这里用到了Oracle的一个树形结构查询函数select *  from record START WITH A.TREE_NODE IN ('COST_CTR_10053')
CONNECT BY PRIOR A.TREE_NODE_NUM = A.PARENT_NODE_NUM.查出指定节点下的所以子节点然后连接leaf表,选出所有的叶子。

/* Formatted on 11/7/2015 11:05:57 PM (QP5 v5.267.14150.38573) */
SELECT B.RANGE_FROM
FROM PSTREELEAF B,
( SELECT DISTINCT A.SETID,
A.SETCNTRLVALUE,
A.TREE_NAME,
A.EFFDT,
A.TREE_NODE_NUM
FROM PSTREENODE A,
(SELECT A.SETID,
A.SETCNTRLVALUE,
A.TREE_NAME,
A.EFFDT
FROM PSTREEDEFN A
WHERE A.SETID = 'SHARE'
AND A.SETCNTRLVALUE = ' '
AND A.TREE_NAME = 'PLD_LOC_COMBO'
AND A.EFFDT =
(SELECT MAX (B.EFFDT)
FROM PSTREEDEFN B
WHERE A.SETID = B.SETID
AND A.SETCNTRLVALUE = B.SETCNTRLVALUE
AND A.TREE_NAME = B.TREE_NAME
AND B.EFF_STATUS = 'A'
AND B.EFFDT <= SYSDATE)) DEFN
WHERE A.SETID = DEFN.SETID
AND A.SETCNTRLVALUE = DEFN.SETCNTRLVALUE
AND A.TREE_NAME = DEFN.TREE_NAME
AND A.EFFDT = DEFN.EFFDT
START WITH A.TREE_NODE IN ('COST_CTR_10053')
CONNECT BY PRIOR A.TREE_NODE_NUM = A.PARENT_NODE_NUM) SA
WHERE SA.SETID = B.SETID
AND SA.SETCNTRLVALUE = B.SETCNTRLVALUE
AND SA.TREE_NAME = B.TREE_NAME
AND SA.EFFDT = B.EFFDT
AND SA.TREE_NODE_NUM = B.TREE_NODE_NUM

************************************************************************************************************************************************

SELECT P2.*
FROM
(SELECT A.SETID ,
A.SETCNTRLVALUE ,
A.TREE_NAME ,
A.EFFDT
FROM PSTREEDEFN A
WHERE A.SETID = 'SHARE'
AND A.SETCNTRLVALUE = ' '
AND A.TREE_NAME = 'PLD_LOC_COMBO'
AND A.EFFDT =
(SELECT MAX (B.EFFDT)
FROM PSTREEDEFN B
WHERE A.SETID = B.SETID
AND A.SETCNTRLVALUE = B.SETCNTRLVALUE
AND A.TREE_NAME = B.TREE_NAME
AND B.EFF_STATUS = 'A'
AND B.EFFDT <= SYSDATE)) P,
PSTREENODE P1 ,
PSTREELEAF P2
WHERE P1.SETID = P.SETID
AND P1.SETCNTRLVALUE = P.SETCNTRLVALUE
AND P1.TREE_NAME = P.TREE_NAME
AND P1.EFFDT = P.EFFDT
AND P1.TREE_NODE = 'COST_CTR_10053'
AND P2.SETID = P1.SETID
AND P2.SETCNTRLVALUE = P1.SETCNTRLVALUE
AND P2.TREE_NAME = P1.TREE_NAME
AND P2.EFFDT = P1.EFFDT
-- AND P2.TREE_NODE_NUM BETWEEN P1.TREE_NODE_NUM AND P1.TREE_NODE_NUM_END
AND P2.TREE_NODE_NUM = P1.TREE_NODE_NUM

AND (
( NVL(LENGTH(REPLACE(TRANSLATE('143','0123456789.',' '),' ','')), 0) <> 0 AND '143' BETWEEN P2.RANGE_FROM AND P2.RANGE_TO)
OR
( NVL(LENGTH(REPLACE(TRANSLATE('143','0123456789.',' '),' ','')), 0) = 0 AND '143' BETWEEN P2.RANGE_FROM AND P2.RANGE_TO AND LENGTH('143') BETWEEN LENGTH(P2.RANGE_FROM) AND LENGTH(P2.RANGE_TO))
)

Select Tree Node的更多相关文章

  1. [置顶] ※数据结构※→☆非线性结构(tree)☆============树结点 链式存储结构(tree node list)(十四)

    结点: 包括一个数据元素及若干个指向其它子树的分支:例如,A,B,C,D等. 在数据结构的图形表示中,对于数据集合中的每一个数据元素用中间标有元素值的方框表示,一般称之为数据结点,简称结点. 在C语言 ...

