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  • HQL recursion, how do I do this?

    - by niklassaers
    Hi guys, I have a tree structure where each Node has a parent and a Set<Node> children. Each Node has a String title, and I want to make a query where I select Set<String> titles, being the title of this node and of all parent nodes. How do I write this query? The query for a single title is this, but like I said, I'd like it expanded for the entire branch of parents. SELECT node.title FROM Node node WHERE node.id = :id Cheers Nik

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  • Coupling/Cohesion

    - by user559142
    Hi All, Whilst there are many good examples on this forum that contain examples of coupling and cohesion, I am struggling to apply it to my code fully. I can identify parts in my code that may need changing. Would any Java experts be able to take a look at my code and explain to me what aspects are good and bad. I don't mind changing it myself at all. It's just that many people seem to disagree with each other and I'm finding it hard to actually understand what principles to follow... package familytree; /** * * @author David */ public class Main { /** * @param args the command line arguments */ public static void main(String[] args) { // TODO code application logic here KeyboardInput in = new KeyboardInput(); FamilyTree familyTree = new FamilyTree(in, System.out); familyTree.start(); } } package familytree; import java.io.PrintStream; /** * * @author David */ public class FamilyTree { /** * @param args the command line arguments */ private static final int DISPLAY_FAMILY_MEMBERS = 1; private static final int ADD_FAMILY_MEMBER = 2; private static final int REMOVE_FAMILY_MEMBER = 3; private static final int EDIT_FAMILY_MEMBER = 4; private static final int SAVE_FAMILY_TREE = 5; private static final int LOAD_FAMILY_TREE = 6; private static final int DISPLAY_ANCESTORS = 7; private static final int DISPLAY_DESCENDANTS = 8; private static final int QUIT = 9; private KeyboardInput in; private Family family; private PrintStream out; public FamilyTree(KeyboardInput in, PrintStream out) { this.in = in; this.out = out; family = new Family(); } public void start() { out.println("\nWelcome to the Family Tree Builder"); //enterUserDetails(); initialise(); while (true) { displayFamilyTreeMenu(); out.print("\nEnter Choice: "); int option = in.readInteger(); if (option > 0 && option <= 8) { if (quit(option)) { break; } executeOption(option); } else { out.println("Invalid Choice!"); } } } //good private void displayFamilyTreeMenu() { out.println("\nFamily Tree Menu"); out.println(DISPLAY_FAMILY_MEMBERS + ". Display Family Members"); out.println(ADD_FAMILY_MEMBER + ". Add Family Member"); out.println(REMOVE_FAMILY_MEMBER + ". Remove Family Member"); out.println(EDIT_FAMILY_MEMBER + ". Edit Family Member"); out.println(SAVE_FAMILY_TREE + ". Save Family Tree"); out.println(LOAD_FAMILY_TREE + ". Load Family Tree"); out.println(DISPLAY_ANCESTORS + ". Display Ancestors"); out.println(DISPLAY_DESCENDANTS + ". Display Descendants"); out.println(QUIT + ". Quit"); } //good private boolean quit(int opt) { return (opt == QUIT) ? true : false; } //good private void executeOption(int choice) { switch (choice) { case DISPLAY_FAMILY_MEMBERS: displayFamilyMembers(); break; case ADD_FAMILY_MEMBER: addFamilyMember(); break; case REMOVE_FAMILY_MEMBER: break; case EDIT_FAMILY_MEMBER: break; case SAVE_FAMILY_TREE: break; case LOAD_FAMILY_TREE: break; case DISPLAY_ANCESTORS: displayAncestors(); break; case DISPLAY_DESCENDANTS: displayDescendants(); break; default: out.println("Not a valid option! Try again."); break; } } //for selecting family member for editing adding nodes etc private void displayFamilyMembers() { out.println("\nDisplay Family Members"); int count = 0; for (FamilyMember member : family.getFamilyMembers()) { out.println(); if (count + 1 < 10) { out.println((count + 1) + ". " + member.getFirstName() + " " + member.getLastName()); out.println(" " + member.getDob()); out.println(" Generation: " + member.getGeneration()); } else { out.println((count + 1) + ". " + member.getFirstName() + " " + member.getLastName()); out.println(" " + member.getDob()); out.println(" Generation: " + member.getGeneration()); } count++; } } private int selectRelative() { out.println("\nSelect Relative"); out.println("1. Add Parents"); out.println("2. Add Child"); out.println("3. Add Partner"); out.println("4. Add Sibling"); out.print("\nEnter Choice: "); int choice = in.readInteger(); if (choice > 0 && choice < 5) { return choice; } return (-1); } private void addFamilyMember() { int memberIndex = selectMember(); if (memberIndex >= 0) { FamilyMember member = family.getFamilyMember(memberIndex); int relative = selectRelative(); if (relative > 0) { out.println("\nAdd Member"); //if choice is valid switch (relative) { case 1: //adding parents if (member.getFather() == null) { FamilyMember mum, dad; out.println("Enter Mothers Details"); mum = addMember(relative, "Female"); out.println("\nEnter Fathers Details"); dad = addMember(relative, "Male"); member.linkParent(mum); member.linkParent(dad); mum.linkPartner(dad); mum.setGeneration(member.getGeneration() - 1); dad.setGeneration(member.getGeneration() - 1); sortGenerations(); } else { out.println(member.getFirstName() + " " + member.getLastName() + " already has parents."); } break; case 2: //adding child if (member.getPartner() == null) { FamilyMember partner; if (member.getGender().equals("Male")) { out.println("Enter Mothers Details"); partner = addMember(1, "Female"); } else { out.println("Enter Fathers Details"); partner = addMember(1, "Male"); } //create partner member.linkPartner(partner); partner.setGeneration(member.getGeneration()); out.println(); } out.println("Enter Childs Details"); FamilyMember child = addMember(relative, ""); child.linkParent(member); child.linkParent(member.getPartner()); child.setGeneration(member.getGeneration() + 1); sortGenerations(); break; case 3: //adding partner if (member.getPartner() == null) { out.println("Enter Partners Details"); FamilyMember partner = addMember(relative, ""); member.linkPartner(partner); partner.setGeneration(member.getGeneration()); } else { out.println(member.getFirstName() + " " + member.getLastName() + " already has a partner."); } break; case 4: //adding sibling FamilyMember mum, dad; if (member.getFather() == null) { out.println("Enter Mothers Details"); mum = addMember(1, "Female"); out.println("\nEnter Fathers Details"); dad = addMember(1, "Male"); member.linkParent(mum); member.linkParent(dad); mum.linkPartner(dad); mum.setGeneration(member.getGeneration() - 1); dad.setGeneration(member.getGeneration() - 1); sortGenerations(); out.println("\nEnter Siblings Details"); } else { out.println("Enter Siblings Details"); } FamilyMember sibling = addMember(relative, ""); //create mum and dad mum = member.getMother(); dad = member.getFather(); sibling.linkParent(mum); sibling.linkParent(dad); sibling.setGeneration(member.getGeneration()); break; } } else { out.println("Invalid Option!"); } } else { out.println("Invalid Option!"); } } private int selectMember() { displayFamilyMembers(); out.print("\nSelect Member: "); int choice = in.readInteger(); if (choice > 0 && choice <= family.getFamilyMembers().size()) { return (choice - 1); } return -1; } private FamilyMember addMember(int option, String gender) { out.print("Enter First Name: "); String fName = formatString(in.readString().trim()); out.print("Enter Last Name: "); String lName = formatString(in.readString().trim()); if (option != 1) { //if not adding parents out.println("Select Gender"); out.println("1. Male"); out.println("2. Female"); out.print("Enter Choice: "); int gOpt = in.readInteger(); if (gOpt == 1) { gender = "Male"; } else if (gOpt == 2) { gender = "Female"; } else { out.println("Invalid Choice"); return null; } } String dob = enterDateOfBirth(); lName = formatString(lName); FamilyMember f = family.getFamilyMember(family.addMember(fName, lName, gender, dob)); f.setIndex(family.getFamilyMembers().size() - 1); return (f); } private String formatString(String s){ String firstLetter = s.substring(0, 1); String remainingLetters = s.substring(1, s.length()); s = firstLetter.toUpperCase() + remainingLetters.toLowerCase(); return s; } private String enterDateOfBirth(){ out.print("Enter Year Of Birth (0 - 2011): "); String y = in.readString(); out.print("Enter Month Of Birth (1-12): "); String m = in.readString(); if (Integer.parseInt(m) < 10) { m = "0" + m; } m += "-"; out.print("Enter Date of Birth (1-31): "); String d = in.readString(); if (Integer.parseInt(d) < 10) { d = "0" + d; } d += "-"; String dob = d + m + y; while(!DateValidator.isValid(dob)){ out.println("Invalid Date. Try Again:"); dob = enterDateOfBirth(); } return (dob); } private void displayAncestors() { out.print("\nDisplay Ancestors For Which Member: "); int choice = selectMember(); if (choice >= 0) { FamilyMember node = family.getFamilyMember(choice ); FamilyMember ms = findRootNode(node, 0, 2, -1); FamilyMember fs = findRootNode(node, 1, 2, -1); out.println("\nPrint Ancestors"); out.println("\nMothers Side"); printDescendants(ms, node, ms.getGeneration()); out.println("\nFathers Side"); printDescendants(fs, node, fs.getGeneration()); } else { out.println("Invalid Option!"); } } private void displayDescendants() { out.print("\nDisplay Descendants For Which Member: "); int choice = selectMember(); if (choice >= 0) { FamilyMember node = family.getFamilyMember(choice); out.println("\nPrint Descendants"); printDescendants(node, null, 0); } else { out.println("Invalid Option!"); } } private FamilyMember findRootNode(FamilyMember node, int parent, int numGenerations, int count) { FamilyMember root; count++; if (node.hasParents() && count < numGenerations) { if (parent == 0) { node = node.getMother(); root = findRootNode(node, 1, numGenerations, count); } else { node = node.getFather(); root = findRootNode(node, 1, numGenerations, count); } return root; } return node; } private int findHighestLeafGeneration(FamilyMember node) { int gen = node.getGeneration(); for (int i = 0; i < node.getChildren().size(); i++) { int highestChild = findHighestLeafGeneration(node.getChild(i)); if (highestChild > gen) { gen = highestChild; } } return gen; } private void printDescendants(FamilyMember root, FamilyMember node, int gen) { out.print((root.getGeneration() + 1) + " " + root.getFullName()); out.print(" [" + root.getDob() + "] "); if (root.getPartner() != null) { out.print("+Partner: " + root.getPartner().getFullName() + " [" + root.getPartner().getDob() + "] "); } if (root == node) { out.print("*"); } out.println(); if (!root.getChildren().isEmpty() && root != node) { for (int i = 0; i < root.getChildren().size(); i++) { for (int j = 0; j < root.getChild(i).getGeneration() - gen; j++) { out.print(" "); } printDescendants(root.getChild(i), node, gen); } } else { return; } } //retrieve highest generation public int getRootGeneration(){ int min = family.getFamilyMember(0).getGeneration(); for(int i = 0; i < family.getFamilyMembers().size(); i++){ min = Math.min(min, family.getFamilyMember(i).getGeneration()); } return Math.abs(min); } public void sortGenerations(){ int amount = getRootGeneration(); for (FamilyMember member : family.getFamilyMembers()) { member.setGeneration(member.getGeneration() + amount); } } //test method - temporary private void initialise() { family.addMember("Bilbo", "Baggins", "Male", "23-06-1920"); } } package familytree; import java.util.ArrayList; import java.util.Date; /** * * @author David */ public class Family { //family members private ArrayList<FamilyMember> family; //create Family public Family() { family = new ArrayList<FamilyMember>(); } //add member to the family public int addMember(String f, String l, String g, String d) { family.add(new FamilyMember(f, l, g, d)); return family.size()-1; } //remove member from family public void removeMember(int index) { family.remove(index); } public FamilyMember getFamilyMember(int index) { return family.get(index); } //return family public ArrayList <FamilyMember> getFamilyMembers() { return family; } public void changeFirstName(int index, String f) { family.get(index).setFirstName(f);//change to setfirstname and others } public void changeLastName(int index, String l) { family.get(index).setLastName(l); } public void changeAge(int index, int a) { family.get(index).setAge(a); } public void changeDOB() { //implement } } package familytree; import java.util.ArrayList; import java.util.Collections; /** * * @author David */ public class FamilyMember extends Person { private FamilyMember mother; private FamilyMember father; private FamilyMember partner; private ArrayList<FamilyMember> children; private int generation; private int index; //initialise family member public FamilyMember(String f, String l, String g, String d) { super(f, l, g, d); mother = null; father = null; partner = null; children = new ArrayList<FamilyMember>(); generation = 0; index = -1; } public void linkParent(FamilyMember parent) { if (parent.getGender().equals("Female")) { this.setMother(parent); } else { this.setFather(parent); } parent.addChild(this); } public void linkPartner(FamilyMember partner) { partner.setPartner(this); this.setPartner(partner); } public boolean hasParents() { if (this.getMother() == null && this.getFather() == null) { return false; } return true; } public FamilyMember getMother() { return mother; } public FamilyMember getFather() { return father; } public FamilyMember getPartner() { return partner; } public FamilyMember getChild(int index) { return children.get(index); } public int getGeneration() { return generation; } public int getIndex() { return index; } public ArrayList<FamilyMember> getChildren() { return children; } public void setMother(FamilyMember f) { mother = f; } public void setFather(FamilyMember f) { father = f; } public void setPartner(FamilyMember f) { partner = f; } public void addChild(FamilyMember f) { children.add(f); //add child if(children.size() > 1){ //sort in ascending order Collections.sort(children, new DateComparator()); } } public void addChildAt(FamilyMember f, int index) { children.set(index, f); } public void setGeneration(int g) { generation = g; } public void setIndex(int i){ index = i; } } package familytree; /** * * @author David */ public class Person{ private String fName; private String lName; private String gender; private int age; private String dob; public Person(String fName, String lName, String gender, String dob){ this.fName = fName; this.lName = lName; this.gender = gender; this.dob = dob; } public String getFullName(){ return (this.fName + " " + this.lName); } public String getFirstName(){ return (fName); } public String getLastName(){ return (lName); } public String getGender(){ return (gender); } public String getDob(){ return dob; } public int getAge(){ return age; } public void setFirstName(String fName){ this.fName = fName; } public void setLastName(String lName){ this.lName = lName; } public void setGender(String gender){ this.gender = gender; } public void setAge(int age){ this.age = age; } }

