SQL: Finden Sie die maximale Anzahl der Zeilen

Question

Ich bin völlig neu in der Codierung so könnte meine Frage ist albern sorry darüber zuerst.

Ich habe eine Datenbank, die darstellen, CUST_NUMBER CUST_REFERRED hat die

Buch jemand bezeichnet

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 

Meine Idee ist, Kunden zu finden, die max Buch bezeichnet. So wie Sie sehen können 3 mal 1003 Nummer verwiesen Buch, der Name ist LEILA SMITH

Ich versuchte einen Code, der;

SELECT 
    CUST_REFERRED, 
    COUNT(*) 
FROM 
    CUSTOMER 
GROUP BY 
    CUST_REFERRED 
    ORDER BY CUST_REFERRED ASC; 

Dieser Code gibt mir:

1003   3 
1006   1 
1010   1 

Zuerst meine Frage ich nicht LIMIT Funktion nutzen könnte maximale Anzahl und die zweite Frage zu finden, wie kann ich mehr Informationen von Kunden hinzufügen?

Answer 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 
                   ) 
                ) 
         ) 
       ) 
; 

Answer 2

versuchen Sie dies:

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) 

Answer 3

Diese Abfrage -

;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 

Answer 4

bearbeitet die angegebenen Kundendaten hinzuzufügen.

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) 
    ; 

Answer 5

Sie können es in einer einzigen Tabelle Scan tun (das heißt ohne Selbst Joins) Funktionen analytisch mit:

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;