[层次结构|树状结构] oracle中用start with...connect by prior子句实现递归查询

oracle 家族树语法:select column from table_name start with column=valueconnect by prior 父主键=子主键 

--自顶向下查询，prior表示上一条记录，比如 connect by prior id=parentid就是说上一条记录的ID是本条记录的parentid，即本记录的父亲是上一条记录。 

以ORACLE中的EMP表为例[例]从顶到底列出各雇员的信息SQL> select lpad(' ',4*(level-1))||ename name,empno,mgr from emp start with mgr is null connect by prior empno=mgr; NAME                                EMPNO       MGR---------                                  ---------           ---------KING                                7839                JONES                           7566            7839          SCOTT                     7788            7566                ADAMS               7876            7788

 ■ In a hierarchical query, one expression in condition must be qualified with thePRIOR operator to refer to the parent row. For example,... PRIOR expr = expror... expr = PRIOR expr

LEVEL Pseudocolumn
For each row returned by a hierarchical query, the LEVEL pseudocolumn returns 1 for
a root row, 2 for a child of a root, and so on. A root row is the highest row within an
inverted tree. A child row is any nonroot row. A parent row is any row that has
children. A leaf row is any row without children.

See Also: "Hierarchical Queries" on page 9-3 for information on
hierarchical queries in general and "IN Condition" on page 7-23 for
restrictions on using the LEVEL pseudocolumn

If the CONNECT BY condition is compound, then only one condition requires thePRIOR operator, although you can have multiple PRIOR conditions. For example:CONNECT BY last_name != 'King' AND PRIOR employee_id = manager_id ...CONNECT BY PRIOR employee_id = manager_id andPRIOR account_mgr_id = customer_id ...

START WITH CONNECT BY PRIOR这个语法主要用于查询数据包中的树型结构关系。先看下原始数据时怎么样的吧！
表中第一行1001是1002的父节点，而第二行1002又是1003的父节点，如此循环。如题：要求给出其中一个数字能找出其最终的根节点！应该要怎么样实现呢？请看如下sql语句，这里我就拿1008为例。
图中第一行第一个num1就是根节点了。再来分析下sql语句：
select num1,num2,level 
from carol_tmp 
start with num2=1008 
connect by  num2=prior num1 order by level desc;
prior放的左右位置决定了检索是自底向上还是自顶向下.很明显以上的sql选择了自底向上，所以最终得到了根节点。
这次要求获得最小的叶节点，看下图：
第一行的num1就是最小的叶节点了，在注意看下sql代码和上面的代码有什么不同之处：
select num1,num2,level 
from carol_tmp 
start with num2=1008 
connect by prior  num2= num1 order by level desc;
这次prior和num2放在了一起，他意思就是从num2开始寻找其下面最小的叶节点。
为什么prior和谁在一起比较重要呢？比如prior和 num2同在等号的一边，这个时候要看num2和num1的关系。若num2一直是parent，那么这个sql找的就是根节点。

 

 

  在oracle中的select语句可以用start with...connect by prior子句实现递归查询，connect by 是结构化查询中用到的，其基本语法是：

//如果只用connect by 而不加 prior 查找的将是 level 为1的一级。
select ... from tablename start with cond1 
connect by prior cond2 
where cond3; 

这里的where是不能加的，我一加一个where就报 sql命令未正确结束的错误。如果有人可以加上去使用，那也请高手我下吧。本人先谢谢了。

不过可以先写 where 然后使用 connect by prior  例如：select * from Sysfunction where nodetype='4' start with parentid ='123'(值) connect by prior functionid=parentid order by level(关键字：级别),parentid, funorder;

简单说来是将一个树状结构存储在一张表里，比如一个表中存在两个字段: 
id,parentid那么通过表示每一条记录的parent是谁，就可以形成一个树状结构。 

用上述语法的查询可以取得这棵树的所有记录。 

其中cond1 是根结点的限定语句，当然可以放宽限定条件，以取得多个根结点，实际就是多棵树。 

cond2是连接条件，其中用prior表示上一条记录，比如 connect by prior id=parentid就是说上一条记录的ID是本条记录的parentid，即本记录的父亲是上一条记录。 
cond3是过滤条件，用于对返回的所有记录进行过滤。 

对于oracle进行简单树查询(递归查询) 
deptid      paredeptid   name 
number    number        char (40 Byte) 
部门id 父部门id(所属部门id) 部门名称 
通过子节点向根节点追朔.