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  • Optimising RSS parsing on App Engine to avoid high CPU warnings

    - by Danny Tuppeny
    I'm pulling some RSS feeds into a datastore in App Engine to serve up to an iPhone app. I use cron to schedule updating the RSS every x minutes. Each task only parses one RSS feed (which has 15-20 items). I frequently get warnings about high CPU usage in the App Engine dashboard, so I'm looking for ways to optimise my code. Currently, I use minidom (since it's already there on App Engine), but I suspect it's not very efficient! Here's the code: dom = minidom.parseString(urlfetch.fetch(url).content) if dom: items = [] for node in dom.getElementsByTagName('item'): item = RssItem( key_name = self.getText(node.getElementsByTagName('guid')[0].childNodes), title = self.getText(node.getElementsByTagName('title')[0].childNodes), description = self.getText(node.getElementsByTagName('description')[0].childNodes), modified = datetime.now(), link = self.getText(node.getElementsByTagName('link')[0].childNodes), categories = [self.getText(category.childNodes) for category in node.getElementsByTagName('category')] ); items.append(item); db.put(items); def getText(self, nodelist): rc = '' for node in nodelist: if node.nodeType == node.TEXT_NODE: rc = rc + node.data return rc There isn't much going on, but the scripts often take 2-6 seconds CPU time, which seems a bit excessive for looping through 20ish items and reading a few attributes. What can I do to make this faster? Is there anything particularly bad in the above code, or should I change to another way of parsing? Are there are any libraries (that work on App Engine) that would be better, or would I be better parsing the RSS myself?

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  • d3.js force layout increase linkDistance

    - by user1159833
    How to increase linkDistance without affecting the node alignment, example: http://mbostock.github.com/d3/talk/20110921/force.html when I try to increase the circle radius and linkDistance the it collapse <script type="text/javascript"> var w = 1280, h = 800, z = d3.scale.category20c(); var force = d3.layout.force() .size([w, h]); var svg = d3.select("#chart").append("svg:svg") .attr("width", w) .attr("height", h); svg.append("svg:rect") .attr("width", w) .attr("height", h); d3.json("flare.json", function(root) { var nodes = flatten(root), links = d3.layout.tree().links(nodes); force .nodes(nodes) .links(links) .start(); var link = svg.selectAll("line") .data(links) .enter().insert("svg:line") .style("stroke", "#999") .style("stroke-width", "1px"); var node = svg.selectAll("circle.node") .data(nodes) .enter().append("svg:circle") .attr("r", 4.5) .style("fill", function(d) { return z(d.parent && d.parent.name); }) .style("stroke", "#000") .call(force.drag); force.on("tick", function(e) { link.attr("x1", function(d) { return d.source.x; }) .attr("y1", function(d) { return d.source.y; }) .attr("x2", function(d) { return d.target.x; }) .attr("y2", function(d) { return d.target.y; }); node.attr("cx", function(d) { return d.x; }) .attr("cy", function(d) { return d.y; }); }); }); function flatten(root) { var nodes = []; function recurse(node, depth) { if (node.children) { node.children.forEach(function(child) { child.parent = node; recurse(child, depth + 1); }); } node.depth = depth; nodes.push(node); } recurse(root, 1); return nodes; } </script>

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  • Undefined referencec to ...

