Tree Limbs on Lines
I was wondering.....will Verizon trim tree limbs of the phone lines (like BGE does) or is the reident responsible for triming the tree?
Hi pc12345,
Your issue has been escalated to a Verizon agent. Before the agent can begin assisting you, they will need to collect further information from you. Please go to your profile page for the forum and look at the top of the middle column where you will find an area titled "My Support Cases". You can reach your profile page by clicking on your name beside your post, or at the top left of this page underneath the title of the board.
Under "My Support Cases" you will find a link to the private board where you and the agent may exchange information. The title of your post is the link. This should be checked on a frequent basis, as the agent may be waiting for information from you before they can proceed with any actions. To ensure you know when they have responded to you, at the top of your support case there is a drop down menu for support case options. Open that and choose "subscribe". Please keep all correspondence regarding your issue in the private support portal.
Similar Messages
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Since Hurricane Sandy, there have been two large tree limbs down on the lines behind our house. They are not on the power lines, only the phone and cable lines. There has been no interruption in service, but the limbs are very large (12-16 inches across) and are hanging on the lines. They must be putting quite a bit of weight on the lines, because they have cause quite a bit of deflection in them already. They also do not appear to be very stable. There's no way that I can remove them safely, and our tree removal service was hesitant because of how the trees are positioned on the lines. What is Verizons policy regarding this situation?
I should add that the limbs are not on the lines running to our house, those are run below ground. The limbs are on the main lines servicing our block. The worst part is, it's not even branches from our tree. It's from our neighbors tree, and I'm not even sure that the lines are on our property (they run along the property line). The majority of the branch is resting in our yard however.Hi EvanBergstrom,
If you're sure that they're Verizon phone lines and not power lines, you can schedule a repair team to come out. Here is the link:
http://www.verizon.com/repair
Please let us know if you can't schedule a repair that way, and we can escalate the issue for you.
Regards,
AnnieS
-
Setting the ALV Hierarchy tree with grid line between the columns and rows
Hi Experts,
I would like to ask if there is any suggestion on setting the ALV hierarchy tree to be separated by grid line between the columns and rows just like how it is display the same way in normal ALV grid.
Thanks in advance.Hi Lin,
The requirement which you have stated is not possible.
Lin,
Also i have a query regarding BADI ZME_PROCESS_REQ_CUST, which you had raised on SDN. You have marked the question as solved/answered.
Changing the data of a customize field in purchase requisition
Could you please let me know, the steps you did to update the screen fields through the BADI.
I would really appreciate your reply, because i am facing exactly the same problem which you have mentioned.
Thanks,
Best regards,
Prashant -
Tree tags selected lines out of order
Why is it that tree controls return the tags of the items selected in the order they are selected, whereas the listbox returns the row number of the items selected in numeric order? Is there any way to have a tree control return the selected items in the order they appear on screen?
MichaelI've attached an example illustrating how listboxes return selected items in order and trees do not. If you click on one of the bottom items, hold control, then click on one of the top items, the listbox will return the row numbers in numeric order whereas the tree returns the tags in the order you clicked them. It would be nice if the tree returned the tags in the order on the screen.
The actual code I am using the tree in is for a plot legend that allows users to select which graphs to make visible. I've already made a workaround that works fine, I was just trying to use a for loop that looked at each item in the listbox or tree and made a plot visible or invisible according to the items selected. However, this required selected items to be in order. I know this probably isn't a very common use of a tree, since I only deal with parents, but some features involving cell coloring when cells are deleted work very well for my application.
If the ordering is just how it is for trees, I guess I'll have to accept it....
Michael -
CVI Tree Control and Line Break
Hi
Is there a way in CVI Treecontrol to do a line break if the column is to short ?
Regards
juergen
=s=i=g=n=a=t=u=r=e= Click on the Star and see what happens :-) =s=i=g=n=a=t=u=r=e=Hi jared,
If the text is to long there should be a line break
Regards
Juergen
=s=i=g=n=a=t=u=r=e= Click on the Star and see what happens :-) =s=i=g=n=a=t=u=r=e=
Attachments:
shoot.jpg 27 KB -
Hi
I have lived in my perfect rural location for just over 2 years now, when I first came the broadband worked beautifully and I was pleasantly surprised at how stable/fast ot was given where I am.