  2. Data Structure Binary Tree: Print ancestors of a given binary tree node without recursion

    http://www.geeksforgeeks.org/print-ancestors-of-a-given-binary-tree-node-without-recursion/ #include ...

  3. Python 解LeetCode:671. Second Minimum Node In a Binary Tree

    题目在这里,要求一个二叉树的倒数第二个小的值.二叉树的特点是父节点的值会小于子节点的值,父节点要么没有子节点,要不左右孩子节点都有. 分析一下,根据定义,跟节点的值肯定是二叉树中最小的值,剩下的只需要 ...

  4. 【easy】671. Second Minimum Node In a Binary Tree

    Given a non-empty special binary tree consisting of nodes with the non-negative value, where each no ...

  5. [Swift]LeetCode671. 二叉树中第二小的节点 | Second Minimum Node In a Binary Tree

    Given a non-empty special binary tree consisting of nodes with the non-negative value, where each no ...

  6. LeetCode算法题-Second Minimum Node In a Binary Tree(Java实现)

    这是悦乐书的第285次更新,第302篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第153题(顺位题号是671).给定非空的特殊二叉树,其由具有非负值的节点组成,其中该树 ...

  7. [LeetCode&Python] Problem 671. Second Minimum Node In a Binary Tree

    Given a non-empty special binary tree consisting of nodes with the non-negative value, where each no ...

  8. 671. Second Minimum Node In a Binary Tree

    /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode ...

  9. Lintcode: Insert Node in a Binary Search Tree

    Given a binary search tree and a new tree node, insert the node into the tree. You should keep the t ...

随机推荐

  1. Bungee Jumping[HDU1155]

    Bungee JumpingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...

  2. Android入门(五):程序架构——MVC设计模式在Android中的应用

    刚刚接触编程的的人,可能会这样认为:只要代码写完了能够跑起来就算完工了.如果只是写一个小程序,“能够跑起来”这样的标准也就可以了,但是如果你是在公司进行程序的开发,那么仅仅让程序成功的跑起来是不行的, ...

  3. Android -- 启动另外一个Activity的方式(2s自动启动)

    1.  使用Handler  并且可以设置进入和退出的动画效果 Class < ? > activityClass; Class [ ] paramTypes = { Integer.TY ...

  4. iOS二维码生成-libqrencode编译报错

    libqrencode使用 1.将libqrencode文件夹整个拖入项目文件夹中 2.在要生成二维码的页面的 .m文件头部添加 #import "QRCodeGenerator.h&quo ...

  5. struts2上传图片的全过程

    struts2上传图片的过程 1.写一个上传的jsp页面upload_image.jsp,内容如下:<body><center>    <font color=" ...

  6. hihoCoder 1184 连通性二·边的双连通分量

    #1184 : 连通性二·边的双连通分量 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 在基本的网络搭建完成后,学校为了方便管理还需要对所有的服务器进行编组,网络所的老 ...

  7. localstorage 的属性改变问题

    localstorage 得到的是对象,我们打算改对象的某个属性的值,方法是 通过临时变量对象得到,改变临时变量,然后把临时变量给localstorage的方法 var localS  = windo ...

  8. cocostudio做出来的界面如何进行分辨率适配,兼论cocos2dx3的多分辨率适配机制,以及retina适配机制

    cocos有很多代码实际上都不再使用了,看代码时反而误导了程序员. 比如一个简单的分辨率适配,我查到了setContentSize,然后调用setContentSize,毫无用处啊!于是乎,我到处查资 ...

  9. linux添加环境变量(centos)

    1.查看当前环境变量 #echo $PATH 2.增加环境变量 #vi /etc/profile export PATH=/usr/path/bin:$PATH 3.生效 #source /etc/p ...

  10. JavaScript:编程改变文本样式

    <!DOCTYPE HTML><html><head><meta http-equiv="Content-Type" Content=&q ...