SQL进阶--用SQL进行集合运算

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.sup
having 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 products
where 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 in
delete 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
;

内容多来自 《SQL进阶教材》，仅做笔记。练习部分代码均为原创。