У меня есть следующий набор результаты от объединения трех таблиц, А.Н. статьи стола, продуктов столика, статьи на продукты отображение таблиц.Проблемы с дубликатами в SQL Регистрация
Я хотел бы, чтобы результаты с дубликатами удалялись аналогично выделенному идентификатору содержимого.
Текущий набор результатов:
[ContendId] [Title] [productId]
1 article one 2
1 article one 3
1 article one 9
4 article four 1
4 article four 10
4 article four 14
5 article five 1
6 article six 8
6 article six 10
6 article six 11
6 article six 13
7 article seven 14
Желаемый набор результатов:
[ContendId] [Title] [productId]
1 article one *
4 article four *
5 article five *
6 article six *
7 article seven *
Вот сгущенное пример соответствующего SQL:
IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'tempdb.dbo.products') AND type = (N'U'))
drop table tempdb.dbo.products
go
CREATE TABLE tempdb.dbo.products (
productid int,
productname varchar(255)
)
go
IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articles') AND type = (N'U'))
drop table tempdb.dbo.articles
go
create table tempdb.dbo.articles (
contentid int,
title varchar(255)
)
IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articles') AND type = (N'U'))
drop table tempdb.dbo.articles
go
create table tempdb.dbo.articles (
contentid int,
title varchar(255)
)
IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articleproducts') AND type = (N'U'))
drop table tempdb.dbo.articleproducts
go
create table tempdb.dbo.articleproducts (
contentid int,
productid int
)
insert into tempdb.dbo.products values (1,'product one'),
(2,'product two'),
(3,'product three'),
(4,'product four'),
(5,'product five'),
(6,'product six'),
(7,'product seven'),
(8,'product eigth'),
(9,'product nine'),
(10,'product ten'),
(11,'product eleven'),
(12,'product twelve'),
(13,'product thirteen'),
(14,'product fourteen')
insert into tempdb.dbo.articles VALUES (1,'article one'),
(2, 'article two'),
(3, 'article three'),
(4, 'article four'),
(5, 'article five'),
(6, 'article six'),
(7, 'article seven'),
(8, 'article eight'),
(9, 'article nine'),
(10, 'article ten')
INSERT INTO tempdb.dbo.articleproducts VALUES (1,2),
(1,3),
(1,9),
(4,1),
(4,10),
(4,14),
(5,1),
(6,8),
(6,10),
(6,11),
(6,13),
(7,14)
GO
select DISTINCT(a.contentid), a.title, p.productid from articles a
JOIN articleproducts ap ON a.contentid = ap.contentid
JOIN products p ON a.contentid = ap.contentid AND p.productid = ap.productid
ORDER BY a.contentid
+1 Включая стол продуктов, вызывал обмана, спасибо! –