Question

具有：

SELECT *, COUNT(MakeModel) as `count`
FROM (`table_a`)
GROUP BY `MakeModel`
ORDER BY `count` DESC

适合我（计数= 2）使用LEFT JOIN扩展此查询，如下所示：

SELECT *, COUNT(distinct b.rating) as ratings
FROM (`table_a`)
LEFT JOIN `model_ratings` AS b ON `b`.`MakeModel` = `table_a`.`MakeModel`
GROUP BY `table_a`.`MakeModel`

也工作正常（评分= 3），但结合这两个计数：

SELECT *, COUNT(table_a.MakeModel) as `count`, COUNT(distinct b.rating) as ratings
FROM (`table_a`)
LEFT JOIN `model_ratings` AS b ON `b`.`MakeModel` = `table_a`.`MakeModel`
GROUP BY `table_a`.`MakeModel`
ORDER BY `count` DESC

然后（count = 6）和（rating = 3）。

因此计数变为（'计数'次'评级'）。

但我需要（count = 2）和（rating = 3）

我在这里缺少什么？

Answer 1

由我自己找到 LEFT JOIN中的COUNT（）和GROUP首先

SELECT *, COUNT(table_a.MakeModel) as `count`, b.rating as ratings
FROM (`table_a`)
LEFT JOIN (
   SELECT MakeModel, COUNT(distinct rating)
   FROM `model_ratings`
   GROUP BY MakeModel
   ) AS b ON `b`.`MakeModel` = `table_a`.`MakeModel`

GROUP BY `table_a`.`MakeModel`
ORDER BY `count` DESC

一个count（）的sql在LEFT JOIN中的另一个count（）

1 个答案: