1

我有一张表（SeekerInfo），喜欢：

--------------------------------------------------
SeekerID -  SeekerName - SeekerSex - SeekerMobile
12121    -  mmmm       - Male      - 067676767
13223    -  ssss       - Female    - 078876767
--------------------------------------------------

和另一个喜欢的表（SeekerCources）（每个搜索者最多有3个课程）：

--------------------------------------------------
SeekerID - CourceName - Duration
12121    - MCSA       - 1 year
12121    - MCPD       - 6 months
13223    - CCNA       - 1 year
13223    - CCNP       - 1 year
13223    - MCTS       - 4 months

我想做一个选择语句，从两个表中预览数据如下所示：

SeekerID -  SeekerName - SeekerSex - SeekerMobile - Cource1 - Cource2 - Cource3
12121    -  mmmm       - Male      - 067676767    - MCSA    - MCPD    - *NULL*
13223    -  ssss       - Female    - 078876767    - CCNA    - CCNP    - MCTS
4

3 回答 3

1

Here's an approach where you assign the number 1-3 to each of the courses and then do three joins 

WITH cte 
     AS (SELECT Row_number() OVER (partition BY SeekerID ORDER BY CourceName) rn 
                , 
                SeekerID, 
                CourceName, 
                Duration 
         FROM   SeekerCources) 
SELECT si.SeekerID, 
       si.SeekerName, 
       si.SeekerSex, 
       si.SeekerMobile, 
       c1.CourceName AS Cource1, 
       c2.CourceName AS Cource2, 
       c3.CourceName AS Cource3 
FROM   SeekerInfo si 
       LEFT JOIN cte c1 
         ON si.SeekerID = c1.SeekerID 
            AND c1.rn = 1 
       LEFT JOIN cte c2 
         ON si.SeekerID = c2.SeekerID 
            AND c2.rn = 2 
       LEFT JOIN cte c3 
         ON si.SeekerID = c3.SeekerID 
            AND c3.rn = 3

DEMO

One unresolved problem is that we don't have a good way to decide which course goes in 1,2,or 3 perhaps if you had a field like date taken you could replace CourceName with that

e.g.
Row_number() OVER (partition BY SeekerID ORDER BY DateTaken)

于 2012-04-27T17:17:26.807 回答
1

It looks like you are going to have a lot of NULL values in the long term which means that you will certainly run into problems sooner or later. Also, you are trying to introduce n number of attributes, n being the largest number of course results, in your result which doesn't make things easier on the client side.

May I suggest a different approach where you will get all results as rows, which means that you will have duplicate data in your result (This doesn't really matter though as you're not storing that).

SELECT i.*, c.CourseName 
FROM SeekerInfo i JOIN SeekerCourses c 
ON i.SeekerID = c.SeekerID;

This will return all courses with the additional seeker information ;)

EDIT: Sorry, I didn't properly read your post - you will always have a maximum of 3 courses. Still, maybe this helps anyway.

于 2012-04-27T17:18:16.573 回答
0

在这里，我试图分三步展示你的答案。

CREATE TABLE [SeekerCources](
    [SeekerID] [nvarchar](50) NULL,
    [CourceName] [nvarchar](50) NULL,
    [Duration] [nvarchar](50) NULL
) 


CREATE TABLE [SeekerInfo](
    [SeekerID] [nvarchar](50) NULL,
    [SeekerName] [nvarchar](50) NULL,
    [SeekerSex] [nvarchar](50) NULL,
    [SeekerMobile] [nvarchar](50) NULL
)

INSERT [SeekerCources] ([SeekerID], [CourceName], [Duration]) VALUES (N'12121', N'MCSA', N'1 year')
INSERT [SeekerCources] ([SeekerID], [CourceName], [Duration]) VALUES (N'12121', N'MCPD', N'6 months')
INSERT [SeekerCources] ([SeekerID], [CourceName], [Duration]) VALUES (N'13223', N'CCNA', N'1 year')
INSERT [SeekerCources] ([SeekerID], [CourceName], [Duration]) VALUES (N'13223', N'CCNP', N'1 year')
INSERT [SeekerCources] ([SeekerID], [CourceName], [Duration]) VALUES (N'13223', N'MCTS', N'4 months')
INSERT [SeekerInfo] ([SeekerID], [SeekerName], [SeekerSex], [SeekerMobile]) VALUES (N'12121', N'mmmm', N'Male', N'067676767')
INSERT [SeekerInfo] ([SeekerID], [SeekerName], [SeekerSex], [SeekerMobile]) VALUES (N'13223', N'ssss', N'Female', N'078876767')

步骤1

Select SeekerID, CourceName, 
'Course' + CAST(ROW_NUMBER() over (partition by SeekerID order by CourceName) as varchar(10)) as CourseSquence
from SeekerCources

第 1 步的结果应如下所示。

步骤 1 结果

演示：链接

第2步 

Select SeekerID, Course1, Course2, Course3
from    
(
    select SeekerID, CourceName, 
    'Course' + CAST(ROW_NUMBER() over (partition by SeekerID order by CourceName) as varchar(10)) as CourseSquence
    from SeekerCources  
) Temp
Pivot
(
    Max(CourceName)
    For CourseSquence in (Course1, Course2, Course3)
) PIV

第 2 步的结果应如下所示。

步骤 2 结果

演示：链接

第 3 步

Select SeekerInfo.*, PIV.SeekerID, Course1, Course2, Course3
from    
(
    select SeekerID, CourceName, 
    'Course' + CAST(ROW_NUMBER() over (partition by SeekerID order by CourceName) as varchar(10)) as CourseSquence
    from SeekerCources  
) Temp
Pivot
(
    Max(CourceName)
    For CourseSquence in (Course1, Course2, Course3)
) PIV
inner join SeekerInfo on SeekerInfo.SeekerID = PIV.SeekerID

第 3 步的结果应如下所示。

步骤 3 结果

演示：链接

解说
视频 1：SQL Server 中的 Pivot - https://youtu.be/h3BtudZehuo
视频 2：将行转换为列 - https://youtu.be/C0mQqDnF7wQ

于 2018-04-16T20:23:01.997 回答