获取sql中的所有重复行

Question

我想获取sql中的所有重复值。我在很多地方搜索过，但是我找不到我想要的东西。我的桌子是这样的：

company_id | supplier_id | company_name | organisation_no | type | sold_date
----------------------------------------------------------------------------
121234     | 934575      | fgdf         | 12345           | sold | 2011-12-2
214365     | 423234      | sdgd         | 5678            | sold | 2011-12-2
546534     | 234234      | bvcv         | 3333            | sold | 2011-12-2
276345     | 243324      | dfgd         | 12345           | sold | 2011-12-2
432642     | 567647      | ghmj         | 12345           | sold | 2011-12-2
846578     | 365356      | egff         | 3333            | sold | 2011-12-2
254334     | 346535      | yuuy         | 7890            | sold | 2011-12-2

我找到的解决方案是这样的：

organisarion_no | count(organisation_no)
----------------------------------------
3333            | 2
12345           | 3

但我想要这样：

company_id | supplier_id | company_name | organisation_no | type | sold_date
----------------------------------------------------------------------------
546534     | 234234      | bvcv         | 3333            | sold | 2011-12-2
846578     | 365356      | egff         | 3333            | sold | 2011-12-2
121234     | 934575      | fgdf         | 12345           | sold | 2011-12-2
276345     | 243324      | dfgd         | 12345           | sold | 2011-12-2
432642     | 567647      | ghmj         | 12345           | sold | 2011-12-2

请帮忙。提前谢谢。

Answer 1

select * from your_table
where organization_no in 
   (select organization_no 
   from your_table
   group by organization_no
   having count(*) > 1)

Answer 2

SELECT * FROM table t 
 WHERE organisation_no IN 
       (SELECT organisation /*from the solution you have found*/)