MySql join 语句返回重复数据

Question

我有一个完全规范化的数据库，并通过 mySql 运行查询。所需的查询之一是返回所需的 Material 以及与工作相关的任何 Material 。我需要将以下 2 个查询的结果组合成一个选择语句(每个查询位于单独的行中，最好使用联接)。下面对此进行了更好的解释:

SELECT m.Material_Name
FROM project_materials pm,materials m
WHERE pm.Fitting_ID =  m.Fitting_ID
AND pm.Project_ID = '2';

这给了我:

Timber
Bricks

还有这个:

SELECT sjt.Job_Name
FROM project_staff ps, staff_job_type sjt
WHERE ps.Staff_job_ID = sjt.Staff_job_ID
AND ps.Project_ID = '2';
;

这给了我:

Brick Layer
Project manager

我尝试在这里将它们合并:

SELECT DISTINCT m.Material_Name, sjt.Job_Name
FROM project_materials pm, materials m, project_staff ps, staff_job_type sjt, projects p
WHERE p.Project_ID = ps.Project_ID
AND p.Project_ID = pm.Project_ID
AND pm.Fitting_ID =  m.Fitting_ID
AND ps.Staff_job_ID = sjt.Staff_job_ID
AND pm.Project_ID = '2';

这根本没有效果，因为我得到的是重复的值。 IE。数据库回复如下:

Timber, Brick Layer
Timber, Project manager
Bricks, Brick Layer
Bricks, Project manager

这会导致回复随着数据的增长而变得越来越糟糕!

我也尝试过使用 JOIN 但得到了相同的结果。这是我尝试过的代码:

SELECT Material_Name, Job_Name
FROM projects INNER JOIN project_staff ON projects.Project_ID = project_staff.Project_ID
              INNER JOIN project_materials ON project_materials.Project_ID = projects.Project_ID
              INNER JOIN materials ON materials.Fitting_ID = project_materials.Fitting_ID
              INNER JOIN staff_job_type ON staff_job_type.Staff_job_ID = project_staff.Staff_job_ID
WHERE project_materials.Project_ID = '2'
;

连接前面的内部、左侧、右侧或没有文本都没有区别。提前致谢，对这么长的帖子表示歉意

编辑我想要这样的东西:

SELECT m.Material_Name
FROM project_materials pm,materials m
WHERE pm.Fitting_ID =  m.Fitting_ID
AND pm.Project_ID = '2'
UNION
SELECT sjt.Job_Name
FROM project_staff ps, staff_job_type sjt
WHERE ps.Staff_job_ID = sjt.Staff_job_ID
AND ps.Project_ID = '2';
;

但我希望将其放在单独的列中，而不是联合

Answer 1

你试过这个吗:

select m.Material_Name, s.JobName
    from project_materials pm 
    join materials m on pm.Fitting_ID = m.Fitting_ID 
        and pm.Project_ID = m.Project_ID
    join project_staff ps on pm.Project_ID = ps.Project_ID
    join staff_job_type sjt on ps.Staff_job_ID = sjt.Staff_job_ID
where pm.Project_ID = '2'

以上查询无法使用，因为 Material 没有 Project_ID。即使色谱柱有 Material 可用，结果仍然是双倍，不符合预期。请忽略它。

我认为我们需要一个列将 Material 记录与staff_job_type 上的记录相关联，例如外键。如果没有那个专栏，我们最终将陷入全面的关系。仅 Project_ID 是不够的，因为它对于 Material 和 Staff_job_type 来说不是唯一的。列staff_job_id 将staff_job_type 与project_staff 关联，列Fitting_ID 将 Material 与project_material 关联。没有列将 Material 与 Staff_job_type 相关。

material_name job_name        sjt.staff_job_id ps.staff_job_id m.fitting_id pm.fitting_id  
------------- --------------- ---------------- --------------- ------------ ------------- 
Timber        Brick Layer     1                1               1            1           
Bricks        Brick Layer     1                1               2            2           
Timber        Project Manager 2                2               1            1           
Bricks        Project Manager 2                2               2            2           

我建议您创建一个表来将 Material 关联/映射到staff_job_type。该表应该是这样的:

create table material_job (
  fitting_id int,
  staff_job_id int
);

该表包含 Material 和staff_job_type关系:

fitting_id  staff_job_id 
----------- ------------ 
1           1            
2           2            

假设 Material 如下:

fitting_id  material_name                                      
----------- -------------------------------------------------- 
1           Timber                                             
2           Bricks                                             

staff_job_type 就像

staff_job_id job_name                                           
------------ -------------------------------------------------- 
1            Brick Layer                                        
2            Project Manager                                    

项目 Material 如

project_id  fitting_id  amount      
----------- ----------- ----------- 
2           1           500         
2           2           600         

项目人员喜欢

project_id  staff_job_id amount      
----------- ------------ ----------- 
2           1            100         
2           2            200         

我们可以使用查询

select m.material_name, sjt.job_name from material_job mj
    join materials m on mj.fitting_id = m.fitting_id
    join staff_job_type sjt on mj.staff_job_id = sjt.staff_job_id;

结果

material_name job_name                                           
------------- ---------------
Timber        Brick Layer                                        
Bricks        Project Manager                                    

如果您还需要project_staff和project_material记录，我们可以使用sql:

select m.material_name, sjt.job_name from material_job mj
    join materials m on mj.fitting_id = m.fitting_id
    join staff_job_type sjt on mj.staff_job_id = sjt.staff_job_id
    join project_materials pm on m.fitting_id = pm.fitting_id
    join project_staff ps on sjt.staff_job_id = ps.staff_job_id;

结果

material_name job_name        pm_amount ps_amount      
------------- --------------- --------- ---------
Timber        Brick Layer     500       100                                                           
Bricks        Project Manager 600       200