Last winter we had some pretty poor weather/high winds and since then my connection has been ropey at best. I have had 2 engineers out so far and am awaiting another hopefully today. I think that the problem is caused by the line going through so many trees until it reaches the pole, the first engineer replaced the line to allow some slack for movement but still have problems. The second engineer thought he had found a fault 100m or more from the house but then reported all ok on investigation. When I called to report the problem (last 3 days no phone or internet!) the automated service kindly told me there was no fault with the line! Argh! But an engineer is due to resolve by today apparentely.....
How do we stand with the whole tree issue and interference caused as a result? The council tell me they don't cut or trim trees for phone lines but I am at the point of cancelling everything as whats the point in paying for something that doesnt work or at best is intermittent!?
Any advice/help please...if it is the trees that are causing the problem then it is the owners of the trees who need to trim them no openreach. obviously if that is your problem then regardless which ISP you use apart from cable you will have the same problem
If you like a post, or want to say thanks for a helpful answer, please click on the Ratings star on the left-hand side of the post.
If someone answers your question correctly please let other members know by clicking on ’Mark as Accepted Solution’. -
How to draw a line on ADF page between two nodes for mapping.
Hi everyone,
Does anyone have a solution that how can I wiring two points by drawing a line on ADF pages.
My scenario is user want to do a mapping between two xml files. We will build an ADF faces page. This page have two parts, left part contains one tree(base on the source xml), right part is the destination tree(target xml). User can drag an node from left and drop to right. Meantime, a line will be created to connect two node, it would be perfect that when user scroll down the page, or extend the tree node, the line between source and target will be remain connection two node.
Does anyone have a solution for this, thanks in advance.
Hongfu.so you want to do something like. xsl mapper in soa
http://www.haertfelder.com/images/pSoaBPEL3.png
i can think of using javascript.. not sure... neeed extra programming.. -
Populating Tree with External XML File
I want to use external files for populating Flex 3
components. I have been successful using code very similar to that
below for a data grid & for a combo box, but it won't work for
a tree. Can you determine wh?
<?xml version="1.0" encoding="utf-8"?>
<mx:WindowedApplication xmlns:mx="
http://www.adobe.com/2006/mxml"
layout="absolute">
<mx:HTTPService id="dp_Tree1" url="Tree1.xml" />
<mx:Tree dataProvider="{dp_Tree1.lastResult.root.node}"
creationComplete="dp_Tree1.send()" width="333"/>
</mx:WindowedApplication>
It's not working & I can't determine why. Attached is my
xml file that I tested it with.Thanks a bunch folks, but my intent is to have as little code
as possible. I want to help people who are not coders create
designs using external xml files to populate controls. I was able
to create very simple code with just two lines to populate a
DataGrid from an external xml file:
<mx:HTTPService id="dp_DataGrid1" url="DataGrid1.xml"
/>
<mx:DataGrid
dataProvider="{dp_DataGrid1.lastResult.component.rows}"
creationComplete="dp_DataGrid1.send()" />
And just two lines for a ComboBox:
<mx:HTTPService id="dp_ComboBox1" url="ComboBox1.xml"
/>
<mx:ComboBox
dataProvider="{dp_ComboBox1.lastResult.component.rows}"
creationComplete="dp_ComboBox1.send()" />
Isn't it possible to populate a tree with two lines like as
it is with a DataGrid & ComboBox? -
Binary search tree in java using a 2-d array
Good day, i have been wrestling with this here question.
i think it does not get any harder than this. What i have done so far is shown at the bottom.
We want to use both Binary Search Tree and Single Linked lists data structures to store a text. Chaining
techniques are used to store the lines of the text in which a word appears. Each node of the binary search
tree contains four fields :
(i) Word
(ii) A pointer pointing to all the lines word appears in
(iii) A pointer pointing to the subtree(left) containing all the words that appears in the text and are
predecessors of word in lexicographic order.
