1

Okay so say I have something like this:

ID | Name | Address
 1 | Bob  | 123 Fake Street
 1 | Bob  | 221 Other Street

done by doing something like:

select p.ID, p.Name a.Address from People p 
inner join Addresses a on a.OwnerID = p.ID

Is there any way to turn that into 

ID | Name |    Address_1    |     Address_2    | etc...
 1 | Bob  | 123 Fake Street | 221 Other street | etc

I've seen things that do comma separated values in one column but I don't want that I want distinct columns. I am querying this using MSSQL and C# I don't know if that changes anything. Also this is a made up scenario that is just similar to what I'm doing so the actual structure of the tables can't be changed.

Anyone have any suggestions?

4

1 回答 1

4

您可以使用 PIVOT 函数来获取结果，但您还必须使用 a 来实现，row_number()这样您就可以将每个人的多个地址转换为列。

如果您有已知数量的地址，那么您将对查询进行硬编码：

select id, name, address_1, address_2
from
(
  select p.id, p.name, a.address,
    'Address_'+cast(row_number() over(partition by p.id 
                                      order by a.ownerid) as varchar(10)) rn
  from people p
  inner join addresses a
    on p.id = a.ownerid
) d
pivot
(
  max(address)
  for rn in (address_1, address_2)
) piv;

请参阅SQL Fiddle with Demo。

但是，如果您的情况是，每个人的地址数量未知，因此您需要使用动态 SQL 并将其放入存储过程中执行：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME('Address_'+d.rn) 
                    from 
                    (
                      select cast(row_number() over(partition by a.ownerid
                                      order by a.ownerid) as varchar(10)) rn
                      from Addresses a
                    ) d
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT id, name, ' + @cols + ' 
            from
            (
              select p.id, p.name, a.address,
                ''Address_''+cast(row_number() over(partition by p.id 
                                                  order by a.ownerid) as varchar(10)) rn
              from people p
              inner join addresses a
                on p.id = a.ownerid
            ) d
            pivot 
            (
                max(address)
                for rn in (' + @cols + ')
            ) p '

execute(@query);

请参阅SQL Fiddle with Demo。这些都给出了结果：

| ID | NAME |         ADDRESS_1 |        ADDRESS_2 | ADDRESS_3 |
----------------------------------------------------------------
|  1 |  Bob |   123 Fake Street | 221 Other Street |    (null) |
|  2 |  Jim | 123 e main street |           (null) |    (null) |
|  3 |  Tim |   489 North Drive |   56 June Street |  415 Lost |
于 2013-06-13T13:20:35.913 回答