    - by Patrick LaChance
    I keep getting this error message every time I try to compile, and I cannot find out what the problem is. any help would be greatly appreciated: C:\DOCUME~1\Patrick\LOCALS~1\Temp/ccL92mj9.o:main.cpp:(.txt+0x184): undefined reference to 'List::List()' C:\DOCUME~1\Patrick\LOCALS~1\Temp/ccL92mj9.o:main.cpp:(.txt+0x184): undefined reference to 'List::add(int)' collect2: ld returned 1 exit status code: //List.h ifndef LIST_H define LIST_H include //brief Definition of linked list class class List { public: /** \brief Exception for operating on empty list */ class Empty : public std::exception { public: virtual const char* what() const throw(); }; /** \brief Exception for invalid operations other than operating on an empty list */ class InvalidOperation : public std::exception { public: virtual const char* what() const throw(); }; /** \brief Node within List */ class Node { public: /** data element stored in this node */ int element; /** next node in list / Node next; /** previous node in list / Node previous; Node (int element); ~Node(); void print() const; void printDebug() const; }; List(); ~List(); void add(int element); void remove(int element); int first()const; int last()const; int removeFirst(); int removeLast(); bool isEmpty()const; int size()const; void printForward() const; void printReverse() const; void printDebug() const; /** enables extra output for debugging purposes */ static bool traceOn; private: /** head of list */ Node* head; /** tail of list */ Node* tail; /** count of number of nodes */ int count; }; endif //List.cpp I only included the parts of List.cpp that might be the issue include "List.h" include include using namespace std; List::List() { //List::size = NULL; head = NULL; tail = NULL; } List::~List() { Node* current; while(head != NULL) { current = head- next; delete current-previous; if (current-next!=NULL) { head = current; } else { delete current; } } } void List::add(int element) { Node* newNode; Node* current; newNode-element = element; if(newNode-element head-element) { current = head-next; } else { head-previous = newNode; newNode-next = head; newNode-previous = NULL; return; } while(newNode-element current-element) { current = current-next; } if(newNode-element <= current-element) { newNode-previous = current-previous; newNode-next = current; } } //main.cpp include "List.h" include include using namespace std; //void add(int element); int main (char** argv, int argc) { List* MyList = new List(); bool quit = false; string value; int element; while(quit==false) { cinvalue; if(value == "add") { cinelement; MyList-add(element); } if(value=="quit") { quit = true; } } return 0; } I'm doing everything I think I'm suppose to be doing. main.cpp isn't complete yet, just trying to get the add function to work first. Any help will be greatly appreciated.

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  • Where is the mistake ?

    - by mr.bio
    Hi ... i am implementing a simple linked list in c++. I have a mistake and i don't see it :( #include <stdexcept> #include <iostream> struct Node { Node(Node *next, int value): next(next), value(value) { } Node *next; int value; }; class List { Node *first; // Erstes Element , 0 falls die Liste leer ist int len; // Laenge der liste Node *nthNode(int index); // Hilfsfunktion : O( index ) public: // Default - Konstruktor ( Laenge 0): O (1) List():first(0),len(0){ } // Copy - Konstruktor : O(other.len) List(const List & other){ }; // Zuweisungs - Operator O(len +other.len) List &operator=(const List &other) { clear(); if(!other.len) return *this; Node *it = first = new Node(0,other.first->value); for (Node *n = other.first->next; n; n = n->next) { it = it->next = new Node(0, n->value); } len = other.len; return *this; } // Destruktor ( gibt den Speicher von allen Nodes frei ): O( len ) ~List(){ }; // Haengt der Liste ein Element hinten an: O( len ) void push_back(int value){ }; // Fuegt am Anfang der Liste ein Element ein : O (1) void push_front(int value){ Node* front = new Node(0,value); if(first){ first = front; front->next = 0; }else{ front->next = first; first = front; } len++; }; // gibt eine Referenz auf das index -te Element zurueck : O( index ) int &at(int index){ int count = 0 ; int ret ; Node *it = first; for (Node *n = first->next; n; n = n->next) { if(count==index) ret = n->value; count++; } return ret ; }; // Entfernt alle Elemente : O(len) void clear(){ }; // Zeigt alle Elemente an: hier : O( len * len ) void show() { std::cout << " List [" << len << " ]:{ "; for (int i = 0; i < len; ++i) { std::cout << at(i) << (i == len - 1 ? '}' : ','); } std::cout << std::endl; } }; /* * */ int main() { List l; // l. push_back(1); // l. push_back(2); l. push_front(7); l. push_front(8); l. push_front(9); l.show(); // List(l). show(); } it works ... but the output is : List [3 ]:{ 0,134520896,9484585}

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  • SQL2k8 T-SQL: Output into XML file

    - by Nai
    I have two tables Table Name: Graph UID1 UID2 ----------- 12 23 12 32 41 51 32 41 Table Name: Profiles NodeID UID Name ----------------- 1 12 Robs 2 23 Jones 3 32 Lim 4 41 Teo 5 51 Zacks I want to get an xml file like this: <graph directed="0"> <node id="1"> <att name="UID" value="12"/> <att name="Name" value="Robs"/> </node> <node id="2"> <att name="UID" value="23"/> <att name="Name" value="Jones"/> </node> <node id="3"> <att name="UID" value="32"/> <att name="Name" value="Lim"/> </node> <node id="4"> <att name="UID" value="41"/> <att name="Name" value="Teo"/> </node> <node id="5"> <att name="UID" value="51"/> <att name="Name" value="Zacks"/> </node> <edge source="12" target="23" /> <edge source="12" target="32" /> <edge source="41" target="51" /> <edge source="32" target="41" /> </graph> Thanks very much!

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  • After drawing circles on C# form how can i know on what circle i clicked?