(iv) A pointer pointing to the subtree(right) containing all the words that appears in the text and are
successors of word in lexicographic order.
Given the following incomplete Java classes BinSrchTreeWordNode, TreeText, you are asked to complete
three methods, InsertionBinSrchTree, CreateBinSrchTree and LinesWordInBinSrchTree. For
simplicity we assume that the text is stored in a 2D array, a row of the array represents a line of the text.
Each element in the single linked list is represented by a LineNode that contains a field Line which represents a line in which the word appears, a field next which contains the address of a LineNode representing the next line in which the word appears.
public class TreeText{
BinSrchTreeWordNode RootText = null;// pointer to the root of the tree
String TextID; // Text Identification
TreeText(String tID){TextID = tID;}
void CreateBinSrchTree (TEXT text){...}
void LinesWordInBinSrchTree(BinSrchTreeWordNode Node){...}
public static void main(String[] args)
TEXT univ = new TEXT(6,4);
univ.textcont[0][0] = "Ukzn"; univ.textcont[0][1] ="Uct";
univ.textcont[0][2] ="Wits";univ.textcont[0][3] ="Rhodes";
univ.textcont[1][0] = "stellenbosch";
univ.textcont[1][1] ="FreeState";
univ.textcont[1][2] ="Johannesburg";
univ.textcont[1][3] = "Pretoria" ;
univ.textcont[2][0] ="Zululand";univ.textcont[2][1] ="NorthWest";
univ.textcont[2][2] ="Limpopo";univ.textcont[2][3] ="Wsu";
univ.textcont[3][0] ="NorthWest";univ.textcont[3][1] ="Limpopo";
univ.textcont[3][2] ="Uct";univ.textcont[3][3] ="Ukzn";
univ.textcont[4][0] ="Mit";univ.textcont[4][1] ="Havard";
univ.textcont[4][2] ="Michigan";univ.textcont[4][3] ="Juissieu";
univ.textcont[5][0] ="Cut";univ.textcont[5][1] ="Nmmu";
univ.textcont[5][2] ="ManTech";univ.textcont[5][3] ="Oxford";
// create a binary search tree (universities)
// and insert words of text univ in it
TreeText universities = new TreeText("Universities");
universities.CreateBinSrchTree(univ);
// List words Universities trees with their lines of appearance
System.out.println();
System.out.println(universities.TextID);
System.out.println();
universities.LinesWordInBinSrchTree(universities.RootText);
public class BinSrchTreeWordNode {
BinSrchTreeWordNode LeftTree = null; // precedent words
String word;
LineNode NextLineNode = null; // next line in
// which word appears
BinSrchTreeWordNode RightTree = null; // following words
BinSrchTreeWordNode(String WordValue)
{word = WordValue;} // creates a new node
BinSrchTreeWordNode InsertionBinSrchTree
(String w, int line, BinSrchTreeWordNode bst)
public class LineNode{
int Line; // line in which the word appears
LineNode next = null;
public class TEXT{
int NBRLINES ; // number of lines
int NBRCOLS; // number of columns
String [][] textcont; // text content
TEXT(int nl, int nc){textcont = new String[nl][nc];}
The method InsertionBinSrchTree inserts a word (w) in the Binary search tree. The method Create-
BinSrchTree creates a binary search tree by repeated calls to InsertionBinSrchTree to insert elements
of text. The method LinesWordInBinSrchTree traverses the Binary search tree inorder and displays the
words with the lines in which each appears.
>>>>>>>>>>>>>>>>>>>>>>
//InsertionBinTree is of type BinSearchTreeWordNode
BinSrchTreeWordNode InsertionBinSrchTree(String w, int line, BinSrchTreeWordNode bst)
//First a check must be made to make sure that we are not trying to //insert a word into an empty tree. If tree is empty we just create a //new node.
If (bst == NULL)
System.out.println(Tree was empty)
For (int rows =0; rows <= 6; rows++)
For (int cols = 0; cols <= 4; cols++)
Textcont[i][j] = wWhat is the purpose of this thread? You are yet to ask a question... Such a waste of time...
For future reference use CODE TAGS when posting code in a thread.
