REF, COUNT(CUST_REFERRED)
MOST_CUS_REF_COUNT
FROM (SELECT CUST_REFERRED
FROM customer
WHERE cust_referred IS NOT NULL
)
GROUP BY CUST_REFERRED
HAVING COUNT(CUST_REFERRED) = (SELECT MAX (cust_ref_num)
FROM (SELECT CUST_REFERRED,
COUNT(CUST_REFERRED) cust_ref_num
FROM (SELECT CUST_REFERRED
FROM customer
WHERE cust_referred IS NOT NULL
)
GROUP BY CUST_REFERRED
)
)
)
)
;
答案 2 :(得分：0)
尝试此查询 -
;WITH CTE
AS (
SELECT CUSTOMER_REFID COUNT(*) AS REF_COUNT
FROM CUSTOMER
GROUP BY CUSTOMER_REFID
)
SELECT TOP 1 C2.CUSTOMER_ID
,C2.FIRST_NAME
,C2.LAST_NAME
,REF_COUNT
FROM CTE C1
INNER JOIN CUSTOMER C2
ON C1.CUSTOMER_REFID = C2.CUSTOMER_ID
ORDER BY REF_COUNT DESC
答案 3 :(得分：0)
编辑以添加推荐的客户详细信息。
with data (cust_num, name_s, name_f, addr, code, cust_referred) as
(
/* begin: test data */
select 1001 ,'MORALES ','BONITA ','P.O. BOX 651 ',32328, null from dual union all
select 1002 ,'THOMPSON ','RYAN ','P.O. BOX 9835 ',90404, null from dual union all
select 1003 ,'SMITH ','LEILA ','P.O. BOX 66 ',32306, null from dual union all
select 1004 ,'PIERSON ','THOMAS ','69821, SOUTH AVENUE ',83707, null from dual union all
select 1005 ,'GIRARD ','CINDY ','P.O. BOX 851 ',98115, null from dual union all
select 1006 ,'CRUZ ','MESHIA ','82 DIRT ROAD ',12211, null from dual union all
select 1007 ,'GIANA ','TAMMY ','9153 MAIN STREET ',78710, 1003 from dual union all
select 1008 ,'JONES ','KENNETH ','P.O. BOX 137 ',82003, null from dual union all
select 1009 ,'PEREZ ','JORGE ','P.O. BOX 8564 ',91510, 1003 from dual union all
select 1010 ,'LUCAS ','JAKE ','114 EAST SAVANNAH ',30314, null from dual union all
select 1011 ,'MCGOVERN ','REESE ','P.O. BOX 18 ',60606, null from dual union all
select 1012 ,'MCKENZIE ','WILLIAM ','P.O. BOX 971 ',02110, null from dual union all
select 1013 ,'NGUYEN ','NICHOLAS ','357 WHITE EAGLE AVE ',34711, 1006 from dual union all
select 1014 ,'LEE ','JASMINE ','P.O. BOX 2947 ',82414, null from dual union all
select 1015 ,'SCHELL ','STEVE ','P.O. BOX 677 ',33111, null from dual union all
select 1016 ,'DAUM ','MICHELL ',',9851231, LONG ROAD ',91508, 1010 from dual union all
select 1017 ,'NELSON ','BECCA ','P.O. BOX 563 ',49006, null from dual union all
select 1018 ,'MONTIASA ','GREG ','1008 GRAND AVENUE ',31206, null from dual union all
select 1019 ,'SMITH ','JENNIFER ','P.O. BOX 1151 ',07962, 1003 from dual union all
select 1020 ,'FALAH ','KENNETH ','P.O. BOX 335 ',08607, null from dual
/* end: test data */
-- replace the above block with your table
-- eg. select * from customers_table
)
,
max_referred as
(
-- just interested in the first row after sorting by
-- the count of referred column values
select rownum, cust_referred, cnt from
(
select cust_referred, count(cust_referred) cnt from data group by cust_referred order by 2 desc
)
where rownum = 1
)
-- joining on cust_referred column in *data* and *max_referred* tables to get the customer details
-- and joining again to the *data* table for fetching the referred customer name
select
cust.cust_num, cust.name_s, cust.name_f, cust.addr, cust.code, cust.cust_referred, ms.name_f || ms.name_s as "Referred Customer"
from
data cust
join
max_referred mr on (cust.cust_referred = mr.cust_referred)
join
data ms
on (mr.cust_referred = ms.cust_num)
;
答案 4 :(得分：0)
您可以使用分析函数在单个表扫描（即没有任何自连接）中执行此操作：
SELECT *
FROM (
SELECT t.*,
MIN( CUST_REFERRED )
KEEP ( DENSE_RANK FIRST ORDER BY num_referrals DESC )
OVER ()
AS best_referrer
FROM (
SELECT c.*,
COUNT( CUST_REFERRED )
OVER ( PARTITION BY CUST_REFERRED )
AS num_referrals
FROM CUSTOMER c
) t
)
WHERE cust_num = best_referrer;