    - by SorinA.
    I have to represent graphically an oriented graph like in the image below. i have a C# form, when i click with the mouse on it i have to draw a node. If i click somewhere on the form where is not already a node drawn it means i cliked with the intetion of drawing a node, if it is a node there i must select it and memorize it. On the next mouse click if i touch a place where there is not already a node drawn it means like before that i want to draw a new node, if it is a node where i clicked i need to draw the line from the first memorized node to the selected one and add road cost details. i know how to draw the circles that represent the nodes of the graph when i click on the form. i'm using the following code: namespace RepGraficaAUnuiGraf { public partial class Form1 : Form { Graphics graphDrawingArea; Bitmap bmpDrawingArea; Graphics graph; public Form1() { InitializeComponent(); } private void Form1_Load(object sender, EventArgs e) { bmpDrawingArea = new Bitmap(Width, Height); graphDrawingArea = Graphics.FromImage(bmpDrawingArea); graph = Graphics.FromHwnd(this.Handle); } private void Form1_Click(object sender, EventArgs e) { DrawCentralCircle(((MouseEventArgs)e).X, ((MouseEventArgs)e).Y, 15); graph.DrawImage(bmpDrawingArea, 0, 0); } void DrawCentralCircle(int CenterX, int CenterY, int Radius) { int start = CenterX - Radius; int end = CenterY - Radius; int diam = Radius * 2; bmpDrawingArea = new Bitmap(Width, Height); graphDrawingArea = Graphics.FromImage(bmpDrawingArea); graphDrawingArea.DrawEllipse(new Pen(Color.Blue), start, end, diam, diam); graphDrawingArea.DrawString("1", new Font("Tahoma", 13), Brushes.Black, new PointF(CenterX - 8, CenterY - 10)); } } } My question is how can i find out if at the coordinates (x,y) on my form i drew a node and which one is it? I thought of representing the nodes as buttons, having a tag or something similar as the node number(which in drawing should be 1 for Santa Barbara, 2 for Barstow etc.)

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  • pointer and reference question (linked lists)

    - by sil3nt
    Hi there, I have the following code struct Node { int accnumber; float balance; Node *next; }; Node *A, *B; int main() { A = NULL; B = NULL; AddNode(A, 123, 99.87); AddNode(B, 789, 52.64); etc… } void AddNode(Node * & listpointer, int a, float b) { // add a new node to the FRONT of the list Node *temp; temp = new Node; temp->accnumber = a; temp->balance = b; temp->next = listpointer; listpointer = temp; } in this here void AddNode(Node * & listpointer, int a, float b) { what does *& listpointer mean exactly.

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  • Can I use relative XPath expressions in libxml2?

    - by brbr
    I am wondering whether it is possible to use relative XPath expressions in libxml2. This is from the javax.xml.xpath API and I would like to do the similar thing using libxml2: Node widgetNode = (Node) xpath.evaluate(expression, document, XPathConstants.NODE); With a reference to the element, a relative XPath expression can now written to select the child element: XPath xpath = XPathFactory.newInstance().newXPath(); String expression = "manufacturer"; Node manufacturerNode = (Node) xpath.evaluate(expression, **widgetNode**, XPathConstants.NODE);

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  • What is the most efficient/elegant way to parse a flat table into a tree?

    - by Tomalak
    Assume you have a flat table that stores an ordered tree hierarchy: Id Name ParentId Order 1 'Node 1' 0 10 2 'Node 1.1' 1 10 3 'Node 2' 0 20 4 'Node 1.1.1' 2 10 5 'Node 2.1' 3 10 6 'Node 1.2' 1 20 What minimalistic approach would you use to output that to HTML (or text, for that matter) as a correctly ordered, correctly intended tree? Assume further you only have basic data structures (arrays and hashmaps), no fancy objects with parent/children references, no ORM, no framework, just your two hands. The table is represented as a result set, which can be accessed randomly. Pseudo code or plain English is okay, this is purely a conceptional question. Bonus question: Is there a fundamentally better way to store a tree structure like this in a RDBMS? EDITS AND ADDITIONS To answer one commenter's (Mark Bessey's) question: A root node is not necessary, because it is never going to be displayed anyway. ParentId = 0 is the convention to express "these are top level". The Order column defines how nodes with the same parent are going to be sorted. The "result set" I spoke of can be pictured as an array of hashmaps (to stay in that terminology). For my example was meant to be already there. Some answers go the extra mile and construct it first, but thats okay. The tree can be arbitrarily deep. Each node can have N children. I did not exactly have a "millions of entries" tree in mind, though. Don't mistake my choice of node naming ('Node 1.1.1') for something to rely on. The nodes could equally well be called 'Frank' or 'Bob', no naming structure is implied, this was merely to make it readable. I have posted my own solution so you guys can pull it to pieces.

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  • How do I detect a loop in this linked list?

    - by jjujuma
    Say you have a linked list structure in Java. It's made up of Nodes: class Node { Node next; // some user data } and each Node points to the next node, except for the last Node, which has null for next. Say there is a possibility that the list can contain a loop - i.e. the final Node, instead of having a null, has a reference to one of the nodes in the list which came before it. What's the best way of writing boolean hasLoop(Node first) which would return true if the given Node is the first of a list with a loop, and false otherwise? How could you write so that it takes a finite amount of space and a reasonable amount of time?

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  • Initialisation of Objects Syntax question

    - by Brock Woolf
    When I initialise a struct in C (Node is the struct): struct Node { /* Non-Relevant code */ }; This works: Node *rootNode = new Node(); but so does this: Node *rootNode = new Node; Is there a difference, and what is the difference between using () or not using the brackets? Just off memory, I think the same applies above for C++ object initialisations. What is happening here?

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  • Queue Data structure app crash with front() method

    - by Programer
    I am implementing queue data strcutre but my app gets crashed, I know I am doing something wrong with Node pointer front or front() method of queue class #include <iostream> using namespace std; class Node { public: int get() { return object; }; void set(int object) { this->object = object; }; Node * getNext() { return nextNode; }; void setNext(Node * nextNode) { this->nextNode = nextNode; }; private: int object; Node * nextNode; }; class queue{ private: Node *rear; Node *front; public: int dequeue() { int x = front->get(); Node* p = front; front = front->getNext(); delete p; return x; } void enqueue(int x) { Node* newNode = new Node(); newNode->set(x); newNode->setNext(NULL); rear->setNext(newNode); rear = newNode; } int Front() { return front->get(); } int isEmpty() { return ( front == NULL ); } }; main() { queue q; q.enqueue(2); cout<<q.Front(); system("pause"); }

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  • CSS style refresh in IE after dynamic removal of style link

    - by rybz
    Hi! I've got a problem with the dynamic style manipulation in IE7 (IE8 is fine). Using javascript I need to add and remove the < link / node with the definition of css file. Adding and removing the node as a child of < head / works fine under Firefox. Unfortunately, after removing it in the IE, although The tag is removed properly, the page style does not refresh. In the example below a simple css (makes background green) is appended and removed. After the removal in FF the background turns default, but in IE stays green. index.html <html> <head> </head> <script language="javascript" type="text/javascript"> var node; function append(){ var headID = document.getElementsByTagName("head")[0]; node = document.createElement('link'); node.type = 'text/css'; node.rel = 'stylesheet'; node.href = "s.css"; node.media = 'screen'; headID.appendChild(node); } function remove(){ var headID = document.getElementsByTagName("head")[0]; headID.removeChild(node); } </script> <body> <div onClick="append();"> add </div> <div onClick="remove();"> remove </div> </body> </html> And the style sheet: s.css body { background-color:#00CC33 } Here is the live example: http://rlab.pl/dynamic-style/ Is there a way to get it working?