But again have a think about how to convey a question to others instead of blabbering on about nothing.
i think it does not get any harder than this.What is so difficult to understand. Google an implementation of a binary tree using a single array. Then you can integrate this into the required 2-dimension array for your linked list implemented as an array in your 2-d array.
Mel -
VB Like tree in Webdynpro Java
We are using Webdynpro Tree UI Control for displaying functional location hierarchy. However the standard webdynpro Tree does not display clear lines associating children's with its parents - VB Tree has these lines which is very useful for the users. We tried to work around this issue by creating icons with vertical lines in it. But we were not able to get continues lines and its hard to display multiple lines using icons.
Have anyone encountered the same issue?. if yes, how did you resolve it?.
is SAP planning to incorporate this requirement in near future?.
Regards,
ArulHi Arul,
I would suggest you use TreeByNestingTable Column element.
You can find it as Master Column in the Table UI element.
This will solve your problem because table row will solve the lines problem that you are facing.
Go through the following tutorial to get familiar with the element.
<a href="https://www.sdn.sap.comhttp://www.sdn.sap.comhttp://www.sdn.sap.com/irj/servlet/prt/portal/prtroot/com.sap.km.cm.docs/library/webdynpro/wd%20java/wd%20tutorials/integration%20of%20a%20tree%20structure%20in%20a%20web%20dynpro%20table.pdf">Integration of Tree and Table</a>
Regards
Amit -
Fillable object from line segments
I've created a tree object from line segments and it looks how I want it to look. I made it a compound path. When I fill it, I get weird results. I used live fill and that works fine, but I still dont get the pointed edges that I would like to see on the outline and I feel like there is a more elegant way to go about this. I have attached the image of the tree. The blue circle is the weird fill. The red circle is the pointed edges that I want. I got this point using the pen tool, but it doesn't act correctly with the rest of the path.
I am probably doing this all completely wrong, but I'm a bit impatient for classes and I learn best through trial and error.Select your object and in your stroke panel increase the miter limit to a higher number
The weird gren or blue fill is probably because your points are not joined. Use the hollow arrow tool to select a point and move it you will see. You can fix by using the hollow arrow tool to select only the 2 points, then using object >> path >> join -
Af|tree::expanded-icon issue ?
af|tree::expanded-icon
content: url(/skins/mycompany/skin_images/expand.gif);
width:16px;
height:16px;
i recieve error at af|tree::expanded-icon line (on second COLON':').Css file does not accept this identifier there ? i have tried many ways but all in vain.
is there any thing special i need to import for new af|tree styles? I am using JDev 10.1.3.3Hi
First of all see if your af:dialog has the contentDelivery="lazyUncached" . This way your popup would not be cached.
If this is not the solution , then try:
tree.getDisclosedRowKeys().removeAll();This will remove all disclosed rowKeys.
A.Gruev -
Tree Nodes has White Outline/Border
Hi,
I'm using an ADF tree located inside a panelGroupLayout.
The panelGroupLayout has the background color "light blue".
At runtime each tree node is displayed with a surrounding white outline and that can only appear when you have a background color different than white (in my case the panelGroupLayout is "light blue").
Does anybody knows how to remove that outline? or to change its color?
Thanks,
Alain.Here are the skin seletors
tr:tree Component
Icon Selectors
Name Description
af|tree::expanded-icon This icon is displayed before the expanded tree node.
af|tree::collapsed-icon This icon is displayed before the collapsed tree node.
af|tree::no-children-icon This icon is displayed instead of the expanded/collapsed icon, when the node has no children
af|tree::line-icon This icon is used as a vertical line between the nodes.
af|tree::line-middle-icon This icon is used as the horizontal line in the background of the expand/collapse icon of the node, in the case the node is not the last sibling of its parent node.
af|tree::line-last-icon This icon is used as the horizontal line in the background of the expand/collapse icon of the node, in the case the node is the last sibling of its parent node.
af|tree::node-icon This icon selector is used in the case the Node class has a getNodeType() method that returns the node type as string. The nodetype gets added to this selector, separated by a ':'. If the node is a container (has children) the following suffixes get added depending on the expanded/collapsed state: '-expanded' / '-collapsed'. e.g. "af|tree::node-icon:container-collapsed", "af|tree::node-icon:container-expanded", "af|tree::node-icon:noncontainer".
