oracle取差值集合

Oracle Minus关键字
SQL中的MINUS关键字
SQL中有一个MINUS关键字，它运用在两个SQL语句上，它先找出第一条SQL语句所产生的结果，然后看这些结果有没有在第二个SQL语句的结果中。如果有的话，那这一笔记录就被去除，而不会在最后的结果中出现。如果第二个SQL语句所产生的结果并没有存在于第一个SQL语句所产生的结果内，那这笔资料就被抛弃，其语法如下：
[SQL Segment 1]
MINUS
[SQL Segment 2]
--------------------------------------------
//创建表1

create table test1
(
name varchar(10),
sex varchar(10),
);insert into test1 values('test','female');
insert into test1 values('test1','female');
insert into test1 values('test1','female');
insert into test1 values('test11','female');
insert into test1 values('test111','female');

//创建表2

create table test2
(
name varchar(10),
sex varchar(10),
);
insert into test1 values('test','female');
insert into test1 values('test2','female');
insert into test1 values('test2','female');
insert into test1 values('test22','female');
insert into test1 values('test222','female');

-------------------------------------------

select * from test1 minus select * from test2;结果：NAME SEX 
---------- ---------- 
test1 female 
test11 female 
test111 female 
-----------------------------------------------------------select * from test2 minus select * from test1;结果:NAME SEX 
---------- ---------- 
test2 female 
test22 female 
test222 female

结论：Minus返回的总是左边表中的数据，它返回的是差集。注意：minus有剃重作用

==========================================================
下面是我做的实验，很明显能够看出MINUS的效率，made_order共23万笔记录，charge_detail共17万笔记录

性能比较:

SELECT order_id FROM made_orderMINUSSELECT order_id FROM charge_detail1.14 secSELECT a.order_id FROM made_order aWHERE NOT exists (SELECT order_idFROM charge_detailWHERE order_id = a.order_id)18.19 secSELECT order_id FROM made_orderWHERE order_id NOT in (SELECT order_idFROM charge_detail)20.05 sec