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oracle 递归查询-个人总结

Jun 07, 2016 pm 03:18 PM
c leo oracle personal Summarize Inquire recursion first

首先,建表: create table T_TEST_WORD( id NUMBER, pid VARCHAR2(20), name VARCHAR2(20)) 插入数据: insert into T_TEST_WORD (id, pid, name)values (1, '-1', '中国');insert into T_TEST_WORD (id, pid, name)values (2, '1', '江苏');insert into T_

首先,建表:

create table T_TEST_WORD
(
  id   NUMBER,
  pid  VARCHAR2(20),
  name VARCHAR2(20)
)
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插入数据:


insert into T_TEST_WORD (id, pid, name)
values (1, '-1', '中国');
insert into T_TEST_WORD (id, pid, name)
values (2, '1', '江苏');
insert into T_TEST_WORD (id, pid, name)
values (3, '1', '浙江');
insert into T_TEST_WORD (id, pid, name)
values (4, '2', '南京');
insert into T_TEST_WORD (id, pid, name)
values (5, '2', '无锡');
insert into T_TEST_WORD (id, pid, name)
values (6, '1', '安徽');
insert into T_TEST_WORD (id, pid, name)
values (7, '4', '雨花台区');
insert into T_TEST_WORD (id, pid, name)
values (8, '-1', '美国');
insert into T_TEST_WORD (id, pid, name)
values (9, '-1', '俄罗斯');
commit;
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查询下表中的数据:

表1

oracle 递归查询-个人总结


其中PID为-1的代表根节点。


开始进行递归查询:

查询结果如下:

表2

oracle 递归查询-个人总结


说明:

select ... from tablename(表名)   where  条件1  start  with  条件2   connect  by  条件3  and  条件4 ...


看下这个语句(由根节点向子节点查询)


select  t.* , level  from T_TEST_WORD  t  start  with pid='-1'  connect  by  prior  id = pid(由根节点向子节点查询)


从T_TEST_WORD表中查询所有记录,从pid=-1的开始查询,且上一次查询记录的id作为本次查询的pid,以上表为例,举例如下:

第一次查询出pid=-1的记录是id为1的那条。

第二次查询时,查询pid=1的(上次查询记录的id是1)记录,可以查询到id为2的那条记录

第三次查询时,查询pid=2的(上次查询记录的id是2)记录,可以查询到id为4的那条记录,以此类推......

由上,查询到了表2的记录。


由树的根节点向子节点查询时,查询节点的顺序是按照树的前序遍历(DLR)进行的(1,2,4,7,5,3,6,8,9),如下图1

图1:

oracle 递归查询-个人总结

(比较丑的图,凑合看吧,嘻嘻)

注意:那个level要有start  with  ......  connect  by ......  才有效哦,不然会报错的。


再看这句(由子节点向根节点查询)

select t.* ,level from T_TEST_WORD t start with pid='2' connect by  id = prior pid
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从pid=2的开始(向上)查询,且上一次查询记录的pid作为本次查询的id


oracle 递归查询-个人总结


prior挨着谁,就将谁作为本次查询的某某关键字,比如connect by  id = prior pid,就是将上次的pid作为本次查询的id。

注意,level是伪列,查询数据所对应的级,或者说深度吧。


小菜的总结,有问题请指出,谢谢咯!


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