누락 된 관계에 대한 대체/기본값이있는보기 만들기

Question

개체의 여러 일대 다 관계가 있고 각각이 별도의 테이블에서 오는 이름 특성을 사용하는 설치에서.

Building(BuildingName), Floor(FloorName) 

건물이 2 명 (2 languageIDs) 만 3 층이 모두 언어 ID의 이름이 5 층에있는 경우 예를

를 들어, 나는 10 개 결과 항목이 여전히 싶다. 언어 이름이 누락되었을 때 누락 된 바닥 이름은 일치하지 않는 층 ID에서 가져옵니다 (기본값).

Answer 1

당신이 겪고있는 일에 맞는 줄이 있습니까? 여기

의 Runnable 예 : http://sqlfiddle.com/#!3/894e9/4

if object_id('[FloorName]') is not null drop table [FloorName] 
if object_id('[BuildingName]') is not null drop table [BuildingName] 
if object_id('[Floor]') is not null drop table [Floor] 
if object_id('[Building]') is not null drop table [Building] 
if object_id('[Language]') is not null drop table [Language] 

create table [Language] 
(
    Id bigint not null identity(1,1) primary key clustered 
    , code nvarchar(5) 
) 
create table [Building] 
(
    Id bigint not null identity(1,1) primary key clustered 
    , something nvarchar(64) 
) 
create table [Floor] 
(
    Id bigint not null identity(1,1) primary key clustered 
    , BuildingId bigint foreign key references [Building](Id) 
    , something nvarchar(64) 
) 
create table [BuildingName] 
(
    Id bigint not null identity(1,1) primary key clustered 
    , BuildingId bigint foreign key references [Building](Id) 
    , LanguageId bigint foreign key references [Language](Id) 
    , name nvarchar(64) 
) 
create table [FloorName] 
(
    Id bigint not null identity(1,1) primary key clustered 
    , FloorId bigint foreign key references [Floor](Id) 
    , LanguageId bigint foreign key references [Language](Id) 
    , name nvarchar(64) 
) 

insert [Language] 
     select 'en-us' 
union select 'en-gb' 
union select 'fr' 

insert [Building] 
     select 'B1' 
union select 'B2' 

insert [Floor] 
     select 1, 'F1.1' 
union select 1, 'F1.2' 
union select 1, 'F1.3' 
union select 1, 'F1.4' 
union select 1, 'F1.5' 
union select 2, 'F2.1' 
union select 2, 'F2.2' 
union select 2, 'F2.3' 
union select 2, 'F2.4' 
union select 2, 'F2.5' 

insert BuildingName 
select b.Id 
, l.id 
, 'BuildingName :: ' + b.something + ' ' + l.code 
from [Building] b 
cross join [Language] l 
where l.code in ('en-us', 'fr') 

insert FloorName 
select f.Id 
, l.Id 
, 'FloorName :: ' + f.something + ' ' + l.code 
from [Floor] f 
cross join [Language] l 
where f.something in ('F1.1', 'F1.2', 'F2.1') 
and l.code in ('en-us', 'fr') 

insert FloorName 
select f.Id 
, l.Id 
, 'FloorName :: ' + f.something + ' ' + l.code 
from [Floor] f 
cross join [Language] l 
where f.something not in ('F1.1', 'F1.2', 'F2.1') 
and l.code in ('en-us') 


declare @defaultLanguageId bigint 
select @defaultLanguageId = id from [Language] where code = 'en-us' --default language is US English 

select b.Id 
, b.something 
, bn.name 
, isnull(bfn.name, bfnDefault.name) 
, bl.code BuildingLanguage 
from [Building] b 
inner join [BuildingName] bn 
    on bn.BuildingId = b.Id 
inner join [Language] bl 
    on bl.Id = bn.LanguageId 
inner join [Floor] bf 
    on bf.BuildingId = b.Id 
left outer join [FloorName] bfn 
    on bfn.FloorId = bf.Id 
    and bfn.LanguageId = bl.Id 
left outer join [Language] bfl 
    on bfl.Id = bfn.LanguageId 
left outer join [FloorName] bfnDefault 
    on bfnDefault.FloorId = bf.Id 
    and bfnDefault.LanguageId = @defaultLanguageId 

편집

이 버전은 기본적으로 어떤 언어 : 나는 당신이 될 수있다 생각하지만, 내가, 당신이 정말로 원하는 것을 아무 생각이

select b.Id 
, b.something 
, bn.name 
, isnull(bfn.name, (select top 1 name from [FloorName] x where x.FloorId=bf.Id)) 
, bl.code BuildingLanguage 
from [Building] b 
inner join [BuildingName] bn 
    on bn.BuildingId = b.Id 
inner join [Language] bl 
    on bl.Id = bn.LanguageId 
inner join [Floor] bf 
    on bf.BuildingId = b.Id 
left outer join [FloorName] bfn 
    on bfn.FloorId = bf.Id 
    and bfn.LanguageId = bl.Id 
left outer join [Language] bfl 
    on bfl.Id = bfn.LanguageId