 

复习一下:start with ...connect by 的用法， start with 后面所跟的就是就是递归的种子。 

递归的种子也就是递归开始的地方 connect by 后面的"prior" 如果缺省：则只能查询到符合条件的起始行，并不进行递归查询； 

connect by prior 后面所放的字段是有关系的，它指明了查询的方向。

 

下面看下几个例子：

create table automobiles(
       part_id number(5)
         constraint pk_auto_part_id primary key,
       parent_id number(5)
         constraint fk_auto_ppid_ references  automobiles(part_id),
       part_cname varchar2(30) not null,
       part_ename varchar2(30) not null,
       mp_cost number(9,2),
       desribe varchar2(20)    
);

--插入数据
insert into automobiles values( 1,null,'汽车','mobile',84321.99,'Assembly');
insert into automobiles values( 2,1,'车身','bodywork',19892.99,'Manufacture');
insert into automobiles values( 3,1,'发送机','engine',42128,'Purchase');
insert into automobiles values( 4,1,'附件','attached',15212,'Assembly');
insert into automobiles values( 5,2,'保险杠','bumper',4812.95,'Purchase');
insert into automobiles values( 6,2,'底盘','chassis',12795.11,'Manufacture');
insert into automobiles values( 7,2,'行李箱','Boot',812.11,'Manufacture');
--分层sql脚本语句练习
select level,part_id,parent_id,part_cname,part_ename,mp_cost,desribe
from automobiles
start with part_id=1
connect by prior part_id=parent_id 
order by level;
--缩进显示
select level,
    lpad(' ',2*level-1)||part_cname||' '||part_ename as partName
from automobiles
start with part_id=1
connect by prior part_id=parent_id
order by level;
--使用子查询  当然也可以在from中加入条件来达到子查询的效果
select level,lpad(' ',2*level-1)||part_cname||' '||part_ename as partName from automobiles
start with part_id=(select part_id from automobiles where part_cname like '%轴%')
connect by prior  part_id=parent_id
order by level;
--自底向上的遍历
select level,lpad(' ',2*level-1)||part_cname||' '||part_ename as partName
from automobiles
start with part_id=(select part_id from automobiles where part_cname like '%轴%')
connect by prior parent_id=part_id
order by level;
--删除指定的节点
select level,lpad(' ',2*level-1)||part_cname||' '||part_ename as partName from automobiles
where part_cname <> '底盘'
start with part_id=1
connect by prior parent_id=part_id
order by level;
--删除分支
select level,lpad(' ',2*level-1)||part_cname||' '||part_ename as partName from automobiles
start with part_id=1
connect by prior parent_id=part_id and part_cname <> '底盘'
order by level;

*************************************** TRY ******************************************************************************

select rownum from dual  connect by rownum <= 3; 1=1


SELECT LEVEL,(TO_DATE('20150101', 'yyyy-mm-dd') + LEVEL - 1) CUR_DATE FROM DUAL             
  CONNECT BY LEVEL <= SYSDATE - TO_DATE('20150101', 'yyyy-mm-dd') + 1
  
  select  SYSDATE - TO_DATE('20150101', 'yyyy-mm-dd')+1 from dual;
  
  select * from dual
  select TO_DATE('20150101', 'yyyy-mm-dd')+32.9 from dual
  
   with  x  as   


    ( select  'aa'  chr  from  dual 


 union  all   


    select  'bb'  chr  from  dual) 


    select  level ,chr,lpad( ' ' ,( level - 1 )* 5 , '-' )||chr other  from  x  connect  by  level <= 3
    
    with t as (select level from dual connect by level<=3)
select rownum from t connect by rownum<=3


with t as (select level l from dual connect by level<=3)
select rownum,l,LEVEL from t connect by rownum<=1


with t as (select level l from dual connect by level<=3)
select rownum,l,LEVEL from t connect by rownum=2 


select distinct ct from (
select d.cal_year,count(*) as ct from td_day d /*where d.calendar_date is null*/ group by d.cal_year ) --365,366


select distinct ct from (
select w.cal_year,count(*) as ct from w_day_d w group by w.cal_year )--372