Question

我有一些数据

id    ref
==   ==========
1    3536757616
1    3536757617
1    3536757618

并希望得到结果

1    3536757616/7/8

所以基本上数据是在id上聚合的，refs连接在一起，用斜杠'/'分隔，但删除了任何公共前缀，所以如果数据就像

id    ref
==   ==========
2    3536757628
2    3536757629
2    3536757630

我想得到结果

2    3536757629/28/30

我知道我可以使用

简单地连接引用

SELECT distinct
    id,
    stuff ( ( SELECT
                  '/ ' + ref 
              FROM
                  tableA tableA_1
              where tableA_1.id = tableA_2.id
    FOR XML PATH ( '' ) ) , 1 , 2 , '' )
from TableA tableA_2

给予

1   3536757616/ 3536757617/ 3536757618
2   3536757628/ 3536757629/ 3536757630

但是它消除了我追求的共同元素......

测试数据代码：

create table tableA (id int, ref varchar(50))

insert into tableA
select 1, 3536757616
union select 1, 3536757617
union select 1, 3536757618
union select 2, 3536757628
union select 2, 3536757629
union select 2, 3536757630

Answer 1

WITH hier(cnt) AS
        (
        SELECT  1
        UNION ALL
        SELECT  cnt + 1
        FROM    hier
        WHERE   cnt <= 100
        )
SELECT  CASE WHEN ROW_NUMBER() OVER (ORDER BY id) = 1 THEN ref ELSE ' / ' + SUBSTRING(ref, mc + 1, LEN(ref)) END 
FROM    (
        SELECT  MIN(common) AS mc
        FROM    (
                SELECT  (
                        SELECT  MAX(cnt)
                        FROM    hier
                        WHERE   SUBSTRING(initref, 1, cnt) = SUBSTRING(ref, 1, cnt)
                                AND cnt <= LEN(ref)
                        ) AS common
                FROM    (
                        SELECT  TOP 1 ref AS initref
                        FROM    tableA
                        ) i,
                        tableA
                ) q
        ) q2, tableA
FOR XML PATH('')

---

3536757616 / 17 / 18 / 28 / 29 / 30

与团体相同：

WITH hier(cnt) AS
        (
        SELECT  1
        UNION ALL
        SELECT  cnt + 1
        FROM    hier
        WHERE   cnt <= 100
        )
SELECT  (
        SELECT  CASE WHEN ROW_NUMBER() OVER (ORDER BY a2.ref) = 1 THEN ref ELSE ' / ' + SUBSTRING(ref, mc + 1, LEN(ref)) END 
        FROM    tableA a2
        WHERE   a2.id = q2.id
        FOR XML PATH('')
        )
FROM    (
        SELECT  id, MIN(common) AS mc
        FROM    (
                SELECT  a.id,
                        (
                        SELECT  MAX(cnt)
                        FROM    hier
                        WHERE   SUBSTRING(i.initref, 1, cnt) = SUBSTRING(a.ref, 1, cnt)
                                AND cnt <= LEN(ref)
                        ) AS common
                FROM    (
                        SELECT  id, MIN(ref) AS initref
                        FROM    tableA
                        GROUP BY
                                id
                        ) i
                JOIN    tableA a
                ON      i.id = a.id
                ) q
        GROUP BY
                id
        ) q2
---
3536757616 / 7 / 8
3536757628 / 29 / 30

Answer 2

我将我的表命名为#T，并使用以下SELECT语句

select id, number, substring(#t.ref, 1, v.number), count(id)
from master.dbo.spt_values v
inner join #t on v.number <= len(#t.ref)
where v.name is null and v.number > 0 
group by id, number, substring(#t.ref, 1, v.number)
order by id, count(id) desc, number desc

你得到一个结果集，其中每个id的第一条记录包含每个id的最大长度和最长的初始字符串。

这不是一个完整的解决方案，而是一个很好的起点：迭代id，发出SELECT TOP 1来检索最长的字符串，并为每个具有相同id的记录连接字符串差异。

Answer 3

declare @tableA table(id int, ref varchar(50))
declare @suffix table(id int, suffix varchar(50))
declare @todo table(id int)

insert into @tableA
select 1, '3536757616'
union select 1, '3536757617'
union select 1, '3536757618'
union select 2, '3536757628'
union select 2, '3536757629'
union select 2, '3536757630'

insert into @suffix
select * from @tableA

insert into @todo
select s1.id
from (
    select id, cnt = count(*)
    from @suffix
    group by id, substring(suffix, 1, 1) 
  ) s1 
  inner join (
    select id, cnt = count(*)
    from @suffix
    group by id
  ) s2 on s2.id = s1.id and s2.cnt = s1.cnt

while exists (select * from @todo)
begin
  update @suffix
  set suffix = substring(suffix, 2, len(suffix) - 1)
  from @suffix s
       inner join @todo t on t.id = s.id

  delete from @todo

  insert into @todo
  select s1.id
  from (
      select id, cnt = count(*)
      from @suffix
      group by id, substring(suffix, 1, 1) 
    ) s1 
    inner join (
      select id, cnt = count(*)
      from @suffix
      group by id
    ) s2 on s2.id = s1.id and s2.cnt = s1.cnt  
end

select * from @suffix

用于连接和删除公共前缀的SQL查询

3 个答案: