SQL의 전이 속성

Question

이것은 내가 알아야 할 질문이지만, 나는 SQL에서 표현하고자하는 것처럼 지루한 시간을 보냈다.

제 질문은 단순한 질문입니다. SQL에서 전이 관계를 표현하기위한 관용구는 무엇입니까? 구체적인 예 :

user(email, name) 
friends(friend1_email, friend2_email) 

나는 다음과 같은 쿼리 표현에 문제가있어 :

찾기 사용자 A, B 및 C 등은 A가 친구임을를

나는 다음과 같은 스키마를 말해봐 B와 B는 C와 친구지만 C는 A와 친구가 아닙니다.

나는 이것이 숙제라고 인정 하겠지만, 나는 표현을 표현하는 데 개념적인 문제가있었습니다. 어떤 도움을 주시면 감사하겠습니다. 감사.

Answer 1

복잡한 쿼리와 내 조언은 항상 간단하게 시작하는 것입니다 : 충분히 간단한

# Find friends A and B 
select A.email as A_email, A.name as A_name, B.email as B_email, B.name as B_name 
from user A 
join friends 
on A.email = friends.friend1_email 
join user B 
on B.email = friends.friend2.email 

,의는 B와 C 다시 그것을 할 수 있습니다 :

# Find friends B and C 
select B.email as B_email, B.name as B_name, C.email as C_email, C.name as C_name 
from user B 
join friends 
on B.email = friends.friend1_email 
join user C 
on C.email = friends.friend2.email 

이제 단일 쿼리에서 A, B 및 C를 결합하여

# Find friends A, B, and C 
select A.email as A_email, A.name as A_name, B.email as B_email, B.name as B_name, C.email as C_email, C.name as C_name 
from user A 
join friends f1 
on A.email = f1.friend1_email 
join user B 
on f1.friend2_email = B.email 
join friends f2 
on B.email = f2.friend1_email 
join user C 
on f2.friend2_email = C.email 

위의 쿼리는 사용자 C와 친구 인 사용자 B와 친구 인 모든 사용자 A를 제공하지만 결과 세트를 A와 C가 친구가 아닌 레코드로 제한하지 않습니다. 결과 집합을 얻으려면 쿼리를 약간 수정해야합니다.

# Find friends A, B, and C 
select A.email as A_email, A.name as A_name, B.email as B_email, B.name as B_name, C.email as C_email, C.name as C_name 
from user A 
join friends f1 
on A.email = f1.friend1_email 
join user B 
on f1.friend2_email = B.email 
join friends f2 
on B.email = f2.friend1_email 
join user C 
on f2.friend2_email = C.email 
left join friends f3 
on A.email = f3.friend1_email 
and C.email = f3.friend2_email 
where 
    f3.friend1_email is null 

Answer 2

뭔가 같은 :

이

select * 
from friends a 

--If this joins, a and b are friends 
left join friends b 
on (a.friend1_email = b.friend1_email or 
    a.friend1_email = b.friend2_email or 
    a.friend2_email = b.friend1_email or 
    a.friend2_email = b.friend2_email) 

--If this joins, b and c are friends 
left join friends c 
on (b.friend1_email = c.friend1_email or 
    b.friend1_email = c.friend2_email or 
    b.friend2_email = c.friend1_email or 
    b.friend2_email = c.friend2_email) 

--If this joins, c and a are friends. 
--Also making sure that a is same person as a2 
left join friends a2 
on (c.friend1_email = a2.friend1_email or 
    c.friend1_email = a2.friend2_email or 
    c.friend2_email = a2.friend1_email or 
    c.friend2_email = a2.friend2_email) 
and (a.friend1_email = a2.friend1_email or 
    a.friend1_email = a2.friend2_email or 
    a.friend2_email = a2.friend1_email or 
    a.friend2_email = a2.friend2_email) 

where b.friend1_email is not null --join was made, a and b are friends 
and c.friend1_email is not null --join was made, b and c are friends 
and a2.friend1_email is null --join was NOT made, a and c are NOT friends