关系
Supplier
- >与orders
的一对多关系
| suppliers |
| ---------- |
| id | name | archived (bool) | user_id |
| orders |
| ---------------- |
| id | supplier_id | creator_id |
期望
现在,我想获得给定的suppliers
和supplier.user_id
的热门orders.creator_id
列表,supplier
不应该是archived
(请在问题的最后找到预期的结果)。
这就是我想要的:
不成功的尝试
这是我的失败,我不知道如何在此查询中添加供应商条件。
select supplier_id, COUNT(*) as count_suppliers
from orders
where creator_id = 2
group by orders.supplier_id
order by count_suppliers desc
所以这就是我想要的
suppliers
| id | user_id | archived |
| --- | --------- | ------- |
| 1 | 2 | false |
| 2 | 2 | false |
| 3 | 2 | false |
| 4 | 2 | false |
| 5 | 2 | true |
orders
| id | creator_id | supplier_id |
| -- | --------- | ------------ |
| 1 | 1 | 1 |
| 2 | 1 | 1 |
| 3 | 1 | 1 |
| 4 | 1 | 1 |
| 5 | 1 | 2 |
| 6 | 1 | 2 |
| 7 | 1 | 3 |
| 8 | 1 | 4 |
| 9 | 1 | 4 |
| 10 | 1 | 4 |
| 11 | 1 | 5 |
expected output
| supplier_id | supplier_count |
| 1 | 4 |
| 4 | 3 |
| 2 | 2 |
| 3 | 1 |
解
最后通过参考Vamsi的答案,在这里你可以找到RAW SQL版本和ActiveRecord(Rails)版本的解决方案来解决这个问题:
RAW SQL VERSION
SELECT o.supplier_id, COUNT(*) AS count_suppliers
FROM suppliers s
JOIN orders o ON s.id=o.supplier_id
WHERE s.user_id=2
AND s.archived=FALSE
AND o.creator_id=2
GROUP BY o.supplier_id
ORDER BY count_suppliers DESC
LIMIT 5
ActiveRecord (Rails) Version
Supplier
.joins(:orders)
.where(user_id: 2, archived: false, orders: { creator_id: 2 })
.group("orders.supplier_id")
.order("count_all DESC")
.limit(limit)
.count
你可以join
表和计数。
select p.supplier_id, COUNT(*) as count_suppliers
from purchase_orders p
join suppliers s on s.id=p.supplier_id
where s.user_id=2 and p.creator_id=1 and s.archived='False'
group by p.supplier_id
order by count_suppliers desc
select top 1 supplier_id, COUNT(*) as count_suppliers
from purchase_orders
where orders.creator_id = 1
And supplier.archived = false
And supplier.user_id = 2
group by purchase_orders.supplier_id
order by COUNT(*) desc