Question

我有三张牌桌：File，Word和WordInstance：

CREATE TABLE [dbo].[File](
    [FileId] [int] IDENTITY(1,1) NOT NULL,
    [FileName] [nchar](10) NOT NULL)

CREATE TABLE [dbo].[Word](
    [WordId] [int] IDENTITY(1,1) NOT NULL,
    [Word] [nchar](10) NOT NULL,
    [LatestFileId] [int] NOT NULL)

CREATE TABLE [dbo].[WordInstance](
    [WordInstanceId] [int] IDENTITY(1,1) NOT NULL,
    [WordId] [int] NOT NULL,
    [FileId] [int] NOT NULL)

请注意，我省略了外键以简化此操作。给定FileId我想返回一个true / false值，告诉我是否有其他文件与指定的FileId具有相同的单词。

从这开始，我知道它不起作用，但按原样提供：

CREATE FUNCTION [dbo].[DoesFileContainLastWord]
(
    @fileId INT
)
RETURNS BIT
AS
BEGIN
    DECLARE @count  INT
    DECLARE @ret    BIT

    SELECT @count = COUNT([tW].[WordId])
    FROM [Word] AS tW
    INNER JOIN [WordInstance] AS tWI 
        ON [tW].[WordId] = [tWI].[WordId]
    INNER JOIN [File] AS tF
        ON [tF].[FileId] = [tW].[LatestFileId]
    WHERE [tF].[FileId] = @fileId

    IF @count > 0
        BEGIN
            SET @ret = 0
        END
    ELSE
        SET @ret = 1

    RETURN @ret

END;

Answer 1

在SQL Server 2005上测试：

declare @file table (fileid int)
declare @instance table (fileid int,wordid int)
insert into @file (fileid)
select 1 union all select 2
insert into @instance (fileid,wordid) 
select 1,1 
union all select 1,2 
union all select 1,3 
union all select 2,1 
union all select 2,2
declare @fileid int
set @fileid=2
;with fvalues as
(
    select distinct wordid from @instance where fileid=@fileid
)
select case when exists
(
    select * 
    from fvalues v
    inner join @instance i on v.wordid = i.wordid
    and i.fileid<>@fileid
    group by i.fileid
    having count(distinct i.wordid) >= (select count(*) from fvalues)
) 
then cast(1 as bit)
else cast(0 as bit) end

@fileid=1返回0，@fileid=2返回1，因为文件1的字集是文件2的正确超集。

在SQL Server中查找多个到多个超集

1 个答案: