mysql的連接表查詢的2個值

-1

我想在我的數據庫運行查詢，以顯示產品，有選擇的屬性mysql的連接表查詢的2個值

表1 Products

ID | Product Name 
---|----------------- 
1 Red Car 
2 Blue Car 
3 Yellow Car

表2 Attributes

Product ID | Attribute ID 
-----------|----------------- 
1   3 
2   3 
3   3 
1   4 
2   4

例如，我只想顯示具有屬性3和4的產品設置它應該只顯示紅色和藍色的車。但不是黃色的車，因爲沒有爲產品ID設置屬性3

來源

2015-10-06 beanieman85

我剛纔已經回答了完全相同的問題在其他SO話題：http://stackoverflow.com/questions/32945964/php-search-mysql-database-using-multiple-select-dropdown-lists – Shadow

我希望重複的東西更好地工作 – Strawberry

你可以在這裏找到類似的東西：http://stackoverflow.com/questions/32948681/mysql-filter-on-many-to-many/32949847 –

有很多方法可以解決這個問題;最直接的可能是使用一對夫婦的exists條款，或加入attributes表兩次，但你也可以使用group by和having子句來完成相同的結果：

-- option 1: using multiple exists clauses 
select p.id, p.productname 
from Products p 
where exists (select 1 from Attributes a where p.ID = a.ProductID and a.AttributeID = 3) 
    and exists (select 1 from Attributes a where p.ID = a.ProductID and a.AttributeID = 4); 

-- option 2: using multiple joins 
select p.id, p.productname 
from Products p 
join Attributes a3 on p.ID = a3.ProductID 
join Attributes a4 on p.ID = a4.ProductID 
where a3.AttributeID = 3 
    and a4.AttributeID = 4; 

-- option 3: using aggregate and having 
select p.id, p.productname 
from Products p 
join Attributes a on p.ID = a.ProductID 
group by p.id, p.productname 
having sum(case when a.AttributeID = 3 then 1 else 0 end) > 0 
    and sum(case when a.AttributeID = 4 then 1 else 0 end) > 0; 

-- option 4: using having and count 
select p.id, p.productname 
from Products p 
join Attributes a on p.ID = a.ProductID 
where a.AttributeID in (3,4) 
group by p.id, p.productname 
having count(distinct a.attributeid) = 2;

哪種方法是最好的你可能會依賴於你需要的結果和索引等等。

Sample SQL Fiddle.

來源

2015-10-06 00:49:37 jpw

mysql的連接表查詢的2個值

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