Question

我有一个看起来像这样的表

Table: Recommendations
Recommended_ID    Item_ID     Recommended_Item_ID    User_id
1                   1                   3                1
2                   1                   3                6
3                   1                   2                7
4                   2                   5                1

我希望我的SQL查询在伪代码中实现类似的东西：

SELECT recommended_item_id as f, COUNT(recommended_id) WHERE recommended_item_id = f AND item_id = 1, FROM `Recommendations` WHERE item_id = 1

To hopefully return:
Recommended_Item_ID   Recommended_Number_of_Times
3                              2
2                              1

基本上，我尝试获取为item_idx推荐的所有项ID，并返回推荐它们的次数以及ID。如何让我的SQL查询执行此操作？我已尝试过GROUP BY和HAVING，但问题是我不知道如何确保recommended_item_id等于我告诉SQL要搜索的initial_recommended_id。

Answer 1

select 
  Recommended_Item_ID,
  COUNT(recommended_id)
from `Recommendations`
where Item_ID = 1
group by Recommended_Item_ID

Answer 2

查看group by和where and group by。

通过

尝试分组
select Recommended_Item_ID, count(*) as Recommended_Number_of_Times from Recommendations where Item_ID = X group by Recommended_Item_ID order by count(*) desc

在一个查询中执行此操作的方法

2 个答案: