Question

我在构建SQL查询时遇到问题。假设我有一个表专家;例如：

+----+------------+-----------+
| id | first_name | last_name |
+----+------------+-----------+
|  6 | Ahmed      | Mouloud   |
|  7 | Alice      | Lanouix   |
+----+------------+-----------+

表expert_competence，允许加入专家的能力：

+----+---------------+-----------+
| id | competence_id | expert_id |
+----+---------------+-----------+
| 28 |             1 |         7 |
| 29 |             3 |         7 |
| 30 |             4 |         7 |
| 31 |             5 |         7 |
| 32 |            19 |         7 |
| 33 |            20 |         7 |
| 34 |             1 |         6 |
| 35 |             5 |         6 |
| 36 |             6 |         6 |
| 37 |            21 |         6 |
+----+---------------+-----------+

和表能力：

+----+-------------------------------------------------------------------------+
| id | name                                                                    |
+----+-------------------------------------------------------------------------+
|  1 | INSPECTEUR OPERATIONS AERIENNES                                         |
|  2 | INSPECTEUR OPERATIONS AERIENNES SAC MONACO                              |
|  3 | INSPECTEUR OPERATIONS AERIENNES DCA MAURICE                             |
|  4 | CONSULTING AUPRES D'UNE CAA      
  ...                                     |                                    
+----+-------------------------------------------------------------------------+

如何让所有掌握能力为1和3（非OR）的专家？

我尝试了这个但是它没有工作（找到掌握能力1和3的所有专家）：

SELECT DISTINCT ex.* 
FROM expert ex 
LEFT JOIN expert_competence ec 
ON ec.expert_id = ex.id 
WHERE ec.competence_id = 1 
AND ec.competence_id = 3

我也尝试过这个查询：

SELECT DISTINCT ex.* 
FROM expert ex 
LEFT JOIN expert_competence ec 
ON ec.expert_id = ex.id 
WHERE ec.competence_id IN (1,3);

结果：

+----+-----------+------------+
| id | last_name | first_name |
+----+-----------+------------+
|  7 | Lanouix   | Alice      |
|  6 | Mouloud   | Ahmed      |
+----+-----------+------------+

预期结果：

+----+------------+-----------+
| id | first_name | last_name |
+----+------------+-----------+
|  7 | Alice      | Lanouix   |
+----+------------+-----------+

Answer 1

请改为尝试：

SELECT DISTINCT
    ex.*
FROM
    expert AS ex
    LEFT JOIN expert_competence AS ec ON ec.expert_id = ex.id
WHERE 
    ec.competence.id IN (1, 8)

通过使用IN，您可以为列指定多个可能的条件。

Answer 2

SELECT DISTINCT ex.* 
FROM expert_competence ec
LEFT JOIN  expert ex
ON ec.expert_id = ex.id 
WHERE ec.competence.id IN (1, 8)

刚刚注意到ec.competence.id它是否真的以表中的名字命名？现在我在查询中修改了一些条件，并在JOIN查询中修改了一些表格。

让我知道它是否适合你。

Answer 3

根据您提供的输入，您不清楚自己的目标是什么，我假设您正在寻找类似的内容，我认为您已经{{{I} 1}}表存储所有主ID。

使用All运算符可以获得所需的结果。 Live Demo

competence_master

Answer 4

试试这个例子：

                     <Path x:Name="ButtonPath"
                              Margin="1"
                              Stroke="Gray"
                              StrokeThickness="1"
                              StrokeStartLineCap="Square"
                              StrokeEndLineCap="Square"
                              Stretch="Uniform"
                              VerticalAlignment="Center"
                              HorizontalAlignment="Center"

                             >

                            <Path.Data>
                            <PathGeometry>
                                <PathGeometry.Figures>
                                    <PathFigure StartPoint="0,100">
                                        <LineSegment Point="0,-100"/>
                                    </PathFigure>
                                    <PathFigure StartPoint="100,0">
                                        <LineSegment Point="-100,0"/>
                                    </PathFigure>
                                </PathGeometry.Figures>
                            </PathGeometry>
                        </Path.Data>
                   </Path>

Answer 5

最后，我建立了一个有效的解决方案：

SELECT ex.*  FROM 
expert ex LEFT JOIN expert_competence ec 
ON ec.expert_id = ex.id 
WHERE ec.competence_id WHERE ec.competence_id = 1 
AND ex.id in (SELECT ex.id  FROM expert ex 
              LEFT JOIN expert_competence ec 
              ON ec.expert_id = ex.id 
              WHERE ec.competence_id = 3);

多个条件AND在JOIN MySQL

5 个答案: