将一个表中的多个列连接到另一个表中的单个列

I am looking to create a view that pulls data from two tables "Schedule" and "Reference".

我期待创建一个视图,从两个表“Schedule”和“Reference”中提取数据。

Schedule has 50+ columns (it's almost completely denormalized -- not my design), most of which contain a value that could be joined to a column in the Reference table.

Schedule有50多列(它几乎完全非规范化 - 不是我的设计),其中大部分都包含一个可以连接到Reference表中的列的值。

How do I write the SQL statement to correctly join each column in Schedules to the single column in Reference?

如何编写SQL语句以将Schedules中的每个列正确连接到Reference中的单个列?

The Schedule table is defined as:

Schedule表定义为:

    CREATE TABLE [dbo].[Schedule](
    [ID] [int] NOT NULL,
    [SCHEDULEWEEK] [datetime] NOT NULL,
    [EMPNO] [numeric](10, 0) NOT NULL,
    [EMPLNAME] [varchar](32) NULL,
    [EMPFNAME] [varchar](32) NULL,
    [EMPSENDATE] [datetime] NULL,
    [EMPHIREDATE] [datetime] NULL,
    [EMPTYPE] [char](1) NULL,
    [EMPSTATUS] [char](1) NULL,
    [SNREFUSALS] [tinyint] NULL,
    [QUALSTRING] [varchar](128) NULL,
    [JOBOVERSHIFTTYPE] [bit] NULL,
    [SHORTNOTICE] [bit] NULL,
    [SHORTNOTICEWAP] [bit] NULL,
    [SHORTNOTICEPHONE] [varchar](32) NULL,
    [LEADHAND] [bit] NULL,
    [DUALCURRENCY] [bit] NULL,
    [MIN100WINDOW] [bit] NULL,
    [STATHOLIDAY] [bit] NULL,
    [AREAOVERHOURS] [bit] NULL,
    [DOUBLEINTERZONES] [bit] NULL,
    [MAXDAYSPERWEEK] [tinyint] NULL,
    [MAXHOURSPERWEEK] [numeric](10, 2) NULL,
    [MAXHOURSPERSHIFT] [numeric](10, 2) NULL,
    [MAXDOUBLESPERWEEK] [tinyint] NULL,
    [ASSIGNEDDAYS] [tinyint] NULL,
    [ASSIGNEDHOURS] [numeric](10, 2) NULL,
    [ASSIGNEDDOUBLES] [tinyint] NULL,
    [ASSIGNEDLOAHOURS] [numeric](10, 2) NULL,
    [SHIFTNO1] [int] NULL,
    [TEXT1_1] [varchar](64) NULL,
    [TEXT2_1] [varchar](64) NULL,
    [DAYFLAG1] [bit] NULL,
    [COMMENT1] [text] NULL,
    [SHIFTNO2] [int] NULL,
    [TEXT1_2] [varchar](64) NULL,
    [TEXT2_2] [varchar](64) NULL,
    [DAYFLAG2] [bit] NULL,
    [COMMENT2] [text] NULL,
    [SHIFTNO3] [int] NULL,
    [TEXT1_3] [varchar](64) NULL,
    [TEXT2_3] [varchar](64) NULL,
    [DAYFLAG3] [bit] NULL,
    [COMMENT3] [text] NULL,
    [SHIFTNO4] [int] NULL,
    [TEXT1_4] [varchar](64) NULL,
    [TEXT2_4] [varchar](64) NULL,
    [DAYFLAG4] [bit] NULL,
    [COMMENT4] [text] NULL,
    [SHIFTNO5] [int] NULL,
    [TEXT1_5] [varchar](64) NULL,
    [TEXT2_5] [varchar](64) NULL,
    [DAYFLAG5] [bit] NULL,
    [COMMENT5] [text] NULL,
    [SHIFTNO6] [int] NULL,
    [TEXT1_6] [varchar](64) NULL,
    [TEXT2_6] [varchar](64) NULL,
    [DAYFLAG6] [bit] NULL,
    [COMMENT6] [text] NULL
-- Snip
) ON [PRIMARY]

And the Reference table is defined as:

参考表定义为:

CREATE TABLE [dbo].[Reference](
    [ID] [int] NOT NULL,
    [CODE] [varchar](21) NOT NULL,
    [LOCATIONCODE] [varchar](4) NOT NULL,
    [SCHAREACODE] [varchar](16) NOT NULL,
    [LOCATIONNAME] [varchar](32) NOT NULL,
    [FLTAREACODE] [varchar](16) NOT NULL
) ON [PRIMARY]

I am trying to join each [TEXT1_]/[TEXT2_] column in Schedule to the [SCHAREACODE] column in reference. All the reference table contains is a list of areas where the employee could work.

