找到一个树形结构表

Question

我有一个这样的表结构的共同祖先（其实有更多的层次）：

------------------------------------------ 
|region1|region2|region3|region4|postcode| 
|-------|-------|-------|-------|--------| 
|a  |x  |i  |  |1  | 
|a  |y  |i  |  |2  | 
|a  |y  |j  |  |2  | 
|a  |z  |k  |  |3  | 
|b  |u  |m  |  |4  | 
|b  |  |n  |  |4  | 
|c  |  |  |  |5  | 
|c  |q  |  |  |6  | 
------------------------------------------ 

因此，例如，a => x => i和a => y => i有不同的地方，但两者在相同的区域1 a。

我想知道每个邮政编码可以覆盖哪个区域。

例如，代码2涵盖区域a => y => i和a => y => j，因此这些共同的祖先是a => y。

这里是关于例如运行查询所需的输出：

------------------------------------------ 
|postcode|region1|region2|region3|region4| 
|--------|-------|-------|-------|-------| 
|1  |a  |x  |i  |  | 
|2  |a  |y  |  |  | 
|3  |a  |z  |k  |  | 
|4  |b  |  |  |  | 
|5  |c  |  |  |  | 
|6  |c  |q  |  |  | 
------------------------------------------ 

我真的不知道如何攻击这个问题。我想通过邮编分区，但这仍然留下找到每个分区内共同的祖先的问题...

Answer 1

这是一个非常混乱的解决方案，但它似乎给出了正确的答案。毫无疑问，需要相当多的工作来满足您的实际需求，但也许这可能会帮助您指引某种方向！

-- Setup a test table 

DECLARE @tbl AS TABLE(R1 NVARCHAR(10), R2 NVARCHAR(10), R3 NVARCHAR(10), R4 NVARCHAR(10), PC NVARCHAR(10)); 

INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('a','x','i',NULL,'1'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('a','y','i',NULL,'2'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('a','y','j',NULL,'2'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('a','z','k',NULL,'3'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('b','u','m',NULL,'4'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('b',NULL,'n',NULL,'4'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('c',NULL,NULL,NULL,'5'); 
INSERT INTO @tbl(R1,R2,R3,R4,PC) VALUES ('c','q',NULL,NULL,'6'); 

-- Calculate the result: 

SELECT 
    PC, 
    CASE WHEN LVL1 = 1 THEN R1 ELSE NULL END AS R1, 
    CASE WHEN LVL2 = 1 THEN R2 ELSE NULL END AS R2, 
    CASE WHEN LVL3 = 1 THEN R3 ELSE NULL END AS R3, 
    CASE WHEN LVL4 = 1 THEN R4 ELSE NULL END AS R4 
FROM 
(
    SELECT 
     PC, 
     MAX(R1) AS R1, 
     MAX(R2) AS R2, 
     MAX(R3) AS R3, 
     MAX(R4) AS R4, 
     COUNT(DISTINCT ISNULL(R1,'.')) AS LVL1, 
     COUNT(DISTINCT ISNULL(R1,'.') + ISNULL(R2,'.')) AS LVL2, 
     COUNT(DISTINCT ISNULL(R1,'.') + ISNULL(R2,'.') + ISNULL(R3,'.')) AS LVL3, 
     COUNT(DISTINCT ISNULL(R1,'.') + ISNULL(R2,'.') + ISNULL(R3,'.') + ISNULL(R4,'.')) AS LVL4 
    FROM @tbl 
    GROUP BY PC 
) A 

最终结果与问题中的表格匹配。

Answer 2

这个问题，而吸引了我，我想出了一个替代方案，这可能对您有用：

-- Setup test table 
DECLARE @InputTable TABLE (region1 varchar(2), region2 varchar(2), region3 varchar(2), region4 varchar(2), postcode varchar(2)) 

INSERT INTO @InputTable (region1, region2, region3, region4, postcode) 
      SELECT 'a','x','i',null,'1' 
UNION ALL SELECT 'a','y','i',NULL,'2' 
UNION ALL SELECT 'a','y','j',NULL,'2' 
UNION ALL SELECT 'a','z','k',NULL,'3' 
UNION ALL SELECT 'b','u','m',NULL,'4' 
UNION ALL SELECT 'b',NULL,'n',NULL,'4' 
UNION ALL SELECT 'c',NULL,NULL,NULL,'5' 
UNION ALL SELECT 'c','q',NULL,NULL,'6' 

-- Find the common ancestors 
;with totals as (
    select postcode, count(*) as postcodeCount from @InputTable group by postcode 
) 
, region4group as (
    select postcode, region1, region2, region3, region4 from @InputTable in1 
    group by postcode, region1, region2, region3, region4 having count(*)=(select postCodeCount from totals where totals.postcode=in1.postcode) 
) 
, region3group as (
    select * from region4group 
    union 
    select in1.postcode, in1.region1, in1.region2, in1.region3, null from @InputTable in1 
    left outer join region4group on region4group.postcode=in1.postcode 
    where region4group.postcode is null 
    group by in1.postcode, in1.region1, in1.region2, in1.region3 
    having count(*)=(select postCodeCount from totals where totals.postcode=in1.postcode) 
) 
, region2group as (
    select * from region3group 
    union 
    select in1.postcode, in1.region1, in1.region2, null, null from @InputTable in1 
    left outer join region3group on region3group.postcode=in1.postcode 
    where region3group.postcode is null 
    group by in1.postcode, in1.region1, in1.region2 
    having count(*)=(select postCodeCount from totals where totals.postcode=in1.postcode) 
) 
, commonancestors as (
    select * from region2group 
    union 
    select in1.postcode, in1.region1, null, null, null from @InputTable in1 
    left outer join region2group on region2group.postcode=in1.postcode 
    where region2group.postcode is null 
    group by in1.postcode, in1.region1 
    having count(*)=(select postCodeCount from totals where totals.postcode=in1.postcode) 
) 
select * from commonancestors