Oracle(递归查询) Connect By简单用法

Oracle Connect By用法

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：

select ... from <TableName>
where <Conditional-1>
start with <Conditional-2>
connect by <Conditional-3>
;

<Conditional-1>：过滤条件，用于对返回的所有记录进行过滤。
<Conditional-2>：查询结果重起始根结点的限定条件。
<Conditional-3>：连接条件

?

简单用法：

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.

Sql代码
select * from persons.dept start with deptid=76 connect by prior paredeptid=deptid??

select * from persons.dept start with deptid=76 connect by prior paredeptid=deptid?

通过根节点遍历子节点.

Sql代码
select * from persons.dept start with paredeptid=0 connect by prior deptid=paredeptid??

select * from persons.dept start with paredeptid=0 connect by prior deptid=paredeptid?

可通过level 关键字查询所在层次.

Sql代码
select a.*,level from persons.dept a start with paredeptid=0 connect by prior deptid=paredeptid??

select a.*,level from persons.dept a start with paredeptid=0 connect by prior deptid=paredeptid?

再次复习一下: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。

练习： 通过子节点获得顶节点

Sql代码
select FIRST_VALUE(deptid) OVER (ORDER BY LEVEL DESC ROWS UNBOUNDED PRECEDING) AS firstdeptid from persons.dept start with deptid=76 connect by prior paredeptid=deptid?


数据组织结构如下图：

?

数据库表结构如下：

create table t2(
root_id number,
id number,
name varchar(5),
description varchar(10)
);

insert into t2(root_id,id,name,description) values(0,1,'a','aaa');
insert into t2(root_id,id,name,description) values(1,2,'a1','aaa1');
insert into t2(root_id,id,name,description) values(1,3,'a2','aaa2');
insert into t2(root_id,id,name,description) values(0,4,'b','bbb');
insert into t2(root_id,id,name,description) values(4,5,'b1','bbb1');
insert into t2(root_id,id,name,description) values(4,6,'b2','bbb2');

获取完整树：
select * from t2 start with root_id = 0 connect by prior id = root_id;
?

获取特定子树：
select * from t2 start with id = 1 connect by prior id = root_id;

?
select * from t2 start with id = 4 connect by prior id = root_id;

?

如果connect by prior中的prior被省略，则查询将不进行深层递归。
如：

select * from t2 start with root_id = 0 connect by id = root_id;

?

select * from t2 start with id = 1 connect by id = root_id;
如：

?CREATE TABLE t (
rid NUMBER(12),
col1 VARCHAR2(30),
col2 VARCHAR2(300));

CREATE SEQUENCE seq_t_rid;

INSERT INTO t
SELECT seq_t_rid.NEXTVAL, dbms_crypto.randombytes(15), dbms_crypto.randombytes(150)
FROM DUAL
CONNECT BY LEVEL <= 1000; SELECT last_name, employee_id, manager_id, LEVEL
FROM employees
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id;

SELECT last_name, employee_id, manager_id, LEVEL
FROM employees
START WITH employee_id = 101
CONNECT BY PRIOR employee_id = manager_id;

set pagesize 0
col last_name format a30

SELECT LEVEL, LPAD(' ', LEVEL*3) || LAST_NAME AS LAST_NAME
FROM employees
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id;

set pagesize 20 SELECT last_name, employee_id, manager_id, LEVEL
FROM employees
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id
ORDER SIBLINGS BY last_name; set linesize 121
col emp format a20
col mgr format a20

SELECT last_name
FROM employees
WHERE department_id = 110;

SELECT last_name, CONNECT_BY_ROOT last_name MGR,
LEVEL-1 PATHLEN, SYS_CONNECT_BY_PATH(last_name, '/') PATH
FROM employees
WHERE LEVEL > 1
AND department_id = 110
CONNECT BY PRIOR employee_id = manager_id
ORDER BY last_name, MGR, PATHLEN, PATH; UPDATE employees SET manager_id = 145
WHERE employee_id = 100;

set linesize 121
col path format a50

SELECT last_name, LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE LEVEL <= 3 AND department_id = 80
START WITH last_name = 'King'
CONNECT BY PRIOR employee_id = manager_id AND LEVEL <= 4;
ERROR:
ORA-01436: CONNECT BY loop in user data

SELECT last_name, CONNECT_BY_ISCYCLE "Cycle", LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE LEVEL <= 3 AND department_id = 80
START WITH last_name = 'King'
CONNECT BY NOCYCLE PRIOR employee_id = manager_id
AND LEVEL <= 4;

ROLLBACK; SELECT last_name "Employee", CONNECT_BY_ISLEAF "IsLeaf",
LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE level <= 3
AND department_id = 80
START WITH last_name = 'King'
CONNECT BY PRIOR employee_id = manager_id
AND LEVEL <= 4; col empname format a20
col cbp format a30

SELECT LPAD(' ', 2*LEVEL, ' ' ) || ename empName, dname, job,
sys_connect_by_path( ename, '/' ) cbp
FROM emp e, dept d
WHERE e.deptno = d.deptno
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
ORDER SIBLINGS BY job;

SELECT LPAD(' ', 2*LEVEL, ' ' ) || ename empName, dname, job,
sys_connect_by_path(empno, '.') cbp
FROM scott.emp emp, scott.dept dept
WHERE emp.deptno = dept.deptno
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
ORDER SIBLINGS BY ename; CREATE OR REPLACE FUNCTION permissions_sub_tree_root (
the_id IN NUMBER,
the_level IN NUMBER)
RETURN NUMBER ISsub_tree_root NUMBER(10);BEGIN
? SELECT id
? INTO sub_tree_root
? FROM hierarchy
? WHERE level = the_level
? -- Connect 'upwards', i.e. find the parent
? CONNECT BY PRIOR PARENT = id
? START WITH ID = the_id;? RETURN sub_tree_root;
END permissions_sub_tree_root;
/SELECT id, name, username
FROM (
? SELECT ID, PARENT, NAME,
? permissions_sub_tree_root(id, LEVEL) ROOT
? FROM hierarchy
? CONNECT BY PRIOR id = PARENT) HIERARCHY, permissions
WHERE ROOT = hierarchy_id; ?