create table test1 ( name varchar(10), sex varchar(10), age int );
insert into test1 values('luxin','female',25); insert into test1 values('tom','female',26); insert into test1 values('mary1','male',27); insert into test1 values('money','male',27); insert into test1 values('tony','male',28); insert into test1 values('tony1','male',19);
create table test2 ( name varchar(10), sex varchar(10), age int );
insert into test2 values('luxin','female',25); insert into test2 values('tom','female',26); insert into test2 values('mary2','male',27); insert into test2 values('money','male',27); insert into test2 values('tony','male',28); insert into test2 values('tony2','male',19);
select * from test1 minus select * from test2;
结果:
NAME SEX AGE
mary1 male 27
tony1 male 19
select * from test2 minus select * from test1;
结果:
NAME SEX AGE
mary2 male 27
tony2 male 19
结论:Minus返回的总是左边表中的数据,它返回的是差集。
用表1-表2中的数据,如果相同,则去掉,否则返回表1中的数据。
insert into test1 values('luxin','female',25); insert into test1 values('tom','female',26); insert into test1 values('mary1','male',27); insert into test1 values('money','male',27); insert into test1 values('tony','male',28); insert into test1 values('tony1','male',19);
create table test2 ( name varchar(10), sex varchar(10), age int );
insert into test2 values('luxin','female',25); insert into test2 values('tom','female',26); insert into test2 values('mary2','male',27); insert into test2 values('money','male',27); insert into test2 values('tony','male',28); insert into test2 values('tony2','male',19);
select * from test1 minus select * from test2;
结果:
NAME SEX AGE
mary1 male 27
tony1 male 19
select * from test2 minus select * from test1;
结果:
NAME SEX AGE
mary2 male 27
tony2 male 19
结论:Minus返回的总是左边表中的数据,它返回的是差集。
用表1-表2中的数据,如果相同,则去掉,否则返回表1中的数据。