Cet article présente principalement l'utilisation de SQL pour les opérations d'ensemble, qui a une certaine valeur de référence. Maintenant, je le partage avec tout le monde. Les amis dans le besoin peuvent s'y référer
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 ;
Trouver directement la différence entre les deux tables
(select * from tbl_a except select * from tbl_b) union all (select * from tbl_b except select * from tbl_a);
Créer un tableau
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');
Idées : Les deux fournisseurs proposent le même type de pièces (il suffit de se connecter en fonction de la colonne des pièces) Le nombre de types de pièces des deux fournisseurs est le même (c'est-à-dire qu'il existe un mappage un à un) (limite de nombre)
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 ;
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