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Stack Overflow用户
提问于 2013-11-09 19:59:42
回答 2查看 2.2K关注 0票数 2

我有两个规范化的 Server 2008表,一个用于名称,另一个用于电子邮件:

代码语言:javascript
复制
create table Name (NameId int, Name varchar(50))
create table Email (NameId int, Email varchar(50))
go

insert into Name values (1, 'JOHN SMITH')
insert into Name values (2, 'MARY SMITH')
insert into Name values (3, 'PHILL TAYLOR')
go

insert into Email values (1, 'john@hotmail.com')
insert into Email values (1, 'john@gmail.com')
insert into Email values (1, 'john.smith@hotmail.com')
insert into Email values (2, 'mary@hotmail.com')
insert into Email values (3, 'phill@hotmail.com')
insert into Email values (3, 'phill@gmail.com')
insert into Email values (3, 'taylor.phill@hotmail.com')
insert into Email values (3, 'taylor.phill@gmail.com')
go

当我加入这些表时,我有几行,其中一行只有一封电子邮件:

代码语言:javascript
复制
Select name, email
from Name inner join Email on name.NameId=email.NameId


NAME            EMAIL
-------------   --------------------------
JOHN SMITH      john@hotmail.com
JOHN SMITH      john@gmail.com
JOHN SMITH      john.smith@hotmail.com
MARY SMITH      mary@hotmail.com
PHILL TAYLOR    phill@hotmail.com
PHILL TAYLOR    phill@gmail.com
PHILL TAYLOR    taylor.phill@hotmail.com
PHILL TAYLOR    taylor.phill@gmail.com

但我需要将所有电子邮件放在同一行,以满足预先定义的布局:

代码语言:javascript
复制
NAME            EMAIL1              EMAIL2          EMAIL3                      EMAIL4
-------------   -----------------   --------------- -----------------------     -----------------
JOHN SMITH      john@hotmail.com    john@gmail.com  john.smith@hotmail.com
MARY SMITH      mary@hotmail.com
PHILL TAYLOR    phill@hotmail.com   phill@gmail.com taylor.phill@hotmail.com    taylor.phill@hotmail.com

是否可以使用单个选择?

任何想法都将不胜感激。

提亚

里卡多·诺格拉

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2013-11-09 20:44:08

这可以在一个选择中完成。如果您只需要前4个电子邮件地址,并将其用作特定列。使用电子邮件表上的RowNumber函数对每个用户的邮件地址进行编号。此示例假定电子邮件地址应按字母顺序排序:

代码语言:javascript
复制
with m as (
  select 
    ROW_NUMBER() over (partition by nameId order by email) as Nr, 
    email.NameId, email.Email 
  from email
)
select 
  name.NameId, 
  name.Name, 
  m1.Email as Mail1,
  m2.Email as Mail2,
  m3.Email as Mail3,
  m4.Email as Mail4
from name
left join m m1 on (m1.Nr=1 and m1.NameId=name.NameId)
left join m m2 on (m2.Nr=2 and m2.NameId=name.NameId)
left join m m3 on (m3.Nr=3 and m3.NameId=name.NameId)
left join m m4 on (m4.Nr=4 and m4.NameId=name.NameId)
票数 2
EN

Stack Overflow用户

发布于 2013-11-09 20:40:39

它不漂亮,也有使用上的问题,等等,但是如果你必须的话。你必须..。所以,使用你的模式,这个:-

代码语言:javascript
复制
create function dbo.emails(@NameId int, @Ordinal int)
returns varchar(1024)
as begin
    declare @returnValue varchar(1024)
    select @returnValue=email
    from (
        select 
            ROW_NUMBER() over (order by e.Email) as ordinal,
            e.Email
        from Email e
        where e.NameId=@NameId
    ) p1 
    where p1.ordinal=@Ordinal
    return @returnValue
end
go
select e.Name,
    dbo.emails(n.NameId,1) as email1,
    dbo.emails(n.NameId,2) as email2,
    dbo.emails(n.NameId,3) as email3,
    dbo.emails(n.NameId,4) as email4
from Name n
go

生产:-

代码语言:javascript
复制
Name          email1                  email2             email3                  email4
JOHN SMITH    john.smith@hotmail.com  john@gmail.com     john@hotmail.com        NULL
MARY SMITH    mary@hotmail.com        NULL               NULL                    NULL
PHILL TAYLOR  phill@gmail.com         phill@hotmail.com  taylor.phill@gmail.com  taylor.phill@hotmail.com
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/19882393

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