Trinidad properties
Name Description
-tr-show-lines Valid values are true or false (default true). Determines whether the tree lines are displayed or not. e.g., af|tree {-tr-show-lines:false} will not show the lines of the tree.Timo -
So I have an assignment to write a program that deciefers basic morse code using a binary tree. A file was provided with the information for the binary tree and I am using the text books author's version of BinaryTree class with a few methods of my own thrown in.
For starters .. why would you ever use a tree for this .. just make a look up table. But my real problem is i keep getting null pointers at 38, 24, 55 and I have tried just about everything I know to do at this point. My mind is exhausted and I think I need a fresh pair of eyes to tell me what I have done wrong.
I am sorry for such sloppy code.. I have been changing and rearranging too much.
import java.io.*;
import java.util.*;
public class MorseCode extends BinaryTree{
static BufferedReader in;
static BinaryTree<String> bT = new BinaryTree<String>();
/* public static void loadTree() throws FileNotFoundException, IOException{
in = new BufferedReader(new FileReader("MorseCode.txt"));
bT = readBinaryTree(in);
public static String decode(Character c) throws IOException{
if(c.equals("null")){
return "";
}else if(c.equals(" ")){
return " ";
}else{
return (":" + find(c));
public static String find(Character c) throws IOException, FileNotFoundException{
in = new BufferedReader(new FileReader("MorseCode.txt"));
bT = readBinaryTree(in);
Queue<BinaryTree> data = new LinkedList<BinaryTree>();
BinaryTree<String> tempTree = bT;
String temp = null;
Character tempChar = null;
data.offer(tempTree);
while(!data.isEmpty()){
tempTree = data.poll();
temp = tempTree.getData();
tempChar = temp.charAt(0);
if(tempChar.equals(c)){
break;
data.offer(tempTree.getRightSubtree());
data.offer(tempTree.getLeftSubtree());
return temp.substring(2);
public static void main(String[] args) throws FileNotFoundException, IOException{
Scanner scan = new Scanner(new FileReader("encode.in.txt"));
String s = "";
String temp = "";
while(scan.hasNextLine()){
temp = scan.nextLine();
for(int i = 0; i < temp.length(); i++){
s = s + decode(temp.charAt(i));
System.out.println(s);
/** Class for a binary tree that stores type E objects.
* @author Koffman and Wolfgang
class BinaryTree <E> implements Serializable
//===================================================
/** Class to encapsulate a tree node. */
protected static class Node <E> implements Serializable
// Data Fields
/** The information stored in this node. */
protected E data;
/** Reference to the left child. */
protected Node <E> left;
/** Reference to the right child. */
protected Node <E> right;
// Constructors
/** Construct a node with given data and no children.
@param data The data to store in this node
public Node(E data) {
this.data = data;
left = null;
right = null;
// Methods
/** Return a string representation of the node.
@return A string representation of the data fields
public String toString() {
return data.toString();
}//end inner class Node
//===================================================
// Data Field
/** The root of the binary tree */
protected Node <E> root;
public BinaryTree()
root = null;
protected BinaryTree(Node <E> root)
this.root = root;
/** Constructs a new binary tree with data in its root,leftTree
as its left subtree and rightTree as its right subtree.
public BinaryTree(E data, BinaryTree <E> leftTree, BinaryTree <E> rightTree)
root = new Node <E> (data);
if (leftTree != null) {
root.left = leftTree.root;
else {
root.left = null;
if (rightTree != null) {
root.right = rightTree.root;
else {
root.right = null;
/** Return the left subtree.
@return The left subtree or null if either the root or
the left subtree is null
public BinaryTree <E> getLeftSubtree() {
if (root != null && root.left != null) {
return new BinaryTree <E> (root.left);
else {
return null;
/** Return the right sub-tree
@return the right sub-tree or
null if either the root or the
right subtree is null.
public BinaryTree<E> getRightSubtree() {
if (root != null && root.right != null) {
return new BinaryTree<E>(root.right);
} else {
return null;
public String getData(){
if(root.data == null){
return "null";
}else{
return (String) root.data;
/** Determine whether this tree is a leaf.
