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• sql 的 join、left join、full join的區別圖解總結，測試，注意事項

1. 結論示意圖

• 對于intersect、minus，oracle支持，mysql不支持，可以變通（in或exists）實現

2.測試

2.1.創建表和數據

-- 建表
drop table if exists student;   -- oralce 不支持 if exists 
create table student (id   int
);
-- 造數據4條
insert into student (id) values (1);
insert into student (id) values (2);
insert into student (id) values (3);
insert into student (id) values (4);-- 查看表數據
select * from student;

2.2.查詢

2.2.1. A

-- A
select * from student where id in (1,2,3);

2.2.1. B

-- B
select * from student where id in (2,3,4);

2.2.3. intersect(A ∩ B)。交集。oracle支持，mysql不支持（可以變通實現）

-- intersect(A ∩ B)。交集
select * from student where id in (1,2,3)
intersect
select * from student where id in (2,3,4);
-- 變通實現
select * from student where id in (1,2,3)
and id in (
select id from student where id in (2,3,4));

2.2.4. minus(A - A ∩ B)。左差集。oracle支持，mysql不支持（可以變通實現）

-- minus(A - A ∩ B)。左差集
select * from student where id in (1,2,3)
minus
select * from student where id in (2,3,4);
-- 變通實現
select * from student where id in (1,2,3)
and id not in (
select id from student where id in (2,3,4));

2.2.5. minus(A - A ∩ B)。右差集。oracle支持，mysql不支持（可以變通實現）

-- minus(A - A ∩ B)。右差集
select * from student where id in (2,3,4)
minus
select * from student where id in (1,2,3);
-- 變通實現
select * from student where id in (2,3,4)
and id not in (
select id from student where id in (1,2,3));

2.2.6. union(A ∪ B)。并集（去重）

-- union(A ∪ B)。并集（去重）
select * from student where id in (1,2,3)
union 
select * from student where id in (2,3,4);

2.2.7. union all(A + B)。和集（不去重）

-- union all(A + B)。和集（不去重）
select * from student where id in (1,2,3)
union all
select * from student where id in (2,3,4);

2.2.8. (A minus B) union (B minus A)[(A - B) + (B - A)]或 (A union B) minus (A intersect B)[(A ∪ B) - (A ∩ B)] 。A ∩ B在A ∪ B的補集。oracle支持，mysql不支持（可以變通實現）

-- 算法1：`(A minus  B) union (B minus A)`[(A - B) + (B - A)]。A ∩ B在A ∪ B的補集。
(
select * from student where id in (1,2,3)
minus
select * from student where id in (2,3,4)
)
union 
(
select * from student where id in (2,3,4)
minus
select * from student where id in (1,2,3)
);
-- 算法1：變通實現
(
select * from student where id in (1,2,3)
and id not in (
select id from student where id in (2,3,4))
)
union 
(
select * from student where id in (2,3,4)
and id not in (
select id from student where id in (1,2,3))
);-- 算法2：`(A union B) minus (A intersect B)`[(A ∪ B) - (A ∩ B)] 
--  `(union) minus (intersect)`[(A ∪ B) - (A ∩ B)]。A ∩ B在A ∪ B的補集。
(
select * from student where id in (2,3,4)
union
select * from student where id in (1,2,3)
)
minus
(
select * from student where id in (2,3,4)
intersect
select * from student where id in (1,2,3)
);
-- 算法2：變通實現
select * from 
(
select * from student where id in (1,2,3)
union 
select * from student where id in (2,3,4)
)
where id not in
(
select id from student where id in (1,2,3)
and id in (
select id from student where id in (2,3,4))
);