多对多查询

Question

我有两个表产品和多个关系的部分和一个连接表products_sections。产品可以是一个或多个部分（新的，汽车，飞机，旧）。<​​/ p>

Products
id    name
-----------------
1     something
2     something_else
3     other_thing


Sections
id    name
-----------------
1     new
2     car

Products_sections
product_id     section_id
--------------------------
1              1
1              2
2              1
3              2

我想提取新车和汽车部分的所有产品。在此示例中，返回的结果应为product 1.获取此结果的正确mysql查询是什么？

Answer 1

SELECT Products.name
FROM Products
WHERE NOT EXISTS (
  SELECT id
  FROM Sections
  WHERE name IN ('new','car')
  AND NOT EXISTS (
    SELECT *
    FROM Products_sections
    WHERE Products_sections.section_id = Sections.id
    AND Products_sections.product_id = Products.id
  )
)

换句话说，选择那些产品的Products_sections表中没有缺少所需的Section.id值的产品。

回答andho的评论：

你可以把

  NOT EXISTS (<select query>)

像任何其他谓词一样进入WHERE子句。如果＆lt; select query＆gt;描述的结果集中没有行，它将评估为TRUE。

逐步地，这里是如何将此查询作为答案：

步骤1.要求是识别“同时在'新'和'汽车'部分中的所有产品”。

步骤2.如果“新”和“汽车”部分都包含产品，则产品位于“新”和“汽车”部分。同样地，如果这些部分都不能包含产品，则产品同时位于“新”和“汽车”部分。 （请注意双重否定： 无法包含。）再次重申，我们希望所有产品都有否必需部分失败包含产品。

所需的部分是：

SELECT id
FROM Sections
WHERE name IN ('new','car')

因此，所需的产品是：

SELECT Products.name
FROM Products
WHERE NOT EXISTS ( -- there does not exist
  SELECT id  -- a section
  FROM Sections
  WHERE name IN ('new','car') -- that is required
   AND (the section identified by Sections.id fails to contain the product identified by Products.id)
)

步骤3.如果给定部分和特定产品的Products_sections中有一行，则给定部分（例如“新”或“汽车”）确实包含特定产品。因此，如果Products_sections中没有这样的行，则给定的部分包含特定产品。

步骤4.如果下面的查询包含一行，则section_id部分 包含product_id产品：

SELECT *
FROM Products_sections
WHERE Products_sections.section_id = Sections.id
AND Products_sections.product_id = Products_id

因此，如果上面的查询不在其结果中产生一行，或者如果上面的查询

>

那么无法包含产品（这就是我们需要表达的内容）不存在（）。

看起来很复杂，但是一旦你掌握了它，它就会坚持：是否所有必需的物品都存在？是的，只要不存在不存在的必需项目。

答案 1 :(得分：2)

我总是这样做：

从您尝试获取的内容开始（products），然后通过查找表（products_sections）查看您尝试过滤的内容（sections） 。通过这种方式，您可以在普通视图中查找您正在寻找的内容，而且您无需记住代理键（这是一件很棒的事情，而不是记忆）。

select distinct
    p.name
from
    products p
    inner join products_sections ps on
        p.product_id = ps.product_id
    inner join sections s1 on
        ps.section_id = s1.section_id
    inner join sections s2 on
        ps.section_id = s2.section_id
where
    s1.name = 'new'
    and s2.name = 'car'

瞧。三个inner join，你有一个很好的，清晰的，简洁的查询，显而易见的是它带来了什么。希望这有帮助！

答案 2 :(得分：0)

SELECT product_id, count(*) AS TotalSection
FROM Products_sections
GROUP BY product_id
WHERE section_id IN (1,2)
HAVING TotalSection = 2;

看看这是否适用于mysql。

答案 3 :(得分：0)

