[转帖]PL/SQL基础:阶层查询_MySQL, Oracle及数据库讨论区_Weblogic技术|Tuxedo技术|中间件技术|Oracle论坛|JAVA论坛|Linux/Unix技术|hadoop论坛_联动北方技术论坛  
网站首页 | 关于我们 | 服务中心 | 经验交流 | 公司荣誉 | 成功案例 | 合作伙伴 | 联系我们 |
联动北方-国内领先的云技术服务提供商
»  游客             当前位置:  论坛首页 »  自由讨论区 »  MySQL, Oracle及数据库讨论区 »
总帖数
1
每页帖数
101/1页1
返回列表
0
发起投票  发起投票 发新帖子
查看: 3592 | 回复: 0   主题: [转帖]PL/SQL基础:阶层查询        下一篇 
derek
注册用户
等级:中校
经验:1550
发帖:209
精华:0
注册:2011-7-21
状态:离线
发送短消息息给derek 加好友    发送短消息息给derek 发消息
发表于: IP:您无权察看 2011-9-14 14:27:12 | [全部帖] [楼主帖] 楼主

ORACLE 10g新增了阶层查询操作符PRIOR,CONNECT_BY_ROOT 

■PRIOR


阶层查询的CONNECY BY condition的条件式需要用到PRIOR来指定父节点, 
作为运算符,PRIOR和加(+)减(-)运算的优先级相同。 

■阶层查询 
语法:START WITH condition CONNECT BY NOCYCLE condition 

START WITH 指定阶层的根 
CONNECT BY 指定阶层的父/子关系 
NOCYCLE 存在CONNECT BY LOOP的纪录时,也返回查询结果。 
condition ... PRIOR expr = expr 或者 ... expr = PRIOR expr 
例: 

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 ...
■CONNECT_BY_ROOT


查询指定根的阶层数据。 

■CONNECT BY子句的例子 
通过CONNECT BY子句定义职员和上司的关系。 

SQL>SELECT employee_id, last_name, manager_id
FROM employees
CONNECT BY PRIOR employee_id = manager_id;
EMPLOYEE_ID LAST_NAME MANAGER_ID
----------- ------------------------- ----------
101 Kochhar 100
108 Greenberg 101
109 Faviet 108
110 Chen 108
111 Sciarra 108
112 Urman 108
113 Popp 108
200 Whalen 101


■LEVEL的例子 
通过LEVEL虚拟列表示节点的关系。 

SQL>SELECT employee_id, last_name, manager_id, LEVEL
FROM employees
CONNECT BY PRIOR employee_id = manager_id;
EMPLOYEE_ID LAST_NAME MANAGER_ID LEVEL
----------- ------------------------- ---------- ----------
101 Kochhar 100 1
108 Greenberg 101 2
109 Faviet 108 3
110 Chen 108 3
111 Sciarra 108 3
112 Urman 108 3
113 Popp 108 3


■START WITH子句的例子 
通过START WITH指定根节点,ORDER SIBLINGS BY保持阶层的顺序。 

SQL>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; LAST_NAME EMPLOYEE_ID MANAGER_ID LEVEL ------------------------- ----------- ---------- ---------- King 100 1 Cambrault 148 100 2 Bates 172 148 3 Bloom 169 148 3 Fox 170 148 3 Kumar 173 148 3 Ozer 168 148 3 Smith 171 148 3 De Haan 102 100 2 Hunold 103 102 3 Austin 105 103 4 Ernst 104 103 4 Lorentz 107 103 4 Pataballa 106 103 4 Errazuriz 147 100 2 Ande 166 147 3 Banda 167 147 3


hr.employees里,Steven King是公司的最高责任者,没有上司,他有一个叫John Russell的下属是部门80的管理者。 
更新employees表,把Russell设置成King的上司,这样就产生了CONNECT BY LOOP。 

SQL>UPDATE employees SET manager_id = 145 WHERE employee_id = 100; SQL>SELECT last_name "Employee", 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; 2 3 4 5 6 7 ERROR: ORA-01436: CONNECT BY loop in user data CONNECT BY NOCYCLE强制返回查询结果。CONNECT_BY_ISCYCLE显示是否存在LOOP。 SQL>SELECT last_name "Employee", 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; Employee Cycle LEVEL Path ------------------------- ------ ------ ------------------------- Russell 1 2 /King/Russell Tucker 0 3 /King/Russell/Tucker Bernstein 0 3 /King/Russell/Bernstein Hall 0 3 /King/Russell/Hall Olsen 0 3 /King/Russell/Olsen Cambrault 0 3 /King/Russell/Cambrault Tuvault 0 3 /King/Russell/Tuvault Partners 0 2 /King/Partners King 0 3 /King/Partners/King Sully 0 3 /King/Partners/Sully McEwen 0 3 /King/Partners/McEwen


■CONNECT_BY_ROOT的例子 
1,查询110部门的职员,上司,职员和上司之间级别差及路径。 

SELECT last_name "Employee", CONNECT_BY_ROOT last_name "Manager", 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; Employee Manager Pathlen Path --------------- ------------ ---------- ----------------------------------- Higgins Kochhar 1 /Kochhar/Higgins Gietz Kochhar 2 /Kochhar/Higgins/Gietz Gietz Higgins 1 /Higgins/Gietz Higgins King 2 /King/Kochhar/Higgins Gietz King 3 /King/Kochhar/Higgins/Gietz



2,使用GROUP BY语句,查询110部门的职员以及该职员下属职员的工资和。 

SELECT name, SUM(salary) "Total_Salary" FROM ( SELECT CONNECT_BY_ROOT last_name as name, Salary FROM employees WHERE department_id = 110 CONNECT BY PRIOR employee_id = manager_id) GROUP BY name; NAME Total_Salary ------------------------- ------------ Gietz 8300 Higgins 20300 King 20300 Kochhar 20300






赞(0)    操作        顶端 
总帖数
1
每页帖数
101/1页1
返回列表
发新帖子
请输入验证码: 点击刷新验证码
您需要登录后才可以回帖 登录 | 注册
技术讨论