查询：员工不具有相同的名称

Question

鉴于以下数据谁：

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 

名单谁不共享同一名称的员工的名字。

我不确定是否必须自己加入表格（或者做其他事情）。

这是我写到目前为止：

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

Answer 1

无需加盟：

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

算多少每个名字的有，那么只返回那些只有一个名字。例如彼得的计数是2，不会显示，而金姆计数为1，并且会显示出来。

Answer 2

如果你想获得只是不同的名称下面的查询应该足够：

SELECT distinct e.name 
FROM employee e 

希望它可以帮助！

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

使用一个内部联接来计算每个emp的记录数。你可以过滤结果，其中count = 1

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