php - 在没有 like 运算符的情况下检索多个单个或引用的数据

Question

我有两个表，如下所示。在员工表中而不是直接提及技能，我使用另一个表作为引用。

问题:

select * from Employee where Skills = "1";

上面的查询将显示只有“Python”技能的员工记录。当然，它不会显示包含Python的其他组合，例如“1,2”(Python，Java)。我如何在不使用 like 运算符的情况下实现这一目标，因为如果我的技能为 10、11、21，那将是个问题，不是吗。

如果您觉得这种使用引用的方式有难度或不推荐，请提出您的想法:-)

员工表:

+-----+-------------+-------------+
| id  | Name        | Skills      |
+-----+-------------+-------------+
| 1   | Xyz         | 1,2,4       |
| 2   | Xyy         | 1,3         |
| 3   | Abc         | 1,2,3       |
| 4   | Asd         | 1           |
+-----+-------------+-------------+

技能表:

+-----+-------------+
| id  | SkillSet    |
+-----+-------------+
| 1   | Python      |
| 2   | Java        |
| 3   | C           |
| 4   | PHP         |
+-----+-------------+

Answer 1

考虑以下内容

mysql> select * from employee ;
+------+------+--------+
| id   | name | skills |
+------+------+--------+
|    1 | xyz  | 1,2,4  |
|    2 | abc  | 1,3    |
|    3 | lmn  | 1,2,3  |
+------+------+--------+
3 rows in set (0.00 sec)

mysql> select * from skillset ;
+------+----------+
| id   | skillset |
+------+----------+
|    1 | Python   |
|    2 | Java     |
|    3 | C        |
|    4 | PHP      |
+------+----------+
4 rows in set (0.00 sec)

这个结构和你的很相似，但是在这些情况下我们可以使用find_in_set来做查询，但是效率很低，这里有几个例子

mysql> select e.id,
e.name, 
group_concat(s.skillset) as skillset 
from employee e join skillset s 
on find_in_set(s.id,e.skills) > 0 
where find_in_set(1,e.skills) > 0 
 group by e.id ;
+------+------+-----------------+
| id   | name | skillset        |
+------+------+-----------------+
|    1 | xyz  | Python,Java,PHP |
|    2 | abc  | C,Python        |
|    3 | lmn  | Java,Python,C   |
+------+------+-----------------+
3 rows in set (0.00 sec)

select e.id,
e.name, 
group_concat(s.skillset) as skillset 
from employee e 
join skillset s on find_in_set(s.id,e.skills) > 0 
where find_in_set(2,e.skills) > 0  group by e.id ;
+------+------+-----------------+
| id   | name | skillset        |
+------+------+-----------------+
|    1 | xyz  | Python,PHP,Java |
|    3 | lmn  | C,Java,Python   |
+------+------+-----------------+

现在适当的归一化会让生活变得简单得多，并且会有以下关联表

mysql> select * from employee_skills; 
+------------+----------+
| idemployee | idskills |
+------------+----------+
|          1 |        1 |
|          1 |        2 |
|          1 |        3 |
|          2 |        1 |
|          2 |        3 |
|          3 |        1 |
|          3 |        2 |
|          3 |        3 |
+------------+----------+

现在在这种情况下进行查询会更有效率

mysql> select e.id,
e.name,
s.skillset from employee e 
join employee_skills es on es.idemployee = e.id 
join skillset s on s.id = es.idskills where s.id = 1 ;
+------+------+----------+
| id   | name | skillset |
+------+------+----------+
|    1 | xyz  | Python   |
|    2 | abc  | Python   |
|    3 | lmn  | Python   |
+------+------+----------+

使用最后一种方法可以很容易地完成更复杂的计算。