找出哪个组ID包含SQL中的所有相关属性 [英] Find out what group id contains all relevant attributes in SQL

问题描述

因此，在这种情况下，我们所拥有的群体是动物群体.

So lets say in this case, the group that we have is groups of animals.

可以说我有以下表格:

animal_id | attribute_id | animal
----------------------------------
 1        |  1           | dog   
 1        |  4           | dog
 2        |  1           | cat
 2        |  3           | cat
 3        |  2           | fish
 3        |  5           | fish


id | attribute
------------------
 1 | four legs 
 2 | no legs
 3 | feline
 4 | canine
 5 | aquatic

第一个表包含定义动物的属性，第二个表跟踪每个属性的含义.现在假设我们对某些数据运行查询并获得以下结果表:

Where the first table contains the attributes that define an animal, and the second table keeps track of what each attribute is. Now lets say that we run a query on some data and get the following result table:

attribute_id 
------------
     1
     4

此数据将描述一条狗，因为它是唯一同时具有属性1和4的animal_id.我希望能够以某种方式基于第三张表获得animal_id(在这种情况下为1)，本质上是已经生成的包含动物属性的表.

This data would describe a dog, since it is the only animal_id that has both attributes 1 and 4. I want to be able to somehow get the animal_id (which in this case would be 1) based on the third table, which is essentially a table that has already been generated that contains the attributes of an animal.

编辑

因此具有1和4的第三个表不必是1和4.它可以返回2和5(对于鱼)，或者返回1和3(猫).我们可以假设它的结果将始终完全匹配一只动物，但是我们不知道哪一只.

So the third table that has 1 and 4 doesn't have to be 1 and 4. It could return 2 and 5 (for fish), or 1 and 3 (cat). We can assume that it's result will always match one animal completely, but we don't know which one.

推荐答案

您可以使用group by和having:

with a as (
      select 1 as attribute_id from dual union all
      select 4 as attribute_id from dual
     )
select t.animal_id, t.animal
from t join
     a
     on t.attribute_id = a.attribute_id
group by t.animal_id, t.animal
having count(*) = (select count(*) from a);

以上内容将找到所有具有和属性的动物.如果您希望动物具有这两个属性:

The above will find all animals that have those attributes and any others. If you want animals that have exactly those 2 attributes:

with a as (
      select 1 as attribute_id from dual union all
      select 4 as attribute_id from dual
     )
select t.animal_id, t.animal
from t left join
     a
     on t.attribute_id = a.attribute_id
group by t.animal_id, t.animal
having count(*) = (select count(*) from a) and
       count(*) = count(a.attribute_id);

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