oracle 11g新特性---WITH语句增强
--with语句增强，11gR2增强
SQL> select * from test;
ID NAME ADDR
---------- -------------------- --------------------
1 wzk chengdu
2 cherry chengdu
3 wzk heilongjiang
4 cherry beijing
--with语句增加了别名，但是个数和顺序与as后面的子查询一一对应。
SQL> with c(id,name) as
2 (select id,name,addr from test where id=1 or id=2)
3 select * from c;
(select id,name,addr from test where id=1 or id=2)
*
ERROR at line 2:
ORA-32038: number of WITH clause column names does not match number of elements
in select list
--也可以看出select查询之后，返回的列名就是我们在with子句中定义的别名。
SQL> with c(no,name,address) as
2 (select id,name,addr from test where id=1 or id=2)
3 select * from c;
NO NAME ADDRESS
---------- -------------------- --------------------
1 wzk chengdu
2 cherry chengdu
--10g允许吗？可以看见10g是不允许的。
ODS>select * from v$version;
BANNER
----------------------------------------------------------------
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
PL/SQL Release 10.2.0.4.0 - Production
CORE 10.2.0.4.0 Production
TNS for IBM/AIX RISC System/6000: Version 10.2.0.4.0 - Productio
NLSRTL Version 10.2.0.4.0 - Production
ODS>with c(no,name,address) as
2 (select id,name,addr from test where id=1 or id=2)
3 select * from c;
with c(no,name,address) as
*
第 1 行出现错误:
ORA-32033: unsupported column aliasing
ODS>with c(id,name,addr) as
2 (select id,name,addr from test where id=1 or id=2)
3 select * from c;
with c(id,name,addr) as
*
第 1 行出现错误:
ORA-32033: unsupported column aliasing
--测试with的递归查询
create table employees
(employee_id number, --雇员编号
employee_name varchar2(10), --雇员名称
manger_id number, --经理编号
manger_name varchar2(10) --经理名称
);
insert into employees values(101,'wzk',100,'www' );
insert into employees values(102,'cherry',100,'www');
insert into employees values(103,'hl',101,'wzk' );
insert into employees values(104,'hw',101,'wzk' );
insert into employees values(105,'dw',103,'hl' );
insert into employees values(106,'wy',102,'cherry' );
insert into employees values(107,'ll',105,'dw' );
commit;
--统计雇员101下的所有雇员信息。
SQL> with c(eid,ename,mid,mname,lv)
as
(select employee_id,employee_name,manger_id,manger_name,0 lv
from employees
where employee_id=101
union all
select e.employee_id,e.employee_name,e.manger_id,e.manger_name,c.lv+1
from c,employees e
where c.eid=e.manger_id
)
select * from c;
EID ENAME MID MNAME LV
---------- ---------- ---------- ---------- ----------
101 wzk 100 www 0
103 hl 101 wzk 1
104 hw 101 wzk 1
105 dw 103 hl 2
107 ll 105 dw 3
--看来with可以用来取代connect by子句了啊。呵呵。
--depth是先返回儿子信息，在返回兄弟信息。（先儿子，后兄弟）
--可以看lv列，看lv为1的列。当查到employee_id为103的时候，他下面既有子女，身边又有兄弟，那么结果集是先返回兄弟数据哪？还是子女数据拿？
--searth自动给with语句添加了一列 order1
SQL> with c(eid,ename,mid,mname,lv)
as
(select employee_id,employee_name,manger_id,manger_name,0 lv
from employees
where employee_id=101
union all
select e.employee_id,e.employee_name,e.manger_id,e.manger_name,c.lv+1
from c,employees e
where c.eid=e.manger_id
)
search depth first by ename set order1
select * from c
order by order1;
EID ENAME MID MNAME LV ORDER1
---------- ---------- ---------- ---------- ---------- ----------
101 wzk 100 www 0 1
103 hl 101 wzk 1 2
105 dw 103 hl 2 3
107 ll 105 dw 3 4
104 hw 101 wzk 1 5
--breadth是先返回兄弟信息，后儿子信息（先兄弟，后儿子）
SQL> with c(eid,ename,mid,mname,lv)
as
(select employee_id,employee_name,manger_id,manger_name,0 lv
from employees
where employee_id=101
union all
select e.employee_id,e.employee_name,e.manger_id,e.manger_name,c.lv+1
from c,employees e
where c.eid=e.manger_id
)
search breadth first by ename set order1
select * from c
order by order1;
EID ENAME MID MNAME LV ORDER1
---------- ---------- ---------- ---------- ---------- ----------
101 wzk 100 www 0 1
103 hl 101 wzk 1 2
104 hw 101 wzk 1 3
105 dw 103 hl 2 4
107 ll 105 dw 3 5
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