下面的查询有点笨拙，但它应该回答你的问题：

select products.id
from products
where products.id in 
  (
    select products_sections.product_id
    from products_sections
    where products_sections.section_id=1
  )
  and products.id in
  (
    select products_sections.product_id
    from products_sections
    where products_sections.section_id=2
  )

答案 4 :(得分：0)

在两个连接表子集上自行连接，然后选择唯一的产品ID。

SELECT DISTINCT car.product_id 
FROM ( SELECT product_id 
         FROM Product_sections 
        WHERE section_id = 2
     ) car JOIN 
     ( SELECT product_id 
         FROM Product_sections 
        WHERE section_id = 1
     ) neww 
     ON (car.product_id = neww.product_id) 

此查询是更一般解决方案的变体：

SELECT DISTINCT car.product_id
  FROM product_sections car join 
       product_sections neww ON (car.product_id = neww.product_id AND 
                                 car.section_id = 2 AND 
                                 neww.section_id = 2)   

---

效率较低但更直接的解决方案是：

SELECT p.name FROM Products p WHERE
 EXISTS (SELECT 'found car' 
           FROM Products_sections ps 
          WHERE ps.product_id = p.id AND ps.section_id = 2)
 AND
 EXISTS (SELECT 'found new' 
           FROM products_sections ps 
          WHERE ps.product_id = p.id AND ps.section_id = 1)

----------------

为了清晰起见，我使用id进行操作。如有必要，请将表达式section_id = 2和section_id = 1替换为

section_id = (SELECT s.id FROM Sections s WHERE s.name = 'car')

section_id = (SELECT s.id FROM Sections s WHERE s.name = 'new')

此外，您可以通过插入上述任何查询来选择产品名称，如下所示：

SELECT Products.name FROM Products 
 WHERE EXISTS (
         SELECT 'found product'
           FROM product_sections car join 
                product_sections neww ON (car.product_id = neww.product_id AND 
                                          car.section_id = 2 AND 
                                          neww.section_id = 2)
          WHERE car.product_id = Products.id
       )

答案 5 :(得分：0)

SELECT p.*
FROM   Products p
       INNER JOIN (SELECT   ps.product_id
                   FROM     Products_sections ps
                            INNER JOIN Sections s
                              ON s.id = ps.section_id
                   WHERE    s.name IN ("new","car")
                   GROUP BY ps.product_id
                   HAVING   Count(ps.product_id) = 2) pp
         ON p.id = pp.product_id

答案 6 :(得分：0)

当您需要搜索更多部分时，此查询将为您提供结果，而无需添加更多内部联接。这里会改变的是：

1. IN（）paranthesis
中的值

2. count的where子句中的值，应该替换为您要搜索的部分的数量

   选择id，name FROM
   （
   SELECT
     products。id，
     products。name，
     sections。name AS section_name，
     COUNT（*）AS count FROM products
   INNER JOIN products_sections
   在products_sections。product_id = products。id INNER JOIN sections
   在sections。id = products_sections。section_id 在哪里sections。name IN（'car'，'new'）
   GROUP BY products。id
   ）AS P
   在哪里count = 2

答案 7 :(得分：0)

select
`p`.`id`,
`p`.`name`
from `Sections` as `s`
join `Products_sections` as `ps` on `ps`.`section_id` = `s`.`id`
join `Products` as `p` on `p`.`id` = `ps`.`product_id`
where `s`.`id` in ( 1,2 )
having count( distinct `s`.`name` = 2 )

将返回......

id    name
-----------------
1     something

那是你在找什么？

Answer 2

我总是这样做：

从您尝试获取的内容开始（products），然后通过查找表（products_sections）查看您尝试过滤的内容（sections） 。通过这种方式，您可以在普通视图中查找您正在寻找的内容，而且您无需记住代理键（这是一件很棒的事情，而不是记忆）。

select distinct
    p.name
from
    products p
    inner join products_sections ps on
        p.product_id = ps.product_id
    inner join sections s1 on
        ps.section_id = s1.section_id
    inner join sections s2 on
        ps.section_id = s2.section_id
where
    s1.name = 'new'
    and s2.name = 'car'