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  • Skip first entry in Drupal Views Query?

    - by RD
    I've created a view that selects all nodes of type "shoot". But I want it to select all nodes of type "shoot", EXCEPT for the first entry. So, if the normal result is: Node 1 Node 2 Node 3 I want Node 2 Node 3 Node 4 Is that possible?

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  • Drupal Ubercart: error in passing values back to the Content Type after checkout

    - by user512826
    I am trying to set up event registration in a drupal site using Ubercart + the UC Node Checkout Module. I have followed the instructions provided in http://drupaleasy.com/blogs/ultimike/2009/03/event-registration-ubercart. However I seem to be unable to pass the Order ID and Payment Status back to the node. I have created a conditional action that on node checkout executes the following PHP code: I am using the following code to update the node on checkout - but nothing happens: if (isset($order)) { foreach ($order->products as $product) { if (isset($product->data['node_checkout_nid'])) { $node = node_load($product->data['node_checkout_nid']); $node->field_status['0']['value'] = 1; $node->field_orderid['0']['value'] = $order->order_id; node_save($node); } } } I know the conditional action is working because it prints dsm('hello world') messages on node checkout - however when I include a dsm($node) or dsm($product) in the PHP code, they return blank. Also when I go back to my product and click the 'Devel' tab, the 'data' string contains the following characters: a:1:{s:13:"form_build_id";s:37:"form-3ccc03345f4832c69666a89c560de940";} In this link http://www.ubercart.org/forum/support/10951/node_checkout_issue I found someone else with the same issue, but I have been unable to replicate his solution. Can anybody please help? Thanks so much!

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  • Jquery nested div .click event

    - by Sylph
    The link #loadContent will loads tree.html. Upon success loading the content, the script reinitialize some functions which is in tree.html. However, I am unable to get the .click event to function in the loaded content. Index.html <a href="#" id="loadContent">Load</a> <script type="text/javascript"> $(function() { $("#loadContent").click(function() { $.ajax({ url: "tree.html" ,success: function(data) { $('#result').html(data); $("#demo_1").tree({ rules : { use_max_children : false, use_max_depth : false }, callback : { onmove : function (NODE,REF_NODE,TYPE,TREE_OBJ,RB) { alert(TREE_OBJ.get_text(NODE) + " " + TYPE + " " + TREE_OBJ.get_text(REF_NODE)); } } }); } }); }); }); </script> <script type="text/javascript" class="source"> $(function () { $.tree.drag_start = function () { $("#log").append("<br />Drag start "); }; $.tree.drag = function () { $("#log").append(" ."); }; $.tree.drag_end = function () { $("#log").append(" Drag end<br />"); }; $("#demo_1").tree({ rules : { use_max_children : false, use_max_depth : false }, callback : { onmove : function (NODE,REF_NODE,TYPE,TREE_OBJ,RB) { alert(TREE_OBJ.get_text(NODE) + " " + TYPE + " " + TREE_OBJ.get_text(REF_NODE)); } } }); $("#demo_2").tree({ rules : { use_max_children : false, use_max_depth : false }, callback : { onmove : function (NODE,REF_NODE,TYPE,TREE_OBJ,RB) { alert(TREE_OBJ.get_text(NODE) + " " + TYPE + " " + TREE_OBJ.get_text(REF_NODE)); } } }); }); </script> <div class="demo" id="demo_2"> <ul> <li id="phtml_1" class="open"><a href="#"><ins>&nbsp;</ins>Root node 1</a> <ul> <li id="phtml_2"><a href="#"><ins>&nbsp;</ins>Child node 1</a></li> <li id="phtml_3"><a href="#"><ins>&nbsp;</ins>Child node 2</a></li> <li id="phtml_4"><a href="#"><ins>&nbsp;</ins>Some other child node with longer text</a></li> </ul> </li> <li id="phtml_5"><a href="#"><ins>&nbsp;</ins>Root node 2</a></li> </ul> </div> <div id="result"></div><br> <div id="log"></div> Tree.html <div class="demo" id="demo_1"> <ul> <li id="phtml_1" class="open"><a href="#"><ins>&nbsp;</ins>Root node 1</a> <ul> <li id="phtml_2"><a href="#"><ins>&nbsp;</ins>Child node 1</a></li> <li id="phtml_3"><a href="#"><ins>&nbsp;</ins>Child node 2</a></li> <li id="phtml_4"><a href="#"><ins>&nbsp;</ins>Some other child node with longer text</a></li> </ul> </li> <li id="phtml_5"><a href="#"><ins>&nbsp;</ins>Root node 2</a></li> <li><a class="preset_text" id="1">model 1</a> </li> <li><a class="preset_text" id="2">model 2</a></li> </ul> </div> <script type="text/javascript"> $(document).ready(function() { $('.preset_text').click(function(){ var target = $(this).attr("id"); alert(target); }); }); </script> In tree.html, I am unable to get the alert(target). However, If I moved this section out from the "div #demo_1" in tree.html, I am able to receive alert(target). <a class="preset_text" id="1">model 1</a> <a class="preset_text" id="2">model 2</a> How can I get to detect the item clicked in the div demo_1 ? Thanks

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  • what is mistakes/errors in this code c++ tell me the correction ??

    - by jeje
    hello all here in this code the compiler print error : 132 C:.... `createlist' undeclared (first use this function) (Each undeclared identifier is reported only once for each function it appears in.) and repeat it again in all calls in main function :( what's the problem ?? plzzzz help me #include<iostream> #include<string> using namespace std; template <typename T> struct Node { T num; struct Node<T> *next; // to craet list nodes void createlist(Node<T> *p) { T data; for( ; ; ) // its containue until user want to stop { cout<<"enter data number or '#' to stop\n"; cin>>data; if(data == '#') { p->next =NULL; break; } else { p->num= data; p->next = new Node<T>; p=p->next; } } } //count list to use it in sort function int countlist (Node<T> *p) { int count=0; while(p->next != NULL) { count++; p=p->next; } return count; } // sort list void sort( Node<T> *p) { Node<T> *p1, *p2; //element 1 & 2 to compare between them int i, j , n; T temp; n= countlist(p); for( i=1; i<n ; i++) { // here every loop time we put the first element in list in p1 and the second in p2 p1=p; p2=p->next; for(j=1; j<=(n-i) ; j++) { if( p1->num > p2->num) { temp=p2->num; p2->num=p1->num; p1->num=temp; } } p1= p1->next; p2= p2->next; } } //add new number in any location the user choose void insertatloc(Node<T> *p) { T n; //read new num int loc; //read the choosen location Node<T> *locadd, *newnum, *temp; cout <<" enter location you want ..! \n"; cin>>loc; locadd=NULL; //make it null to checked if there is location after read it from user ot not while(p->next !=NULL) { if( p->next==loc) { locadd=p; break; } p=p->next; } if (locadd==NULL) {cout<<" cannot find the location\n";} else //if location is right {cout<<" enter new number\n"; // new number to creat also new location for it cin>>n; newnum= new Node/*<T>*/; newnum->num=n; temp= locadd->next; locadd->next=newnum; newnum->next=temp; } locadd->num=sort(locadd); // call sort function } // display all list nodes void displaylist (Node<T> *p) { while (p->next != NULL) { cout<<" the list contain:\n"; cout<<p->num<<endl; p=p->next; } } };//end streuct int main() { cout<<"*** Welcome in Linked List Sheet 2****\n"; // defined pointer for structer Node // that value is the address of first node struct Node<int>*mynodes= new struct Node<int>; // create nodes in mynodes list cout<<"\nCreate nodes in list"; createlist(mynodes); // insert node in location insertatloc(mynodes); /* count the number of all nodes nodescount = countlist(mynodes); cout<<"\nThe number of nodes in list is: "<<nodescount;*/ // sort nodes in list sort(mynodes); // Display nodes cout<<"\nDisplay all nodes in list:\n"; displaylist(mynodes); system("pause"); return 0; }

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  • Understanding linear linked list

    - by ArtWorkAD
    Hi, I have some problems understanding the linear linked list data structure. This is how I define a list element: class Node{ Object data; Node link; public Node(Object pData, Node pLink){ this.data = pData; this.link = pLink; } } To keep it simple we say that a list are linked nodes so we do not need to define a class list (recursion principle). My problem is that I am really confused in understanding how nodes are connected, more precisely the sequence of the nodes when we connect them. Node n1 = new Node(new Integer(2), null); Node n2 = new Node(new Integer(1), n1); What is link? Is it the previous or the next element? Any other suggestions to help me understanding this data structure?

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  • adding elements in to the doubly linked list

    - by user329820
    Hi this is my code for main class and doubly linked class and node class but when I run the program ,in the concole will show this"datastructureproject.DoublyLinkedList@19ee1ac" instead of the random numbers .please help me thanks! main class: public class Main { public static int getRandomNumber(double min, double max) { Random random = new Random(); return (int) (random.nextDouble() * (max - min) + min); } public static void main(String[] args) { int j; int i = 0; i = getRandomNumber(10, 10000); DoublyLinkedList listOne = new DoublyLinkedList(); for (j = 0; j <= i / 2; j++) { listOne.add(getRandomNumber(10, 10000)); } System.out.println(listOne); } } doubly linked list class: public class DoublyLinkedList { private Node head ; private Node tail; private long size = 0; public DoublyLinkedList() { head= new Node(0, null, null); tail = new Node(0, head, null); } public void add(int i){ head.setValue(i); Node newNode = new Node(); head.setNext(newNode); newNode.setPrev(head); newNode = head; } } and the node class is like the class that you have seen before (Node prev,Node next,int value)

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  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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  • Understanding of a C Code Required..

    - by RBA
    int Size(struct node* node) { if(node == NULL) { return 0; } else if(node != NULL) { return (Size(node->left) + 1 + Size(node->right)); } } Hi, Can Anybody please post the stack trace for the following piece of code. Lets say if we insert the values 2, 1, 10, 5... Then what could be the stack representation during the recursion process.. Please, its very urgent, and its very confusing too...

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  • Haskell graph data type representation

    - by John Retallack
    I want to represent a graph in Haskell in the following manner: For each node I want to store it's value and a list of adjacent nodes,the problem which i'm having difficulties with is that I want the adjacent nodes to be stored as references to other nodes. For example: I want node ny to be stored as („NY“ (l p)) where l and p are adjacent nodes,and not as („NY“ („London“ „Paris“)). I tried something like this : data Node a = Node { value :: a , neighbors :: [Node a] }deriving (Show) let n1 = Node {value=1, neighbors=[n2]} let n2 = Node {value=1, neighbors=[n1 n3]} let n3 = Node {value=1, neighbors=[n2]} But i get en error in let,What am I doing wrong ?

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  • Following the Thread in OSB

    - by Antony Reynolds
    Threading in OSB The Scenario I recently led an OSB POC where we needed to get high throughput from an OSB pipeline that had the following logic: 1. Receive Request 2. Send Request to External System 3. If Response has a particular value   3.1 Modify Request   3.2 Resend Request to External System 4. Send Response back to Requestor All looks very straightforward and no nasty wrinkles along the way.  The flow was implemented in OSB as follows (see diagram for more details): Proxy Service to Receive Request and Send Response Request Pipeline   Copies Original Request for use in step 3 Route Node   Sends Request to External System exposed as a Business Service Response Pipeline   Checks Response to Check If Request Needs to Be Resubmitted Modify Request Callout to External System (same Business Service as Route Node) The Proxy and the Business Service were each assigned their own Work Manager, effectively giving each of them their own thread pool. The Surprise Imagine our surprise when, on stressing the system we saw it lock up, with large numbers of blocked threads.  The reason for the lock up is due to some subtleties in the OSB thread model which is the topic of this post.   Basic Thread Model OSB goes to great lengths to avoid holding on to threads.  Lets start by looking at how how OSB deals with a simple request/response routing to a business service in a route node. Most Business Services are implemented by OSB in two parts.  The first part uses the request thread to send the request to the target.  In the diagram this is represented by the thread T1.  After sending the request to the target (the Business Service in our diagram) the request thread is released back to whatever pool it came from.  A multiplexor (muxer) is used to wait for the response.  When the response is received the muxer hands off the response to a new thread that is used to execute the response pipeline, this is represented in the diagram by T2. OSB allows you to assign different Work Managers and hence different thread pools to each Proxy Service and Business Service.  In out example we have the “Proxy Service Work Manager” assigned to the Proxy Service and the “Business Service Work Manager” assigned to the Business Service.  Note that the Business Service Work Manager is only used to assign the thread to process the response, it is never used to process the request. This architecture means that while waiting for a response from a business service there are no threads in use, which makes for better scalability in terms of thread usage. First Wrinkle Note that if the Proxy and the Business Service both use the same Work Manager then there is potential for starvation.  For example: Request Pipeline makes a blocking callout, say to perform a database read. Business Service response tries to allocate a thread from thread pool but all threads are blocked in the database read. New requests arrive and contend with responses arriving for the available threads. Similar problems can occur if the response pipeline blocks for some reason, maybe a database update for example. Solution The solution to this is to make sure that the Proxy and Business Service use different Work Managers so that they do not contend with each other for threads. Do Nothing Route Thread Model So what happens if there is no route node?  In this case OSB just echoes the Request message as a Response message, but what happens to the threads?  OSB still uses a separate thread for the response, but in this case the Work Manager used is the Default Work Manager. So this is really a special case of the Basic Thread Model discussed above, except that the response pipeline will always execute on the Default Work Manager.   Proxy Chaining Thread Model So what happens when the route node is actually calling a Proxy Service rather than a Business Service, does the second Proxy Service use its own Thread or does it re-use the thread of the original Request Pipeline? Well as you can see from the diagram when a route node calls another proxy service then the original Work Manager is used for both request pipelines.  Similarly the response pipeline uses the Work Manager associated with the ultimate Business Service invoked via a Route Node.  This actually fits in with the earlier description I gave about Business Services and by extension Route Nodes they “… uses the request thread to send the request to the target”. Call Out Threading Model So what happens when you make a Service Callout to a Business Service from within a pipeline.  The documentation says that “The pipeline processor will block the thread until the response arrives asynchronously” when using a Service Callout.  What this means is that the target Business Service is called using the pipeline thread but the response is also handled by the pipeline thread.  This implies that the pipeline thread blocks waiting for a response.  It is the handling of this response that behaves in an unexpected way. When a Business Service is called via a Service Callout, the calling thread is suspended after sending the request, but unlike the Route Node case the thread is not released, it waits for the response.  The muxer uses the Business Service Work Manager to allocate a thread to process the response, but in this case processing the response means getting the response and notifying the blocked pipeline thread that the response is available.  The original pipeline thread can then continue to process the response. Second Wrinkle This leads to an unfortunate wrinkle.  If the Business Service is using the same Work Manager as the Pipeline then it is possible for starvation or a deadlock to occur.  The scenario is as follows: Pipeline makes a Callout and the thread is suspended but still allocated Multiple Pipeline instances using the same Work Manager are in this state (common for a system under load) Response comes back but all Work Manager threads are allocated to blocked pipelines. Response cannot be processed and so pipeline threads never unblock – deadlock! Solution The solution to this is to make sure that any Business Services used by a Callout in a pipeline use a different Work Manager to the pipeline itself. The Solution to My Problem Looking back at my original workflow we see that the same Business Service is called twice, once in a Routing Node and once in a Response Pipeline Callout.  This was what was causing my problem because the response pipeline was using the Business Service Work Manager, but the Service Callout wanted to use the same Work Manager to handle the responses and so eventually my Response Pipeline hogged all the available threads so no responses could be processed. The solution was to create a second Business Service pointing to the same location as the original Business Service, the only difference was to assign a different Work Manager to this Business Service.  This ensured that when the Service Callout completed there were always threads available to process the response because the response processing from the Service Callout had its own dedicated Work Manager. Summary Request Pipeline Executes on Proxy Work Manager (WM) Thread so limited by setting of that WM.  If no WM specified then uses WLS default WM. Route Node Request sent using Proxy WM Thread Proxy WM Thread is released before getting response Muxer is used to handle response Muxer hands off response to Business Service (BS) WM Response Pipeline Executes on Routed Business Service WM Thread so limited by setting of that WM.  If no WM specified then uses WLS default WM. No Route Node (Echo functionality) Proxy WM thread released New thread from the default WM used for response pipeline Service Callout Request sent using proxy pipeline thread Proxy thread is suspended (not released) until the response comes back Notification of response handled by BS WM thread so limited by setting of that WM.  If no WM specified then uses WLS default WM. Note this is a very short lived use of the thread After notification by callout BS WM thread that thread is released and execution continues on the original pipeline thread. Route/Callout to Proxy Service Request Pipeline of callee executes on requestor thread Response Pipeline of caller executes on response thread of requested proxy Throttling Request message may be queued if limit reached. Requesting thread is released (route node) or suspended (callout) So what this means is that you may get deadlocks caused by thread starvation if you use the same thread pool for the business service in a route node and the business service in a callout from the response pipeline because the callout will need a notification thread from the same thread pool as the response pipeline.  This was the problem we were having. You get a similar problem if you use the same work manager for the proxy request pipeline and a business service callout from that request pipeline. It also means you may want to have different work managers for the proxy and business service in the route node. Basically you need to think carefully about how threading impacts your proxy services. References Thanks to Jay Kasi, Gerald Nunn and Deb Ayers for helping to explain this to me.  Any errors are my own and not theirs.  Also thanks to my colleagues Milind Pandit and Prasad Bopardikar who travelled this road with me. OSB Thread Model Great Blog Post on Thread Usage in OSB

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