Question

我有三张桌子。

组件：

id | name | storage | bilance |
   |      |         |         |

商店

id | name | price | componentId |
   |      |       |             |

和订单

id | item | amount | ... |

组件表包含产品。商店桌上有售卖该产品的商店

所以例如

component( 1 , "wheel" , 15 , 15 );

和出售轮子的商店

Shops(1 , "wheelShopone" , 150 , 1);
Shops(2 , "Shoptwo" , 100 , 1 )

这基本上意味着你可以购买更多商店的轮子，例如1：n关系

订单包含要订购的内容，例如

orders(1 , "wheel" , 5 )

我想要做的是找到在某些商店出售的表格订单中的所有元素（例如组件）。

我尝试使用join例如

> Select components.name as comp_name , components.id as comp_id ,
> shops.name as shop_name , shops.id as shop_id , orders.item as item ,
> orders.amount as order_amount from components join shops join orders
> WHERE shop_name = some name

我期望它做的是跟随，imageine表

 components                                       shops
 id | name  | storage | bilance |       id | name | price | componentId|
 1  | wheel | 15      | 15      |       1  | One  | 15    |  1          
 2  | mouse | 1       | 1       |       2  | two  | 5     |  1
                                        3  | three| 5     | 2

使用

加入这两个表

> Select components.name as comp_name , components.id as comp_id ,
> shops.name as shop_name , shops.id as shop_id FROM Components join Shops

应该导致

comp_name | comp_id | shop_name | shop_id 
wheel     |  1      | one       | 1
wheel     |  1      | two       | 2
mouse     |  2      | three     | 3

然后最终加入订单，例如

id | item  | amount | ... 
1  | wheel | 5      | ...

使用我之前提到的命令

> Select components.name as comp_name , components.id as comp_id ,
> shops.name as shop_name , shops.id as shop_id , orders.item as order_item , order.amount as order_amount FROM Components join Shops
> join Orders on components.name = orders.item

应该导致

comp_name | comp_id | shop_name | shop_id | item | amount
wheel     |  1      | one       | 1       | wheel| 5
wheel     |  1      | two       | 2       | wheel| 5

但是使用这个命令，它只是抛出随机表，它加入了看起来像crossjoin的东西，因为我得到了数百行数据。

我对连接的理解是否正确？如果没有我错在哪里，我怎么能做这个工作？谢谢你的帮助！

Answer 1

你可能想要这样的东西：

SELECT components.name as comp_name,
       components.id as comp_id,
       shops.name as shop_name,
       shops.id as shop_id,
       orders.item as item,
       orders.amount as order_amount
FROM shops JOIN components ON shops.componentId=components.id
           JOIN orders ON components.id=orders.item
WHERE shop_name = some name

你缺少的主要是连接条件。通常，在进行连接时，必须指定一个条件，该条件将用于匹配两个表中的正确行。例如，条件shops.componentId=components.id表示来自shops的行与来自components的行匹配，其中componentId中的id与组件的import sys from PyQt5.QtWidgets import QHBoxLayout, QAction, QApplication, QMainWindow class menudemo(QMainWindow): def __init__(self, parent = None): super(menudemo, self).__init__(parent) bar = self.menuBar() file = bar.addMenu("File") file.addAction("New") save = QAction("Save",self) save.setShortcut("Ctrl+S") file.addAction(save) edit = file.addMenu("Edit") edit.addAction("copy") edit.addAction("paste") quit = QAction("Quit",self) file.addAction(quit) file.triggered[QAction].connect(self.processtrigger) self.setWindowTitle("menu demo") def processtrigger(self, q): print(q.text()+" is triggered") def main(): app = QApplication(sys.argv) ex = menudemo() ex.show() sys.exit(app.exec_()) if __name__ == '__main__': main()相同。

Answer 2

如果要进行连接，则必须告诉数据库两个表中的哪些行匹配：

SELECT ...
FROM Components
JOIN Shops ON Components.id = Shops.componentId;

即使您声明了外键约束，也不会自动推断出这种关系。

如果没有连接条件，您确实会得到交叉连接。

在sqlite中连接表

2 个答案: