用SQL进行集合运算

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 ;

(select * from tbl_a except select * from tbl_b) union all (select * from tbl_b except select * from tbl_a);

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);

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');

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) ;

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 );

-- 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 ;