计数索引存在使用包含多个存在（）语句，如果他们在一个单独的表索引中存在

Question

我的查询获取这些产品基于所述结果的查询的次数。我正在尝试对所有存在的实例进行计数，以便按相关性对结果进行排序。我所尝试的一切似乎都会将变量@priority返回为0.任何想法？

也许是更好地使用加入语句？

谢谢你的帮助。这里是我的MySQL查询：

SELECT `products` . * , @priority 
    FROM `products` 
    LEFT JOIN productstypes_index ON productstypes_index.product_id = products.id 
WHERE (

EXISTS (

SELECT * 
    FROM `productstypes_index` 
    WHERE `productstypes_index`.`product_id` = `products`.`id` 
    AND `productstypes_index`.`_type_id` = '1' 
) 
AND (
(
(

EXISTS (

SELECT @priority := COUNT(*) 
    FROM `producthashtags_index` 
    WHERE `producthashtags_index`.`product_id` = `products`.`id` 
    AND `producthashtags_index`.`producthashtag_id` = '43' 
) 
) 
AND (

EXISTS (

SELECT @priority := COUNT(*) 
    FROM `producthashtags_index` 
    WHERE `producthashtags_index`.`product_id` = `products`.`id` 
    AND `producthashtags_index`.`producthashtag_id` = '11' 
) 
) 
) 
) 
) 
ORDER BY `updated_at` DESC; 

Answer 1

你可以这样做没有那些exists，没有变数。另外，如果连接表上有exists条件，left join就没有意义。那么你不妨做更高效的inner join，并把额外的类型条件放在连接条件中。

优先级可以通过计数在哈希标签来计算，但只有那些与id in ('43', '11')。

SELECT  products.* 
      count(distinct producthashtags_index.producthashtag_id) priority 
FROM  products 
INNER JOIN productstypes_index 
     ON productstypes_index.product_id = products.id 
     AND productstypes_index._type_id = '1' 
INNER JOIN producthashtags_index 
     ON producthashtags_index.product_id = products.id 
     AND producthashtags_index.producthashtag_id in ('43', '11') 
GROUP BY products.id 
ORDER BY updated_at DESC; 

Answer 2

的MySQL忽略了SELECT列表中是否存在子查询，所以它没有什么区别，你键入那里。这被记录在here。

的一种方法使用连接看起来象下面这样：

SELECT p.id, 
     COUNT(case when phi.product_id is not null then 1 end) AS instances 
FROM products p 
INNER JOIN productstypes_index pti ON pti.product_id = p.id AND pti.`_type_id` = 1 
LEFT JOIN producthashtags_index phi ON phi.product_id = p.id AND phi.producthashtag_id IN (11,43) 
GROUP BY p.id 
ORDER BY instances DESC; 

我已删除的附加反引号，我相信他们是不会neccessary并且如果在表中你id列是整数，则不需要引号。