@return true if the root has no children
public boolean isLeaf() {
return (root.left == null && root.right == null);
public String toString() {
StringBuilder sb = new StringBuilder();
preOrderTraverse(root, 1, sb);
return sb.toString();
/** Perform a preorder traversal.
@param node The local root
@param depth The depth
@param sb The string buffer to save the output
private void preOrderTraverse(Node <E> node, int depth,
StringBuilder sb) {
for (int i = 1; i < depth; i++) {
sb.append(" ");
if (node == null) {
sb.append("null\n");
else {
sb.append(node.toString());
sb.append("\n");
preOrderTraverse(node.left, depth + 1, sb);
preOrderTraverse(node.right, depth + 1, sb);
/** Method to read a binary tree.
pre: The input consists of a preorder traversal
of the binary tree. The line "null" indicates a null tree.
@param bR The input file
@return The binary tree
@throws IOException If there is an input error
public static BinaryTree<String> readBinaryTree(BufferedReader bR) throws IOException
// Read a line and trim leading and trailing spaces.
String data = bR.readLine().trim();
if (data.equals("null")) {
return null;
else {
BinaryTree < String > leftTree = readBinaryTree(bR);
BinaryTree < String > rightTree = readBinaryTree(bR);
return new BinaryTree < String > (data, leftTree, rightTree);
}//readBinaryTree()
/*Method to determine the height of a binary tree
pre: The line "null" indicates a null tree.
@param T The binary tree
@return The height as integer
public static int height(BinaryTree T){
if(T == null){
return 0;
}else{
return 1 +(int) (Math.max(height(T.getRightSubtree()), height(T.getLeftSubtree())));
public static String preOrderTraversal(BinaryTree<String> T){
String s = T.toString();
String s2 = "";
String temp = "";
Scanner scan = new Scanner(s);
while(scan.hasNextLine()){
temp = scan.nextLine().trim();
if(temp.equals("null")){
s2 = s2;
}else{
s2 = s2 + " " + temp;
return s2;
}//class BinaryTreeAs well as warnerja's point, you say you keep getting these errors. Sometimes it's helpful when illustrating a problem to replace the file based input with input that comes from a given String. That way we all see the same behaviour under the same circumstances.
public static void main(String[] args) throws FileNotFoundException, IOException{
//Scanner scan = new Scanner(new FileReader("encode.in.txt"));
Scanner scan = new Scanner("whatever");The following isn't the cause of an NPE, but it might be allowing one to "slip through" (ie you think you've dealt with the null case when you haven't):
public static String decode(Character c) throws IOException{
if(c.equals("null")){
c is a Character so it will never be the case that it is equal to the string n-u-l-l.
Perhaps the behaviour of each of these methods needs to be (documented and) tested. -
I now have a Tree Control populated from an ArrayCollection
and using the defaultDataDescriptor.
At this point in the program the data exists only in the
myTree.dataProvider and the root node is at myTree.dataProvider[0]
or myTree.dataProvider.source[0] and I can see from the variables
window that the correct data is located here.
What do I have to put for 'item' in
myTree.expandChildrenOf(??item??, true). Everything I can think of
returns:
"Cannot access a property or method of a null object
reference."
Does anyone have any idea how to make this work.
DougThe error occurs in the Flex file Tree.as at line 1383.
I can now see that the null object is not item -
dataProvider[0], this contains the correct data. The problem is
that iterator is null and therefore iterator.view is a null
reference.
Is there any way to solve this problem?
My ArrayCollection is built in pretty much the same way as
shown in:
Flex 2 Developer's Guide > Building User Interfaces for
Flex Applications > Using Data Providers and Collections
> Using hierarchical data providers > Data descriptors and
hierarchical data provider structure > Creating a custom data
descriptor
But I have used the defaultDataDescriptor and my arrays are
not built from literals.
Doug
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