Question

鉴于以下数据：

EMPID   NAME    SALARY  DID
1       kevin   32000   2
2       joan    42000   1
3       brian   37000   3
4       larry   82000   5
5       harry   92000   4
6       peter   45000   2
7       peter   68000   3
8       smith   39000   4
9       chen    71000   1
10      kim     46000   5
11      smith   46000   1

列出不具有相同名称的员工的姓名。

我不确定我是否必须自己加入桌子（或做其他事情）。

这是我到目前为止写的：

SELECT distinct e.name
FROM employee e, employee ee
WHERE e.name <> ee.name

Answer 1

无需加入：

SELECT name
FROM employee
GROUP BY name
HAVING COUNT(name) = 1

计算每个名称的数量，然后只返回只有一个名称的名称。例如peter的数量为2，不会出现，而kim的数量为1并且会出现。

Answer 2

如果您想获得不同的名称，以下查询就足够了：

SELECT distinct e.name
FROM employee e

希望它有所帮助!!!

Answer 3

mysql> select * from employee;
+-------+-------+--------+-----+
| empid | name  | salary | did |
+-------+-------+--------+-----+
|     1 | kevin |  32000 |   2 |
|     2 | joan  |  42000 |   1 |
|     3 | brian |  37000 |   3 |
|     4 | larry |  82000 |   5 |
|     5 | harry |  92000 |   4 |
|     6 | peter |  45000 |   2 |
|     7 | peter |  68000 |   3 |
|     8 | smith |  39000 |   4 |
|     9 | chen  |  71000 |   1 |
|    10 | kim   |  46000 |   5 |
|    11 | smith |  46000 |   1 |
+-------+-------+--------+-----+
11 rows in set (0.00 sec)


mysql> select distinct name from employee;
+-------+
| name  |
+-------+
| kevin |
| joan  |
| brian |
| larry |
| harry |
| peter |
| smith |
| chen  |
| kim   |
+-------+
9 rows in set (0.00 sec)

Answer 4

使用内连接计算每个emp的记录数。你可以过滤count = 1的结果

SELECT 
    T.EMPID,
    T.NAME,
    T.SALARY,
    T.DID 
FROM 
    employee  AS T 
    INNER JOIN (SELECT COUNT(*) AS CNT , NAME FROM employee  GROUP BY NAME) AS A ON T.NAME = A.NAME
WHERE 
    CNT = 1

查询：不具有相同名称的员工

4 个答案: