oracle_sql递归查询积累

一、start with.....connect by递归查询

建表语句：

CREATE TABLE D_ZONECODE
(
	ID VARCHAR2(36) NOT NULL UNIQUE,
	ZONECODE VARCHAR2(6) NOT NULL,
	SUPERCODE VARCHAR2(6) NOT NULL,
	ZONELLEVEL VARCHAR2(2) NOT NULL,
	ZONENAME VARCHAR2(60) NOT NULL
);

 插入数据语句：

INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('1', '370000', '000000', '01', '山东省');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('2', '370100', '370000', '02', '济南市');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('3', '370102', '370100', '03', '历下区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('4', '370103', '370100', '03', '市中区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('5', '370104', '370100', '03', '槐荫区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('6', '370105', '370100', '03', '天桥区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('7', '370112', '370100', '03', '历城区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('8', '370113', '370100', '03', '长清区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('9', '370114', '370100', '03', '高新区');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('10', '370124', '370100', '03', '平阴县');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('11', '370125', '370100', '03', '济阳县');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('12', '370126', '370100', '03', '商河县');
INSERT INTO D_ZONECODE (ID, ZONECODE, SUPERCODE, ZONELLEVEL, ZONENAME) VALUES ('13', '370181', '370100', '03', '章丘市');

 

root向树末梢查询：

 

select * from d_zonecode
 start with id=’370000’
 connect by prior zonecode = supercode;

  

 树末梢向ROOT查询：

 

select * from d_zonecode
 start with zonecode = '370100'
 connect by prior supercode = zonecode;

 

 

附：

start with指明从哪里开始遍历树

connect by 就是指明父子关系,注重PRIOR位置

CONNECT_BY_ROOT: 提供获取根节点记录的字段信息。

 

二、 今天客户突然要一个表格，这里面包括两个表的总分关系，比如我要显示部门同时和该部门下的所有人员用一条记录显示，在网上搜到一个例子，记录下方便使用。

 

 

 

建表语句：

 

/* Create Tables */
CREATE TABLE DEPT
(
	ID VARCHAR2(36) NOT NULL UNIQUE,
	DEPTID VARCHAR2(4) NOT NULL UNIQUE,
	DEPTNAME VARCHAR2(60) NOT NULL
);
CREATE TABLE D_USER
(
	ID VARCHAR2(36) NOT NULL UNIQUE,
	USERID VARCHAR2(6) NOT NULL UNIQUE,
	USERNAME VARCHAR2(60) NOT NULL,
	DEPTID VARCHAR2(4) NOT NULL
);

 

插入数据语句：

 

-- 部门表
INSERT INTO dept (id, deptid, deptname) VALUES ('1', '0001', '市场部');
INSERT INTO dept (id, deptid, deptname) VALUES ('2', '0002', '开发部');
INSERT INTO dept (id, deptid, deptname) VALUES ('3', '0003', '项目部');
-- 用户表
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('1', '100001', '张肃宁', '0001');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('2', '100002', '王济南', '0002');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('3', '100003', '赵临沂', '0001');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('4', '100004', '金淄博', '0003');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('5', '100005', '李德州', '0002');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('6', '100006', '周济宁', '0001');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('7', '100007', '姜潍坊', '0003');
INSERT INTO D_USER (id, userid, username, deptid) VALUES ('8', '100008', '万青岛', '0001');

 

查询：

 

select * from dept t;
select * from d_user t;

 

 

 
 测试sql如下：

 

select username from (
select row_number() over(order by lv desc) id,username,deptid from (
select level lv, replace(sys_connect_by_path(username,','),',',',') username,deptid from(
select deptid,username,row_number() over(order by username) id from
(
select a.deptid,a.deptname,b.username
from dept a,d_user b
where a.deptid = b.deptid
)
) connect by prior id = id-1 )) where id = 1;

 

 

要是需要和部门连接查询，我采用建立一个方法，然后查询部门记录时调用该方法，传部门id这个参数进去。

 

--方法建立：
create or replace function getUsername(oc_deptid in varchar2
                                  ) return varchar2 is
  oc_result varchar(300);
  oc_username varchar2(300); -- 取值
begin
--查询用户名称
  select username into oc_result from (
select row_number() over(order by lv desc) id,username,deptid from (
select level lv, replace(sys_connect_by_path(username,','),',',',') username,deptid from(
select deptid,username,row_number() over(order by username) id from
(
select a.deptid,a.deptname,b.username
from dept a,d_user b
where a.deptid = b.deptid and a.deptid = oc_deptid
)
) connect by prior id = id-1 )) where id = 1;
  oc_username:=oc_result;
  if oc_result is NULL then
     oc_username := '';
  else
     oc_username := SUBSTR(oc_result,INSTR(oc_result,',')+1,LENGTH(oc_result)-1);
  end if;
  return oc_username;
end;

 

查询sql:

 

 

select t.deptid,t.deptname,getUsername(t.deptid) as username from dept t;

 

 

附（CONNECT_BY_ROOT使用）：

 

select zonecode,zonename,CONNECT_BY_ROOT(zonecode) as root_code  from d_zonecode
start with zonecode = '370100'
connect by prior zonecode = supercode and zonecode != supercode;
select zonecode,CONNECT_BY_ROOT(zonecode) as root_code from d_zonecode
start with zonecode = '370114'
connect by prior supercode = zonecode and zonecode != supercode;
select CONNECT_BY_ROOT(t.zonecode) as ROOT from d_zonecode t
where t.zonecode = '370100'
start with t.zonecode = t.supercode
connect by prior t.zonecode = t.supercode and t.zonecode != t.supercode;

 

 

 

 

评论

2 楼 wang8662350 2014-07-28  

很不错啊，学习了

1 楼 datawarehouse 2012-07-12  

收藏啊