日期:2014-05-17  浏览次数:20676 次

有这么个变态需求,能否用一句SQL实现
假设有100个客户,通过分组统计得到每个客户的营业额,现在想求,第一名比第二名高出多少,第二名比第三名高出多少....以此类推,最终结果显示为一个客户一行,同时显示营业额以及超出下一客户的量。

------解决方案--------------------
SQL code

with t1 as
(
    --此处写你的分组统计汇总的语句
    select 'user1' as username, 100 as sale from dual
     union all
    select 'user2' as username, 150 as sale from dual
),
t2 as
(
    --根据上面的语句修改字段名
    select username, sale, row_number() over(order by sale desc) as isort from t1
)select isort,username, sale, 
        nvl(sale - (select sale from t2 b where b.isort = a.isort + 1), 0) as dif
  from t2 a;

------解决方案--------------------
SQL code

--利用lead函数就可以了,考虑到可能会有的分组情况,模拟如下
----如果你没有分组情况,直接去掉over()里面的partition by 字段即可。
----另:由于排名最后的一个客户下面没人了因此给了他一个默认值LEAD(SALE,1,0)为0,相减之后还是客户的SALE

[SYS@myorcl] SQL>WITH T1 AS(
  2    SELECT DATE'2011-01-01' d_time,'user1' AS USERNAME, 100 AS SALE FROM DUAL
  3    UNION ALL
  4    SELECT DATE'2011-01-01' d_time,'user2' AS USERNAME, 990 AS SALE FROM DUAL
  5    UNION ALL
  6    SELECT DATE'2011-01-01' d_time,'user3' AS USERNAME, 11 AS SALE FROM DUAL
  7    UNION ALL
  8    SELECT DATE'2011-01-01' d_time,'user4' AS USERNAME, 130 AS SALE FROM DUAL
  9    UNION ALL
 10    SELECT DATE'2011-01-02' d_time,'user1' AS USERNAME, 120 AS SALE FROM DUAL
 11    UNION ALL
 12    SELECT DATE'2011-01-02' d_time,'user2' AS USERNAME, 50 AS SALE FROM DUAL
 13    UNION ALL
 14    SELECT DATE'2011-01-02' d_time,'user3' AS USERNAME, 650 AS SALE FROM DUAL
 15  )SELECT d_time,
 16          USERNAME,
 17          SALE,
 18          SALE-LEAD(SALE,1,0)OVER(PARTITION BY d_time ORDER BY SALE DESC) AS diff
 19     FROM t1
 20  ;

D_TIME              USERN       SALE       DIFF
------------------- ----- ---------- ----------
2011-01-01 00:00:00 user2        990        860
2011-01-01 00:00:00 user4        130         30
2011-01-01 00:00:00 user1        100         89
2011-01-01 00:00:00 user3         11         11
2011-01-02 00:00:00 user3        650        530
2011-01-02 00:00:00 user1        120         70
2011-01-02 00:00:00 user2         50         50

已选择7行。

------解决方案--------------------
with t as 
( select 'a' c1,800 c2 from dual union all
select 'b' c1,1800 c2 from dual union all
select 'c' c1,2200 c2 from dual union all
select 'd' c1,5800 c2 from dual 
)
select c1,c2,lead(c2,1) over(order by c2) - c2 as diff from t
------解决方案--------------------
SQL code



CREATE TABLE T_CLIENT_VB
(
v_id NUMBER(4),
v_all_vb NUMBER(10)
);
INSERT INTO T_CLIENT_VB VALUES(1,100);
INSERT INTO T_CLIENT_VB VALUES(2,200);
INSERT INTO T_CLIENT_VB VALUES(3,300);
INSERT INTO T_CLIENT_VB VALUES(4,400);
INSERT INTO T_CLIENT_VB VALUES(5,500);
INSERT INTO T_CLIENT_VB VALUES(6,550);
INSERT INTO T_CLIENT_VB VALUES(7,580);
COMMIT;

SELECT A.V_ID, A.V_ALL_VB - B.V_ALL_VB, A.V_ALL_VB, B.V_ALL_VB, A.RK, B.RK
  FROM (SELECT V_ID, V_ALL_VB, RANK() OVER(ORDER BY V_ALL_VB DESC) AS RK
          FROM T_CLIENT_VB
         ORDER BY RK DESC) A,
       (SELECT V_ID, V_ALL_VB, RANK() OVER(ORDER BY V_ALL_VB DESC) AS RK
          FROM T_CLIENT_VB
         ORDER BY RK DESC) B
 WHERE A.RK + 1 = B.RK