如何选择所有联接记录都不符合条件的记录

时间:2018-10-29 13:28:51

标签: mysql sql relational-division

我使用以下表格进行了设置(使用MySQL):

  • orders,其中有很多:
  • 联接表order_items,其中有一个联接表:
  • products

我写了一个查询来选择orders,其中所有他们的products都属于某个type

SELECT orders.* FROM orders 
INNER JOIN order_items ON order_items.order_id = orders.id   
INNER JOIN products ON products.id = order_items.product_id     
WHERE products.type = 'FooProduct'
AND (
  NOT EXISTS (
    SELECT null
    FROM products
    INNER JOIN order_items ON order_items.product_id = products.id
    WHERE order_items.order_id = orders.id
    AND products.type != 'FooProduct'
  )
 )

我也有几次类似的运行:首先要获得包含所有FooProduct的订单,然后要获得包含所有BarProduct的订单。

我的症结一直是生成第三个查询以获取所有其他订单,即,其所有产品的类型不是独占FooProduct还是独占BarProduct(又称为两者的组合)或其他产品类型。

因此,我的问题是,如何获得所有所有产品类型不完全是FooProduct或完全不是BarProduct的记录。


这里有一些示例数据,我想从中返回ID为3和4的订单:

- orders
id
 1
 2
 3
 4

-- order_items

id order_id product_id
 1        1          1
 2        1          1
 3        2          2
 4        2          2
 5        3          3
 6        3          4
 7        4          1
 8        4          2

-- products
id type
 1 'FooProduct'
 2 'BarProduct'
 3 'OtherProduct'
 4 'YetAnotherProduct'

我已经尝试过这样做,因此非常抱歉,因此请使用以下内容代替现有的AND(甚至是语法):

NOT HAVING COUNT(order_items.*) = (
  SELECT null
        FROM products
        INNER JOIN order_items ON  order_items.product_id = products.id
        WHERE order_items.order_id = orders.id
        AND products.type IN ('FooProduct', 'BarProduct')
)

5 个答案:

答案 0 :(得分:1)

可以使用Having和基于条件聚合函数的过滤来代替使用相关子查询。

如果

products.type IN ('FooProduct', 'BarProduct')的产品类型都不是,则返回0。我们可以在其上使用Sum()函数,以进行进一步的过滤。

请尝试以下操作:

SELECT orders.order_id 
FROM orders 
INNER JOIN order_items ON order_items.order_id = orders.id   
INNER JOIN products ON products.id = order_items.product_id 
GROUP BY orders.order_id 
HAVING SUM(products.type IN ('FooProduct', 'BarProduct')) < COUNT(*)

在这种情况下,如果要查找仅具有FooProduct类型的订单,则可以改用以下内容:

SELECT orders.order_id 
FROM orders 
INNER JOIN order_items ON order_items.order_id = orders.id   
INNER JOIN products ON products.id = order_items.product_id 
GROUP BY orders.order_id 
HAVING SUM(products.type <> 'FooProduct') = 0

另一种可能的方法是:

SELECT orders.order_id 
FROM orders 
INNER JOIN order_items ON order_items.order_id = orders.id   
INNER JOIN products ON products.id = order_items.product_id 
GROUP BY orders.order_id 
HAVING SUM(products.type = 'FooProduct') = COUNT(*)

答案 1 :(得分:1)

您可以为此使用聚合和having子句:

SELECT o.*
FROM orders o INNER JOIN
     order_items oi
     ON oi.order_id = o.id INNER JOIN
     products p
     ON p.id = oi.product_id   
GROUP BY o.id  -- OK assuming `id` is the primary key
HAVING SUM(p.type NOT IN ('FooProduct', 'BarProduct')) > 0;  -- at least one other product 

实际上,这不太正确。这会获得具有其他产品的订单,但不会选择仅由foo和bar混合而成的订单。我认为这会吸引其他人:

HAVING SUM(p.type = 'FooProduct') < COUNT(*) AND
       SUM(p.type = 'BarProduct') < COUNT(*) 

答案 2 :(得分:1)

这是一个关系划分问题。
查找所有产品都是给定类型的订单的一种解决方案是:

SELECT *
FROM orders
INNER JOIN order_items ON order_items.order_id = orders.id
INNER JOIN products ON products.id = order_items.product_id
WHERE orders.id IN (
    SELECT order_items.order_id
    FROM order_items
    INNER JOIN products ON products.id = order_items.product_id
    GROUP BY order_items.order_id
    HAVING COUNT(CASE WHEN products.type = 'FooProduct' THEN 1 END) = COUNT(*)
)

略微调整上面的内容,以找到所有产品都来自给定类型列表的订单是

HAVING COUNT(CASE WHEN products.type IN ('FooProduct', 'BarProduct') THEN 1 END) = COUNT(*)

要查找所有产品与给定列表中的所有类型匹配的所有订单

HAVING COUNT(CASE WHEN products.type IN ('FooProduct', 'BarProduct') THEN 1 END) = COUNT(*)
AND    COUNT(DISTINCT products.type) = 2

DB Fiddle with tests

答案 3 :(得分:1)

这是一个基本解决方案,效率不高,但很简单:

SELECT * FROM orders WHERE id NOT IN (
    SELECT orders.id FROM orders 
    INNER JOIN order_items ON order_items.order_id = orders.id   
    INNER JOIN products ON products.id = order_items.product_id     
    WHERE products.type = 'FooProduct'
    AND (
      NOT EXISTS (
        SELECT null
        FROM products
        INNER JOIN order_items ON order_items.product_id = products.id
        WHERE order_items.order_id = orders.id
        AND products.type != 'FooProduct'
      )
 )
) AND id NOT IN (
    SELECT orders.id FROM orders 
    INNER JOIN order_items ON order_items.order_id = orders.id   
    INNER JOIN products ON products.id = order_items.product_id     
    WHERE products.type = 'BarProduct'
    AND (
      NOT EXISTS (
        SELECT null
        FROM products
        INNER JOIN order_items ON order_items.product_id = products.id
        WHERE order_items.order_id = orders.id
        AND products.type != 'BarProduct'
      )
 )
)

答案 4 :(得分:1)

我建议像这样在连接的子选择中使用count(distinct):

SELECT orders.*
FROM orders 
inner join (
    SELECT orderid, max(products.type) as products_type
    FROM order_items
    INNER JOIN products ON products.id = order_items.product_id
    GROUP BY orderid
    -- distinct count of different products = 1 
    --    -> all order items are for the same product type
    HAVING COUNT(distinct products.type ) = 1 
    -- alternative is:
    -- min(products.type )=max(products.type )
) as tmp on tmp.orderid=orders.orderid 
WHERE 1=1
-- if you want only single type product orders for some specific product
and tmp.products_type = 'FooProduct'