选择在列中具有相同值且在另一列中具有不同值的记录

Question

在我的表INVENTORY中有3列varchar代码，版本，说明。 e.g

code || version || description 
-------------------------- 
AS100 || 01 || description 
AS100 || 02 || description 
AS100 || 02 || description 
AS100 || 02 || description 
AS200 || 01 || description 
AS300 || 01 || description 
AS300 || 01 || description 

我要选择那些具有相同的产品和多个版本的产品。在上表中，该查询返回会：

AS100 - 01 - description
AS100 - 02 - description

我试图用HAVING声明，但我没有得到上述结果..

例如，我尝试这样做：

SELECT code, version, Count(*) FROM INVENTORY 
GROUP BY version 
HAVING Count(*) =1 

，但它会返回出现一次的所有代码。

我创建了一个SQLFIDDLE这里（我希望它扮演，有时页面冻结） 我的DB

CREATE TABLE "INVENTORY" (
    "code" VARCHAR, 
    "version" VARCHAR, 
    "description" VARCHAR 
) 


INSERT INTO INVENTORY (code, version, description) VALUES("AS100", "01", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS100", "02", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS100", "02", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS100", "02", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS200", "02", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS300", "01", "description"); 
INSERT INTO INVENTORY (code, version, description) VALUES("AS300", "01", "description"); 

Answer 1

的小转储也许这就是你想要什么？

select *, count(*) from (select * from INVENTORY group by code, version) 
group by code 
having count(*)>1 

也就是说，首先要由感兴趣的两列汇总表，然后选择从该表中，其中也有代码至少两个不同值的行。