Question

我有这张桌子：

表1

--------------------
id  |  category
--------------------
1   | Cat1
2   | Cat2
3   | Cat3

对于每个类别，我都有包含子类别的表格和相应的类别ID，如下所示：

表2

--------------------------------
id  | subcategory  | category_id
--------------------------------
1   | subcat1      |     1
2   | subcat2      |     1
3   | subcat3      |     1
4   | subcat4      |     1

表3

--------------------------------
id  | subcategory  | category_id
--------------------------------
1   | subcat5      |     2
2   | subcat6      |     2
3   | subcat7      |     2
4   | subcat8      |     2

我想查询我的数据，所以我得到以下内容：

-----------------------------------------------------
category  | subcategory ( from Table2 and Table3)  
-----------------------------------------------------
Cat1      | subcat1      
Cat1      | subcat2      
Cat1      | subcat3      
Cat1      | subcat4
Cat2      | subcat5      
Cat2      | subcat6      
Cat2      | subcat7      
Cat2      | subcat8
Cat3      | NULL

我尝试过类似的东西，但是它将每个表中的数据放在一个新列上。

SELECT Table1.category, Table2.subcategory  , Table3.subcategory  
FROM Table1
LEFT JOIN Table2 ON Table1.id=Table2 .category_id
LEFT JOIN Table3 ON Table1.id=Table3.category_id

我是初学者。

谢谢！

Answer 1

每个子类别都不需要新表。如果你改变它，你只有两个表，只需加入它们。

Answer 2

SELECT c.Category,s.SubCategory
  FROM Table1 c
 LEFT JOIN 
       (
         SELECT * FROM Table2
         UNION ALL
         SELECT * FROM Table3
       ) s
   ON c.id = s.id;

Answer 3

当我从你的问题中读到子类别在不同的表中时，你可以为这些表创建一个UNION（ALL），然后加入你的Table1，就像这样

SELECT Table1.category, Table2.subcategory, Table3.subcategory  
FROM Table1
LEFT JOIN (SELECT Table2.category_id, Table2.subcategory
           FROM   Table2
           UNION ALL
           SELECT Table3.category_id, Table3.subcategory
           FROM   Table3
          ) sub
ON Table1.id=sub.category_id

希望这有助于你

Answer 4

建议：创建一些VIEW以确保您具有合理且一致的数据项，这将使查询更容易编写，例如

CREATE VIEW Subcategories
AS 
SELECT subcategory, category_id
  FROM Table2
UNION
SELECT subcategory, category_id
  FROM Table3;


CREATE VIEW Categories
AS 
SELECT id AS category_id, category
  FROM Table1;


SELECT category, subcategory
  FROM Categories NATURAL JOIN Subcategories
UNION
SELECT category, 0 AS subcategory
  FROM Categories
 WHERE category_id NOT IN ( SELECT category_id 
                              FROM Subcategories );

在同一列上加入包含结果的多个表

4 个答案: