加入非独特和不同的表

时间:2015-06-02 04:42:53

标签: sql sql-server tsql

使用MS SQL SERVER 2008

我有以下情况:

===================
|     Table A     |
===================
|id      |desc      |code 
| 1      |ballpen   |1010
| 2      |pencil    |1010
| 3      |stabilo   |1010
| 4      |pins      |1011
| 5      |clips     |1011

===================
|      Table B    |
===================
| id     |code
| 1010   |AAA
| 1011   |BBB
| 1013   |CCC

我需要完成的事情如下:

===================
|      Table C    |
===================
|id   |desc    |code  |code 2
| 1   |ballpen |1010  |AAA
| 2   |pencil  |1010  |AAA
| 3   |stabilo |1010  |AAA
| 4   |pins    |1011  |BBB
| 5   |clips   |1011  |BBB

这是我到目前为止的情况,不幸的是LEFT JOIN导致最终查询输出膨胀

select *
from table a
  left join table b on a.code = b.id

结果:

========
Table D
========
|id |desc    |code |code2
|1  |ballpen |1010 |AAA
|1  |ballpen |1010 |AAA
|1  |ballpen |1010 |AAA
|2  |pencil  |1010 |AAA
|2  |pencil  |1010 |AAA
|2  |pencil  |1010 |AAA
|3  |stabilo |1010 |AAA
|3  |stabilo |1010 |AAA
|3  |stabilo |1010 |AAA
|4  |pins    |1011 |BBB
|4  |pins    |1011 |BBB
|5  |clips   |1011 |BBB
|5  |clips   |1011 |BBB

3 个答案:

答案 0 :(得分:0)

注意:我认为您在join.let中使用了相同的表名,请尝试使用不同的表名
我使用了Left Join它有效..

create table #master
(
ID int,
descs varchar(15),
code int
)
create table #codes
(
ids int,
codes varchar(15)
)


insert into #master values(1,'ballpen',1010)
insert into #master values(2,'pencil',1010)
insert into #master values(3,'stabilo',1010)
insert into #master values(4,'pins',1011)
insert into #master values(5,'Clips',1011)

insert into #codes values(1010,'AAA')
insert into #codes values(1011,'BBB')
insert into #codes values(1013,'CCC')


select ID,descs,code,codes from #master a
 left join #codes b
 on a.code = b.ids

输出:

ID  descs   code  codes
1   ballpen 1010   AAA
2   pencil  1010   AAA
3   stabilo 1010   AAA
4   pins    1011   BBB
5   Clips   1011   BBB

如果左连接不起作用,请使用内连接:

select ID,descs,code,codes from #master a,#codes b
where a.code=b.ids

答案 1 :(得分:0)

你应该尝试内部加入 的

       SELECT 
       distinct  a.id,a.desc1,a.code,b.code as code2
           from  a
           inner join  b
           on a.code = b.id

答案 2 :(得分:0)

我想Plus.PeopleApi.loadVisible(mGoogleApiClient, null) .setResultCallback(this); 不包含唯一的table B,因此您需要对所选数据集code进行操作

这里是测试:

distinct

下面的结果截图:

enter image description here

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