獲取沒有連接的記錄

Question

我有3個表格：Products,ProductProperties,Properties。我需要獲得所有沒有「物業B」的產品。表：

# Products   # Join table     # Properties 
+----+-----------+ +------------+-------------+ +-------------+---------------+ 
| id | name  | | product_id | property_id | | property_id | property_name | 
+----+-----------+ +------------+-------------+ +-------------+---------------+ 
| 1 | Product 1 | |   1 |   1 | |   1 | Propeprty A | 
| 2 | Product 2 | |   1 |   2 | |   2 | Propeprty B | 
| 3 | Product 3 | |   2 |   1 | +-------------+---------------+ 
+----+-----------+ |   2 |   2 | 
        |   3 |   1 | 
        +------------+-------------+ 

在這種特殊情況下我希望產品3要返回。

是否有可能在單個數據庫查詢中獲得所有必需的產品？什麼最小的查詢可能實現呢？

編輯。考慮子查詢的查詢不好。

Answer 1

好像你想要一個OUTER JOIN，然後一個WHERE帶來子句後面的非匹配屬性。這是未經測試，但：

SELECT Products.* FROM 
Products RIGHT OUTER JOIN Join_Table 
    ON Products.ID = Join_Table.Product_ID 
    AND Join_Table.Property_ID = 2 
WHERE Products.ID IS NULL 

Answer 2

select * from Products P where not exists (select * from ProductProperties 
inner join Properties on ProductProperties .property_id = Properties property_id 
where P.product_id = ProductProperties.product_id and property_name = 'Propeprty B') -- *or whatever* 

Answer 3

SELECT Products.ProductID, Products.Name, Properties.property_name FROM Products 
INNER JOIN ProductProperties ON Products.product_ID = ProductProperties.product_ID 
INNER JOIN Properties ON ProductProperties.Property_ID = Properties.Property_ID 
WHERE Properties.Property_ID <> 2 

Answer 4

是（修正去除潛艇），

Select * from 
    Products pr 
    left join jointable jt on pr.id = jt.product_id 
    join properties pr on jt.property_id = pr.property_id and pr.property_name = 'Property B' 
where jt.product_id is null 

Answer 5

我需要把所有的產品，不具有 「物業B」

逆問題。找到與PropertyB，然後否定它。

開始與所有PropertyB屬性：

SELECT 
    Property_Id 
FROM Properties 
WHERE 
    Property_Name = 'Property B' 

然後，找到ProductId s表示做有那些Property_Id S：

SELECT 
    ProductId 
FROM ProductProperties 
JOIN (
    --1st query 
    SELECT 
     Property_Id 
    FROM Properties 
    WHERE 
     Property_Name = 'Property B' 
) as PropertyB ON 
    ProductProperties.Property_Id = PropertyB.Property_Id 

然後，找到所有的Product不是該集合：

SELECT 
    ProductId 
FROM Product 
LEFT OUTER JOIN (
    --2nd query 
    SELECT 
     ProductId 
    FROM ProductProperties 
    JOIN (
     --1st query 
     SELECT 
      Property_Id 
     FROM Properties 
     WHERE 
      Property_Name = 'Property B' 
    ) as PropertyB ON 
     ProductProperties.Property_Id = PropertyB.Property_Id 
) as ProductsWithPropertyB ON 
    Products.ProductId = ProductsWithPropertyB.ProductId 
WHERE 
    ProductsWithPropertyB.ProductId IS NULL 

然後，您可以簡化一下：

SELECT 
    ProductId 
FROM Product 
LEFT OUTER JOIN (
    SELECT 
     ProductId 
    FROM ProductProperties 
    JOIN Properties ON 
     ProductProperties.PropertyId = Properties.PropertyId 
    WHERE 
     Properties.Name = 'Property B' 
) as ProductsWithPropertyB ON 
    Products.ProductId = ProductsWithPropertyB.ProductId 
WHERE 
    ProductsWithPropertyB.ProductId IS NULL 

或者，如果你喜歡IN條款（服務器可能不關心）：

SELECT 
    ProductId 
FROM Products 
WHERE 
    ProductId NOT IN (
      SELECT ProductId FROM ProductProperties WHERE PropertyId IN (
       SELECT PropertyId FROM Properties WHERE PropertyName = 'Property B' 
     ) 
    )