如何强制查询仅返回一条记录？

Question

我收到以下错误：

'子查询返回的值超过1。 当子查询遵循=，！=，...时，这是不允许的

这是我的查询：

let getLinkTopicsSql = "SELECT       Topic.Id, 
                                     Topic.Name, 
                                     isnull( (select 1 from FeaturedTopic where TopicId = Topic.Id),0) as IsFeatured 

                        FROM         Topic
                        INNER JOIN   LinkTopic
                                        ON   LinkTopic.TopicId = Topic.Id
                        INNER JOIN   Link
                                        ON   LinkTopic.LinkId =  Link.Id
                        WHERE        Link.Id = @LinkId"

我试过了：

SELECT TOP 1

像这样：

let getLinkTopicsSql = "SELECT TOP 1 Topic.Id, 
                                     Topic.Name, 
                                     isnull( (select 1 from FeaturedTopic where TopicId = Topic.Id),0) as IsFeatured 

                        FROM         Topic
                        INNER JOIN   LinkTopic
                                        ON   LinkTopic.TopicId = Topic.Id
                        INNER JOIN   Link
                                        ON   LinkTopic.LinkId =  Link.Id
                        WHERE        Link.Id = @LinkId"

但是我仍然收到同样的异常。

我不确定该如何解决...

Answer 1

我想你想要exists ：

(case when exists (select 1 from FeaturedTopic where TopicId = Topic.Id)
      then 1 else 0
 end) as IsFeatured 

您可以使用isnull()来完成此操作，如下所示：

isnull( (select top (1) 1 from FeaturedTopic where TopicId = Topic.Id), 0) as IsFeatured 

是否：

select coalesce(max(1), 0) from FeaturedTopic where TopicId = Topic.Id) as IsFeatured 

但我认为exists是最清晰的形式。