Question

这对我来说很难解释所以如果需要进一步澄清，请随时告诉我。我有5张桌子：

艺术品{artwork_id，artwork_title，artwork_artist_id}
Artwork_Insurance {artwork_id，insurance_id}
保险{insurance_id，insurance_name}
Artwork_Exhibit {artwork_id，exhibit_id}
展览{exhibit_id，exhibit_name，exhibit_start_date，exhibit_end_date}

每件艺术品可以有几种保险，但每件艺术品只能有一个有效展品（确定当前日期是否在展览的开始和结束日期之间）。我的问题是如何检索所有艺术作品行（包括所有艺术作品表的列），应用于该作品的所有保险（该艺术品逗号分隔的insurance_name列中的值）以及该艺术作品的exhibit_name如果一个当前处于活动状态，则返回null值。

我的SQL知识非常有限，而且我希望实现的很多东西都没有被刷新，但到目前为止我还有：

select distinct 
    dbo.Artwork.artwork_id, 
    dbo.Artwork.artwork_title,
    dbo.Artwork.artwork_artist_id,
    dbo.Insurance.insurance_name, 
    dbo.Exhibit.exhibit_name
from
    dbo.Artwork_Insurance
left join 
    dbo.Artwork on dbo.Artwork_Insurance.artwork_id = dbo.Artwork.artwork_id
left join 
    dbo.Insurance on dbo.Insurance.insurance_id = dbo.Artwork_Insurance.insurance_id
left join 
    dbo.Artwork_Exhibit on dbo.Artwork_Exhibit.artwork_id = dbo.Artwork.artwork_id
left join 
    dbo.Exhibit on dbo.Exhibit.exhibit_id = dbo.Artwork_Exhibit.exhibit_id;

但是，这不仅会返回与运行select * from dbo.Artwork相同的行数，而且旧的展品也不会被过滤掉，保险名称也不会在同一个单元格中以逗号分隔。我已经对左连接做了一些阅读，但我不认为我正确地使用它们来完成我之后的事情。我已经在这方面工作了好几天，可以真正使用一些帮助。任何帮助赞赏。谢谢！

Answer 1

首先获得逗号分隔保险。它可以通过多种方式完成。我使用过for xml。但是还有其他方法可以使用pivot或group + case。然后加入其他表。我没有检查查询，但它应该正常工作并返回您的预期输出

with cte_insurance as (
    select
        distinct Artwork_Insurance.artwork_id, stuff((
            select
                ', ' + Insurance.insurance_name
            from
                Insurance
            where
                Insurance.insurance_id = Artwork_Insurance.insurance_id
        ), 1, 2, '') insurance_name
    from
        Artwork_Insurance
)

select
    Artwork.artwork_id, Artwork.artwork_title, Artwork.artwork_artist_id
    , cte_insurance.insurance_name, Exhibit.exhibit_name
from
    Artwork
    left join cte_insurance on Artwork.artwork_id = cte_insurance.artwork_id
    left join Artwork_Exhibit on Artwork.artwork_id = Artwork_Exhibit.artwork_id
    left join Exhibit 
        on Artwork_Exhibit.exhibit_id = Exhibit.exhibit_id 
            and getdate() between Exhibit.exhibit_start_date and Exhibit.exhibit_end_date

SQL Server 2014通过两个关系表获取行，以使用主表中的id分隔表

1 个答案: