[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=value
connect 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 the
PRIOR operator to refer to the parent row. For example,
... PRIOR expr = expr
or
... 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 the
PRIOR 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 and
PRIOR 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