我正在尝试将Schedule中的每个[TEXT1 _] / [TEXT2_]列加入参考中的[SCHAREACODE]列。所有参考表都包含员工可以工作的区域列表。

5 个解决方案

#1

I think he means to join on the Reference table multiple times:

我认为他意味着多次加入Reference表:

SELECT *
  FROM Schedule AS S
 INNER JOIN Reference AS R1 
         ON R1.ID = S.FirstID 
 INNER JOIN Reference AS R2 
         ON R2.ID = S.SecondID 
 INNER JOIN Reference AS R3 
         ON R3.ID = S.ThirdID 
 INNER JOIN Reference AS R4 
         ON R4.ID = S.ForthID

#2

Your description is a bit lacking, so I'm going to assume that

你的描述有点缺乏,所以我会假设

Schedule has 50+ columns (it's almost completely denormalized -- not my design), most of which contain a value that could be joined to a column in the Reference table.

Schedule有50多列(它几乎完全非规范化 - 不是我的设计),其中大部分都包含一个可以连接到Reference表中的列的值。

means that 1 of the 50+ columns in Schedule is a ReferenceId. So, given a table design like:

表示Schedule中50个列中的1列是ReferenceId。所以,给定一个表格设计:

Schedule ( MaybeReferenceId1, MaybeReferenceId2, MaybeReferenceId3, ... )
Reference ( ReferenceId )

Something like:

SELECT *
FROM Schedule
JOIN Reference ON
     Schedule.MaybeReferenceId1 = Reference.ReferenceId
     OR Schedule.MaybeReferenceId2 = Reference.ReferenceId
     OR Schedule.MaybeReferenceId3 = Reference.ReferenceId
     OR Schedule.MaybeReferenceId4 = Reference.ReferenceId
     ...

would work. You could simplify it by using IN if your RDBMS supports it:

会工作。如果您的RDBMS支持它,您可以使用IN来简化它:

SELECT *
FROM Schedule
JOIN Reference ON
     Reference.ReferenceId IN (
        Schedule.MaybeReferenceId1,
        Schedule.MaybeReferenceId2,
        Schedule.MaybeReferenceId3,
        Schedule.MaybeReferenceId4,
        ...
     )

#3

From updated question

从更新的问题

Perhaps something like this? It will be messy no matter what you do.

也许是这样的?无论你做什么,它都会很混乱。

SELECT S.ID
  S.TEXT1_1,
  TEXT1_1_RID = COALESCE((SELECT MAX(R.ID) FROM Reference R WHERE R.SCHAREACODE = S.TEXT1_1), 0),
  S.TEXT1_2,
  TEXT1_2_RID = COALESCE((SELECT MAX(R.ID) FROM Reference R WHERE R.SCHAREACODE = S.TEXT1_2), 0),
  ...
FROM Schedule S

#4

Agree with TheSoftwareJedi, but can I just suggest using LEFT JOINs so that failures-to-match don't cause your Schedule row to disappear?

同意TheSoftwareJedi,但我可以建议使用LEFT JOIN,以便匹配失败不会导致您的Schedule行消失吗?

Of course, doing 28 JOINs is going to be a bit cumbersome whatever the details.

当然,无论细节如何,做28个JOIN都会有点麻烦。

I'm not sure I'd call this "denormalized", more "abnormalized" ... :-)

我不确定我会称之为“非规范化”,更“异常”...... :-)

#5

Try a query like this:

尝试这样的查询:

select s.*, r.schareacode from schedule s, 
where 
s.text1_1 = s.schareacode
or s.text2_1 = s.schareacode
or s.textx_x = s.schareacode
..

You should be able to get the same results with traditional joins so I recommend you experiment with that as well.

您应该能够使用传统联接获得相同的结果,因此我建议您也尝试使用它。

#1