我有一个包含以下列的用户表:

| User_ID  (int) | Name (varchar) |   Age (int)  |  Experience_Level (int) |

我想创建一个SQL查询,以输出在年龄和经验中并非独特的人的所有ID。

到目前为止,我的代码:

SELECT Count(*), User_ID FROM Users 
GROUP BY Age,Experience_Level
HAVING Count(*) > 1

显然,这是不完整的,这将对非唯一的用户进行分组,但不会告诉我他们的所有用户。

在此先感谢您的帮助!

有帮助吗?

解决方案

这是一个否定的逻辑查询:

SELECT *
FROM Users
WHERE UserID not in
(
SELECT MIN(UserID)
FROM Users
GROUP BY Age, Experience_Level
HAVING COUNT(*) = 1
)

其他提示

既然您想要有关多个用户组的信息,那么您希望如何返回此数据?在包含comma分隔user_id值列表的字符串中?

您没有使用使用的SQL数据库品牌标记问题。

如果使用mySQL或SQLite,则可以使用内置 GROUP_CONCAT() 功能:

SELECT Count(*), GROUP_CONCAT(User_ID) AS User_List FROM Users 
GROUP BY Age,Experience_Level
HAVING Count(*) > 1

默认情况下, GROUP_CONCAT() 用逗号分开值。如果您希望该手册以另一种方式格式化,请参考该手册。

还有其他针对其他SQL供应商的解决方案。这个问题已经在堆栈溢出上出现了很多次:

SELECT t.User_ID, t.Age, t.Experience_Level
FROM Users t INNER JOIN 
    (SELECT Age, Experience_Level
    FROM Users
    GROUP BY Age, Experience_Level
    HAVING Count(*) > 1) d ON t.Age = d.Age AND t.Experience_Level = d.Experience_Level

测试脚本:

create table Users (
User_ID int,
Name varchar(50),
Age int,
Experience_Level int
)

insert into Users (User_ID, Name, Age, Experience_Level) values (1, 'A', 33, 1)
insert into Users (User_ID, Name, Age, Experience_Level) values (2, 'B', 37, 1)
insert into Users (User_ID, Name, Age, Experience_Level) values (3, 'C', 33, 1)
insert into Users (User_ID, Name, Age, Experience_Level) values (4, 'D', 35, 2)
insert into Users (User_ID, Name, Age, Experience_Level) values (5, 'E', 33, 1)
insert into Users (User_ID, Name, Age, Experience_Level) values (6, 'F', 35, 2)
insert into Users (User_ID, Name, Age, Experience_Level) values (7, 'G', 18, 1)

从理论上讲,您想要的是这样的东西,但不幸的是SQL Server不允许它:

SELECT * FROM Users 
WHERE (Age, Experience_Level) IN
(
    SELECT Age, Experience_Level
    FROM Users  
    GROUP BY Age,Experience_Level 
    HAVING Count(*) > 1 
)

因此,相反,您必须满足于加入子查询:

SELECT Users.* FROM Users 
INNER JOIN 
(
    SELECT Age, Experience_Level
    FROM Users  
    GROUP BY Age,Experience_Level 
    HAVING Count(*) > 1 
) subq 
    ON Users.Age = subq.Age
    AND Users.Experience_Level = subq.Experience_Level
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