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. Java 集合系列04之 fail-fast总结(通过ArrayList来说明fail-fast的原理、解决办法)

    概要 前面,我们已经学习了ArrayList.接下来,我们以ArrayList为例,对Iterator的fail-fast机制进行了解.内容包括::1 fail-fast简介2 fail-fast示例 ...

  2. hud 5876 2016 ACM/ICPC Asia Regional Dalian Online

    题意:给一个图 给定一个点s 求补图中s点到达各个点的最短路 思路:从s点开始bfs 在图中与s点有连接的都是在补图中不能直接到达的点 反之在补图中都是可以直接到达的点 由此bfs ((( 诡异的写法 ...

  3. fnc.tld学习编写

    使用 el 的过程中,需要使用到后端代码处理逻辑,这个时候我们就需要自定义 方法. 如我们后端代码定义如下: package com.rhythmk.common; public class FncH ...

  4. 第一章-第一题(小学生四则运算)--By郭青云

    1.项目需求 a) 除了整数以外,还要支持真分数的四则运算. (例如:  1/6 + 1/8 = 7/24) b) 让程序能接受用户输入答案,并判定对错. 最后给出总共 对/错 的数量. c) 逐步扩 ...

  5. AADC安装指南

    可选功能中,”密码哈希同步“可以将本地域账号的密码默认每三小时同步到O365:”密码回写“则是反过来,但是世纪互联的答复是”此功能需要Auzre AD Service功能,目前国内版还不支持使用,国际 ...

  6. php多线程详解

    在说明多线程的题前,需要弄清楚以下几个问题 1,ts 和 nts的区别 Thread Safe和NoneThread Safe 先说windows的,在php官网,在windows区域有在文件下在有 ...

  7. java并发编程-基础

    线程带来的风险 安全性:多线程操作执行顺序的不可预测性 -- 永远不发生糟糕的事情: 活跃性:代码无法得到执行,死锁.饥饿问题 -- 某件正确的事情最终会发生: 性能问题:活跃性只意味着某件事最终会发 ...

  8. 让javascript显原型!

    相信以下的javascript让你读起来痛苦不已,告诉你一下简单的办法,就可以让它显出原型!将第一个单词,即eval换成document.write,然后再运行一下,它立即就原形毕露了! eval(f ...

  9. Thinking in Java——笔记(11)

    Holding Your Objects In general, your programs will always be creating new objects based on some cri ...

  10. shudupoj2676

    #include<stdio.h> int num,v[100][2],map[10][10]; bool judge(int x,int y,int k) {  int i,j,it,j ...