瞧。三个inner join，你有一个很好的，清晰的，简洁的查询，显而易见的是它带来了什么。希望这有帮助！

Answer 3

SELECT product_id, count(*) AS TotalSection
FROM Products_sections
GROUP BY product_id
WHERE section_id IN (1,2)
HAVING TotalSection = 2;

看看这是否适用于mysql。

Answer 4

下面的查询有点笨拙，但它应该回答你的问题：

select products.id
from products
where products.id in 
  (
    select products_sections.product_id
    from products_sections
    where products_sections.section_id=1
  )
  and products.id in
  (
    select products_sections.product_id
    from products_sections
    where products_sections.section_id=2
  )

Answer 5

在两个连接表子集上自行连接，然后选择唯一的产品ID。

SELECT DISTINCT car.product_id 
FROM ( SELECT product_id 
         FROM Product_sections 
        WHERE section_id = 2
     ) car JOIN 
     ( SELECT product_id 
         FROM Product_sections 
        WHERE section_id = 1
     ) neww 
     ON (car.product_id = neww.product_id) 

此查询是更一般解决方案的变体：

SELECT DISTINCT car.product_id
  FROM product_sections car join 
       product_sections neww ON (car.product_id = neww.product_id AND 
                                 car.section_id = 2 AND 
                                 neww.section_id = 2)   

---

效率较低但更直接的解决方案是：

SELECT p.name FROM Products p WHERE
 EXISTS (SELECT 'found car' 
           FROM Products_sections ps 
          WHERE ps.product_id = p.id AND ps.section_id = 2)
 AND
 EXISTS (SELECT 'found new' 
           FROM products_sections ps 
          WHERE ps.product_id = p.id AND ps.section_id = 1)

----------------

为了清晰起见，我使用id进行操作。如有必要，请将表达式section_id = 2和section_id = 1替换为

section_id = (SELECT s.id FROM Sections s WHERE s.name = 'car')

section_id = (SELECT s.id FROM Sections s WHERE s.name = 'new')

此外，您可以通过插入上述任何查询来选择产品名称，如下所示：

SELECT Products.name FROM Products 
 WHERE EXISTS (
         SELECT 'found product'
           FROM product_sections car join 
                product_sections neww ON (car.product_id = neww.product_id AND 
                                          car.section_id = 2 AND 
                                          neww.section_id = 2)
          WHERE car.product_id = Products.id
       )

Answer 6

SELECT p.*
FROM   Products p
       INNER JOIN (SELECT   ps.product_id
                   FROM     Products_sections ps
                            INNER JOIN Sections s
                              ON s.id = ps.section_id
                   WHERE    s.name IN ("new","car")
                   GROUP BY ps.product_id
                   HAVING   Count(ps.product_id) = 2) pp
         ON p.id = pp.product_id

Answer 7

当您需要搜索更多部分时，此查询将为您提供结果，而无需添加更多内部联接。这里会改变的是：

1. IN（）paranthesis
中的值

2. count的where子句中的值，应该替换为您要搜索的部分的数量

   选择id，name FROM
   （
   SELECT
     products。id，
     products。name，
     sections。name AS section_name，
     COUNT（*）AS count FROM products
   INNER JOIN products_sections
   在products_sections。product_id = products。id INNER JOIN sections
   在sections。id = products_sections。section_id 在哪里sections。name IN（'car'，'new'）
   GROUP BY products。id
   ）AS P
   在哪里count = 2

Answer 8

select
`p`.`id`,
`p`.`name`
from `Sections` as `s`
join `Products_sections` as `ps` on `ps`.`section_id` = `s`.`id`
join `Products` as `p` on `p`.`id` = `ps`.`product_id`
where `s`.`id` in ( 1,2 )
having count( distinct `s`.`name` = 2 )

将返回......

id    name
-----------------
1     something

那是你在找什么？