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如何用SQL进行集合运算

发表于:2024-10-04 作者:千家信息网编辑
千家信息网最后更新 2024年10月04日,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 exis
千家信息网最后更新 2024年10月04日如何用SQL进行集合运算

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;

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