今天就跟大家聊聊有关如何用SQL进行集合运算 ，可能很多人都不太了解，为了让大家更加了解，小编给大家总结了以下内容，希望大家根据这篇文章可以有所收获。
1、比较表和表
drop table if exists tbl_a;create table tbl_a(
key1 varchar(10),
col_1 int4,
col_2 int4,
col_3 int4
);insert into tbl_a values('A', 2, 3, 4);
insert into tbl_a values('B', 0, 7, 9);
insert into tbl_a values('c', 5, 1, 6);
drop table if exists tbl_b;create table tbl_b(
key1 varchar(10),
col_1 int4,
col_2 int4,
col_3 int4
);
insert into tbl_b values('A', 2, 3, 4);
insert into tbl_b values('B', 0, 7, 9);
insert into tbl_b values('c', 5, 1, 6);-- ## 如果union a b 行数一致则两张表相等 select count(1) row_cnt from ( select *
from tbl_A union
select * from tbl_b
) tmp
;直接求两表的不同之处
(select * from tbl_a except
select * from tbl_b) union all
(select * from tbl_b except
select * from tbl_a);2、用差集实现关系除法运算
建表
drop table if exists skills;create table skills(
skill varchar(10)
);insert into skills values('oracle');
insert into skills values('unix');insert into skills values('java');drop table if exists empskills;create table empskills(
emp varchar(10),
skill varchar(10)
);insert into empskills values('相田','oracle');
insert into empskills values('相田','unix');
insert into empskills values('相田','java');
insert into empskills values('相田','c#');
insert into empskills values('神奇','oracle');
insert into empskills values('神奇','unix');
insert into empskills values('神奇','java');
insert into empskills values('平井','oracle');
insert into empskills values('平井','unix');
insert into empskills values('平井','PHP');
insert into empskills values('平井','Perl');
insert into empskills values('平井','C++');
insert into empskills values('若田部','Perl');
insert into empskills values('度来','oracle');--把除法变成减法select distinct emp from empskills es1 where not exists
(select skill from skills
expect select skill from empskills es2 where es1.emp = es2.emp);3、寻求相等的子集
drop table if exists supparts;create table supparts(
sup varchar(10),
part varchar(10)
);insert into supparts values('A', '螺丝');
insert into supparts values('A', '螺母');
insert into supparts values('A', '管子');
insert into supparts values('B', '螺丝');
insert into supparts values('B', '管子');
insert into supparts values('C', '螺丝');
insert into supparts values('C', '螺母');
insert into supparts values('C', '管子');
insert into supparts values('D', '螺丝');
insert into supparts values('D', '管子');
insert into supparts values('E','保险丝');
insert into supparts values('E', '螺母');
insert into supparts values('E', '管子');
insert into supparts values('F','保险丝');思路： 两个供应商都经营同种类型的零件 （简单的按照零件列进行连接） 两个供应商的零件类型数相同（即存在一一映射）（count限定）
select a.sup s1, b.sup s2 from supparts a, supparts b where a.sup < b.sup -- 生成供应商的全部组合
and a.part = b.part -- 条件1：经营同种类型的零件
group by a.sup, b.suphaving count(*) = (select count(1) -- 条件2：经营的零件的数量种类相同 a = 中间数 from supparts c where c.sup = a.sup) and count(*) = (select count(1) -- 条件2：经营的零件的数量种类相同 b = 中间数 from supparts d where d.sup = b.sup)
;4、删除重行
drop table if exists products;create table products(
rowid int4,
name1 varchar(10),
price int4
);insert into products values(1,'苹果',50);insert into products values(2,'橘子',100);
insert into products values(3,'橘子',100);insert into products values(4,'橘子',100);
insert into products values(5,'香蕉',80);-- 删除重行高效SQL语句（1）：通过EXCEPT求补集delete from productswhere rowid in (select rowid -- 全部rowid from products
except -- 减去 select max(rowid) -- 要留下的rowid from products group by name1, price
);-- 删除重行高效SQL语句（2）：通过not indelete from products where rowid not in (select max(rowid) from products group by name1, price
);练习
-- 改进中用union的比较select
case when count(1) = (select count(1) from tbl_A)
and count(1) = (select count(1)+1 from tbl_b)
then count(1) else '不相等' end row_cnt from ( select * from tbl_A union
select * from tbl_b
) tmp
;看完上述内容，你们对如何用SQL进行集合运算 有进一步的了解吗？如果还想了解更多知识或者相关内容，请关注亿速云行业资讯频道，感谢大家的支持。
