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_cntfrom ( 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_bexceptselect * from tbl_a);

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 empfrom empskills es1 where not exists(select skill from skills expect select skill from empskills es2where 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');

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 s2from 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 cwhere 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 dwhere 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 rowidin (select rowid -- 全部rowid from productsexcept -- 减去 select max(rowid)-- 要留下的rowid from productsgroup 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_cntfrom ( select * from tbl_Aunionselect * from tbl_b) tmp;