SQL |如果存在另一个具有相等（b，c）的元组，则列出所有元组（a，b，c）

我有三个表，其中粗体属性是主键

• 餐厅（餐厅ID，名称，…）

• 标识符（餐厅ID，食品ID）

• 食品（食品ID，名称，…）

使用postgres，我想列出所有与至少一家其他餐厅共享同一套食物的餐厅（餐厅id和名称-每家餐厅1行）

比如说

• ID为“1”的餐厅只有相关的食物ID为1和4，如

  标识符所示

• ID为“3”的餐厅只有相关的食物ID为4和1，如

  标识符所示

• ID为“7”的餐厅只有相关食物ID为6，如

  标识符所示

• ID为“9”的餐厅只有相关食物ID为6，如

  标识符所示

• 然后输出

任何帮助都将不胜感激

---

谢谢你

我理解你的问题，你希望所有餐厅的食物清单与餐厅1相同

如果是这样，那就是关系划分问题。下面是一种使用联接和聚合的方法：

select r.name
from identifier i1
inner join identifier i2 on i2.food_id = i1.food_id
inner join restaurant r on r.restaurant_id = i2.restaurant_id
where i1.restaurant_id = 1
group by r.restaurant_id
having count(*) = (select count(*) from identifier i3 where i3.restaurant_id = 1)

---

使用聚合函数

string\u agg（）

获取每个餐厅的完整食物列表：

with cte as (
  select restaurant_ID,
         string_agg(food_ID::varchar(10),',' order by food_ID) foods
  from identifier
  group by restaurant_ID
)
select r.* 
from Restaurants r inner join cte c
on c.restaurant_ID = r.restaurant_ID
where exists (select 1 from cte where restaurant_ID <> c.restaurant_ID and foods = c.foods)

---

请参阅。

这里有一种方法可以获得一套独特的餐厅，它们拥有完全相同的食物。这使用了array_agg（）和array_to_string（）函数

 With cte as
(select T.restaurant_id, array_to_string(array_agg(food_id), ',') as food_list
from
(select *
  from Identifier t1
  order by restaurant_id, food_id) T
  group by T.restaurant_id)

select 
   concat(r1.name,',',r2.name) as resturant_names,
   t1.restaurant_id as restaurant_id1,
   r1.name as restaurant_1,
   t2.restaurant_id as restaurant_id2,
   r2.name as restaurant_2,
   t1.food_list as common_food_ids
from cte t1
join cte t2
on t1.restaurant_id < t2.restaurant_id
and t1.food_list = t2.food_list
left join Restaurants r1
on t1.restaurant_id = r1.restaurant_id
left join Restaurants r2
on t2.restaurant_id = r2.restaurant_id;

---

你的意思是你想在你的例子中列出id为1和3的两家餐馆，而不仅仅是3家。是吗？1.你的示例数据和解释不匹配？2.为什么只列出餐厅

3

而不列出

1

？您应该在问题中包含您试图解决此问题的代码。很抱歉，该示例中有错误。我现在已经修好了。如果有另外两家餐厅的食物收集完全相同，但与1号和3号餐厅的食物收集不同，该怎么办？在这种情况下，你的预期结果是什么？嘿，对不起，我意识到我写的那个问题非常糟糕。我想列出所有与至少一家其他餐厅共享同一套食物的餐厅。哇。。。非常感谢你抽出时间帮助一个陌生人。我以后一定会把我的问题弄清楚的！非常感谢您抽出时间回答！

Restaurant_id      name
1                  name1
3                  name3
7                  ...
9                  ...

select r.name
from identifier i1
inner join identifier i2 on i2.food_id = i1.food_id
inner join restaurant r on r.restaurant_id = i2.restaurant_id
where i1.restaurant_id = 1
group by r.restaurant_id
having count(*) = (select count(*) from identifier i3 where i3.restaurant_id = 1)

with cte as (
  select restaurant_ID,
         string_agg(food_ID::varchar(10),',' order by food_ID) foods
  from identifier
  group by restaurant_ID
)
select r.* 
from Restaurants r inner join cte c
on c.restaurant_ID = r.restaurant_ID
where exists (select 1 from cte where restaurant_ID <> c.restaurant_ID and foods = c.foods)

with cte as (
  select restaurant_ID,
         string_agg(food_ID::varchar(10),',' order by food_ID) foods
  from identifier
  group by restaurant_ID
)
select string_agg(r.name, ',') restaurants
from Restaurants r inner join cte c
on c.restaurant_ID = r.restaurant_ID
group by foods
having count(*) > 1

 With cte as
(select T.restaurant_id, array_to_string(array_agg(food_id), ',') as food_list
from
(select *
  from Identifier t1
  order by restaurant_id, food_id) T
  group by T.restaurant_id)

select 
   concat(r1.name,',',r2.name) as resturant_names,
   t1.restaurant_id as restaurant_id1,
   r1.name as restaurant_1,
   t2.restaurant_id as restaurant_id2,
   r2.name as restaurant_2,
   t1.food_list as common_food_ids
from cte t1
join cte t2
on t1.restaurant_id < t2.restaurant_id
and t1.food_list = t2.food_list
left join Restaurants r1
on t1.restaurant_id = r1.restaurant_id
left join Restaurants r2
on t2.restaurant_id = r2.restaurant_id;

With cte as
(select T.restaurant_id, array_to_string(array_agg(food_id), ',') as food_list
from
(select *
  from Identifier t1
  order by restaurant_id, food_id) T
  group by T.restaurant_id)

select 
   --concat(r1.name,',',r2.name) as resturant_names,
   t1.restaurant_id as restaurant_id,
   r1.name as restaurant--,
  --t2.restaurant_id as restaurant_id2,
  --r2.name as restaurant_2,
  --t1.food_list as common_food_ids
from cte t1
join cte t2
on t1.restaurant_id = t2.restaurant_id
and t1.food_list = t2.food_list
left join Restaurants r1
on t1.restaurant_id = r1.restaurant_id
left join Restaurants r2
on t2.restaurant_id = r2.restaurant_id;