Question

我想在没有项目的情况下获取所有员工，没有员工的项目。我不知道我在哪里做错了。

表格结构：

employee：employee_id（PK），employee_name //所有员工

项目：project_code（PK），project_title //所有项目

project_team：project_code（FK），employee_id（FK）//正在使用项目和员工的桥接表。

这是我的疑问：

SELECT *
FROM employee e
    LEFT JOIN project_team pt USING (employee_id) WHERE pt.`project_code` IS NULL
    INNER JOIN project p USING (project_code) WHERE (e.`employee_id` IS NULL)

Answer 1

你可能想要这样的东西：

SELECT employee_id AS id_or_code, employee_name AS name_or_title FROM employee e LEFT JOIN project_team pt ON e.employee_id = pt.employee_id WHERE pt.project_code is NULL

UNION ALL

SELECT project_code AS id_or_code, project_title AS name_or_title FROM project p LEFT JOIN project_team pt ON p.project_code = pt.project_code WHERE pt.employee_id is NULL;

这样，您可以将查询结果分为两列，第一列作为员工或代码或项目的ID，第二列作为员工的名称或项目的标题。

Answer 2

你需要全部加入，但MySQL没有。

在这里你可以找到联合解决方案https://stackoverflow.com/a/4796911/3970250

Answer 3

您正在尝试将2个不同的选择记录合并到一个结果集中，一种方法是使用union all，然后将每个记录类型标记为

select 
e.employee_id as id , 
e.employee_name as name, 
'Employee' as type 
from employee e 
left join project_team pt on pt.employee_id = e.employee_id 
where pt.employee_id is null
union all
select 
p.project_code as id ,
p.project_title as name, 
'Project' as type 
from project p 
left join project_team pt on pt.project_code = p.project_code 
where pt.project_code is null ;

这是一个测试用例

mysql> select * from employee ;
+-------------+---------------+
| employee_id | employee_name |
+-------------+---------------+
|           1 | A             |
|           2 | B             |
|           3 | C             |
|           4 | D             |
|           5 | E             |
+-------------+---------------+
5 rows in set (0.00 sec)

mysql> select * from project ;
+--------------+---------------+
| project_code | project_title |
+--------------+---------------+
|            1 | P1            |
|            2 | P2            |
|            3 | P3            |
|            4 | P4            |
+--------------+---------------+
4 rows in set (0.00 sec)

mysql> select * from project_team ;
+--------------+--------------+
| project_code | employee_id |
+--------------+--------------+
|            1 |            1 |
|            1 |            2 |
|            2 |            3 |
|            2 |            4 |
|            3 |            2 |
|            3 |            1 |
|            3 |            3 |
+--------------+--------------+

运行上述查询将为您提供

+------+------+----------+
| id   | name | type     |
+------+------+----------+
|    5 | E    | Employee |
|    4 | P4   | Project  |
+------+------+----------+

MySQL：没有项目的员工和没有员工的项目

3 个答案: