Question

我正在试图找出如何查询我的表以查看一组user_id是否与conversation_id匹配。

查询1应返回以下结果： user_id 1正在查看是否有任何conversation_id，其中只有user_id = 2和user_id = 1。（应根据SQL Fiddle示例为每个conversation_id = 1,2,4,5返回一行）

conversation_id
1
2
4
5

查询2应返回以下结果： user_id 1正在查看是否有任何conversation_id与user_id = 2，user_id = 1，和user_id = 4 。（应返回0行，因为它在SQL Fiddle示例中不存在）

表格设置位于 SQL Fiddle

Answer 1

您可以使用group by ... having和相关exists子查询的组合来获得所需的结果：

-- Query 1:

SELECT 
  conversation_id
FROM 
  users_conversations uc
where not exists (
  select 1 from users_conversations
  where conversation_id = uc.conversation_id
  and user_id not in (1,2)
)
group by conversation_id
having count(distinct user_id) = 2;

-- Query 2: same query, only different numbers.

SELECT 
  conversation_id
FROM 
  users_conversations uc
where not exists (
  select 1 from users_conversations
  where conversation_id = uc.conversation_id
  and user_id not in (1,2,4))
group by conversation_id
having count(distinct user_id) = 3;

Sample SQL Fiddle

请注意，第一个查询不会返回1,2,4,5，而是返回2,5，但在您的示例数据中，1或4都只有user_id 1和2作为参与者（conversation 1具有{{1} }}，1,2,3,4有conversation 4）。

Answer 2

如果我理解正确的话应该是他的。

Q1：

SELECT
CASE 
WHEN
count(distinct CASE WHEN user_id in ('1','2') THEN user_id END)>='2' 
THEN `conversation_id` 
END 'test'
FROM 
users_conversations
where 1
group by `conversation_id`

Q2：

SELECT
CASE 
WHEN 
count(distinct CASE WHEN user_id in ('1','2','4') THEN user_id END)>='3' 
THEN `conversation_id` 
END 'test'
FROM 
users_conversations
where 1
group by `conversation_id`

http://sqlfiddle.com/#!9/fb29d/9

MySQL检查逗号分隔值是否与表

2 个答案: