Question

有人可以告诉我如何获得如下结果。

使用dense_rank函数，其中rank＆lt; = 2将给出前2个优惠。

我也希望得到＆total; of_offer＆＃39;这应该是＆quot; offer1＆＃39;的总和。和＆＃39; offer2＆＃39;。当没有offer2（例如：taurus）＆＃39;总报价＆＃39;应该是＆＃39; offer1＆＃39;和＆＃39; null＆＃39;为＆＃39; offer2＆＃39;

输入：

customer    make    zipcode offer
mark            focus   101 250  
mark             focus  101 2500  
mark             focus  101 1000  
mark             focus  101 1500  
henry   520i    21405   500  
henry   520i    21405   100  
henry   520i    21405   750  
henry   520i    21405   100  
mark            taurus  48360   250    
mark            mustang 730 500  
mark            mustang 730 1000  
mark            mustang 730 1250

期望的输出：

 | CUSTOMER | MAKE | ZIPCODE | TOP_OFFER1 | TOP_OFFER2 | Total_offer |  
 | henry | 520i | 21405 | 750 | 500 | 1250  
 | mark | focus | 101 | 2500 | 1500 | 4000  
 | mark | mustang | 730 | 1250 | 1000 | 2250  
 | mark | taurus | 48360 | 250 | NULL| 250 |

Answer 1

试试这个......

select * from (
select customer,make,zipcode,offer, dense_rank() over (PARTITION by customer order by customer,make, zipcode,offer desc) Rank from tablename)
where Rank <4;

Answer 2

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE TEST ( customer, make, zipcode, offer ) AS
          SELECT 'mark',  'focus',    101,  250 FROM DUAL
UNION ALL SELECT 'mark',  'focus',    101, 2500 FROM DUAL
UNION ALL SELECT 'mark',  'focus',    101, 1000 FROM DUAL
UNION ALL SELECT 'mark',  'focus',    101, 1500 FROM DUAL
UNION ALL SELECT 'henry', '520i',   21405,  500 FROM DUAL
UNION ALL SELECT 'henry', '520i',   21405,  100 FROM DUAL
UNION ALL SELECT 'henry', '520i',   21405,  750 FROM DUAL
UNION ALL SELECT 'henry', '520i',   21405,  100 FROM DUAL
UNION ALL SELECT 'mark',  'taurus', 48360,  250 FROM DUAL
UNION ALL SELECT 'mark',  'mustang',  730,  500 FROM DUAL
UNION ALL SELECT 'mark',  'mustang',  730, 1000 FROM DUAL
UNION ALL SELECT 'mark',  'mustang',  730, 1250 FROM DUAL;

查询1 - 如果每组最多需要3行：：

WITH ranks AS (
  SELECT t.*,
         ROW_NUMBER() OVER ( PARTITION BY CUSTOMER, MAKE, ZIPCODE ORDER BY OFFER DESC ) AS RANK
  FROM   TEST t
)
SELECT *
FROM   RANKS
WHERE  RANK <= 3

<强> Results ：

| CUSTOMER |    MAKE | ZIPCODE | OFFER | RANK |
|----------|---------|---------|-------|------|
|    henry |    520i |   21405 |   750 |    1 |
|    henry |    520i |   21405 |   500 |    2 |
|    henry |    520i |   21405 |   100 |    3 |
|     mark |   focus |     101 |  2500 |    1 |
|     mark |   focus |     101 |  1500 |    2 |
|     mark |   focus |     101 |  1000 |    3 |
|     mark | mustang |     730 |  1250 |    1 |
|     mark | mustang |     730 |  1000 |    2 |
|     mark | mustang |     730 |   500 |    3 |
|     mark |  taurus |   48360 |   250 |    1 |

查询2 - 如果您想要前三个等级，包括关系：

WITH ranks AS (
  SELECT t.*,
         DENSE_RANK() OVER ( PARTITION BY CUSTOMER, MAKE, ZIPCODE ORDER BY OFFER DESC ) AS RANK
  FROM   TEST t
)
SELECT *
FROM   RANKS
WHERE  RANK <= 3

<强> Results ：

| CUSTOMER |    MAKE | ZIPCODE | OFFER | RANK |
|----------|---------|---------|-------|------|
|    henry |    520i |   21405 |   750 |    1 |
|    henry |    520i |   21405 |   500 |    2 |
|    henry |    520i |   21405 |   100 |    3 |
|    henry |    520i |   21405 |   100 |    3 |
|     mark |   focus |     101 |  2500 |    1 |
|     mark |   focus |     101 |  1500 |    2 |
|     mark |   focus |     101 |  1000 |    3 |
|     mark | mustang |     730 |  1250 |    1 |
|     mark | mustang |     730 |  1000 |    2 |
|     mark | mustang |     730 |   500 |    3 |
|     mark |  taurus |   48360 |   250 |    1 |

使用pl-sql过程或光标选择前3名

2 个答案: