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  • What Can Search Engine Optimization Do For Your Business?

    When you are first jumping into the business world it can be a bit confusing. There are many things that you need to learn. One of the first things is how to generate clients and sales. In order to do that you need to draw people to your website. The best way to do this is with search engine optimization.

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  • Where should you store variables for a search program in java?

    - by Bored915
    I'm wondering which is a more effective storage method in java. Would be better to save variables that will not change in a class or a resource? They're going to be variables that contain a set amount so that later a search program will go through the list of variables so as to recognize possible options. Also if there is a more efficient method of storing them please say so or if i doesn't matter where i store them.

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  • How Does Link Building Help in SEO (Search Engine Optimization) Rankings?

    Search engine optimization (SEO) is a widely used marketing tool to increase awareness of a particular company. Online businesses can range from small scale companies to large scale companies. Companies want to make sure that they can get their products or services to as many people that they can reach. The internet is one such tool that is used, and proven to be quite effective.

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  • Jquery hiding all descendents of a ul tag...showing child elements as tree menu...

    - by Ronedog
    I want to hide all the descendents of the "ul" for my tree menu when the page loads up, then as each "main" "li" link is clicked display the direct child, and if the direct child has children (grandchild), when the the "Child" is clicked I want it to show the "grandchild" elements. should be simple, but some how I screwed things up and when i click on the main "li" (Heading 1) it displays all of the descendents (Including the "Sub page A - 1"), instead of just the direct children ("Sub Page A"). Which I think means the children, grandchildren, etc. were never hidden to begin with with the .hide(). What I really want to happen is to hide all the descendents (except the main top-level headings) and as I walk down the tree display the children as needed. Any tips on how to make this work? Here's the HTML: <ul id="nav"> <li>Heading 1 <ul> <li>Sub page A <ul> <li>Sub page A - 1</li> <li>Sub page A - 3</li> <li>Sub page A - 2</li> </ul> </li> <li>Sub page B</li> <li>Sub page C</li> </ul> </li> <li>Heading 2 <ul> <li>Sub page D</li> <li>Sub page E</li> <li>Sub page F</li> </ul> </li> <li>Heading 3 <ul> <li>Sub page G</li> <li>Sub page H</li> <li>Sub page I</li> </ul> </li> Here's my Jquery: $(function(){ $('#nav ul').hide(); //Supposed to Hide all <ul> tags for all descendents, but doesn't work $('#nav>li').mouseover(function(){ $(this).addClass("a_hand") }); //Add the class that displays the hand $('#nav>li').toggle(function() { $(this).find('ul').slideDown(200); }, function() { $(this).find('ul').slideUp(200); });//END TOGGLE });//END MAIN FUNCTION thanks.

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  • Android Eclipse test projects cannot be used with a project being built in an Android build tree

    - by orospakr
    An Android Java project placed in a git repository and built in an Android tree in /packages/apps needs to have the project files located at the root of the git repository. This is problematic for creating a complementary Test project, which should ideally be included in the same git repository so commits are atomic for both code and tests. Eclipse gets very unhappy if you include the Test project as a subdirectory. Is there an appropriate approach for dealing with this other than creating a second repository?

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  • How to write a recursive function that returns a linked list of nodes, when given a binary tree of n

    - by Jian Lin
    I was once asked of this in an interview: How to write a recursive function that returns a linked list of nodes, when given a binary tree of nodes? (flattening the data) For some reason, I tend to need more than 3 to 5 minutes to solve any recursive problem. Usually, 15 to 20 minutes will be more like it. How could we attack this problem, such as a very systematic way of reaching a solution, so that they can be solved in 3 to 5 minute time frame?

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  • Calculate the Hilbert value of a point for use in a Hilbert R-Tree?

    - by wrt
    I have an application where a Hilbert R-Tree (wikipedia) (citeseer) would seem to be an appropriate data structure. Specifically, it requires reasonably fast spatial queries over a data set that will experience a lot of updates. However, as far as I can see, none of the descriptions of the algorithms for this data structure even mention how to actually calculate the requisite Hilbert Value; which is the distance along a Hilbert Curve to the point. So any suggestions for how to go about calculating this?

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  • In Ant, copy all files from a tree of folders into a single folder?

    - by Sam Washburn
    Is it possible to use Ant to copy all the files (not folders) from a hierarchy of folders into one destination folder? For instance, I have a tree like this: res |-images | |-fg.png | +-bg.png +-sounds +-music.mp3 And I would like a result like this: data |-fg.png |-bg.png +-music.mp3 The way things are named, filename conflicts are not an issue. Is this possible to do with Ant?

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  • Where should I initialize variables for an OO Recursive Descent Parse Tree?

    - by Vasto
    I'd like to preface this by stating that this is for a class, so please don't solve this for me. One of my labs for my cse class is creating an interpreter for a BNF that was provided. I understand most of the concepts, but I'm trying to build up my tree and I'm unsure where to initialize values. I've tried in both the constructor, and in the methods but Eclipse's debugger still only shows the left branch, even though it runs through completely. Here is my main procedure so you can get an idea of how I'm calling the methods. public class Parser { public static void main(String[] args) throws IOException { FileTokenizer instance = FileTokenizer.Instance(); FileTokenizer.main(args); Prog prog = new Prog(); prog.ParseProg(); prog.PrintProg(); prog.ExecProg(); } Now here is My Prog class: public class Prog { private DeclSeq ds; private StmtSeq ss; Prog() { ds = new DeclSeq(); ss = new StmtSeq(); } public void ParseProg() { FileTokenizer instance = FileTokenizer.Instance(); instance.skipToken(); //Skips program (1) // ds = new DeclSeq(); ds.ParseDS(); instance.skipToken(); //Skips begin (2) // ss = new StmtSeq(); ss.ParseSS(); instance.skipToken(); } I've tried having Prog() { ds = null; ss = null; } public void ParseProg() { FileTokenizer instance = FileTokenizer.Instance(); instance.skipToken(); //Skips program (1) ds = new DeclSeq(); ds.ParseDS(); ... But it gave me the same error. I need the parse tree built up so I can do a pretty print and an execute command, but like I said, I only get the left branch. Any help would be appreciated. Explanations why are even more so appreciated. Thank you, Vasto

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  • C++ linked list based tree structure. Sanely move nodes between lists.

    - by krunk
    The requirements: Each Node in the list must contain a reference to its previous sibling Each Node in the list must contain a reference to its next sibling Each Node may have a list of child nodes Each child Node must have a reference to its parent node Basically what we have is a tree structure of arbitrary depth and length. Something like: -root(NULL) --Node1 ----ChildNode1 ------ChildOfChild --------AnotherChild ----ChildNode2 --Node2 ----ChildNode1 ------ChildOfChild ----ChildNode2 ------ChildOfChild --Node3 ----ChildNode1 ----ChildNode2 Given any individual node, you need to be able to either traverse its siblings. the children, or up the tree to the root node. A Node ends up looking something like this: class Node { Node* previoius; Node* next; Node* child; Node* parent; } I have a container class that stores these and provides STL iterators. It performs your typical linked list accessors. So insertAfter looks like: void insertAfter(Node* after, Node* newNode) { Node* next = after->next; after->next = newNode; newNode->previous = after; next->previous = newNode; newNode->next = next; newNode->parent = after->parent; } That's the setup, now for the question. How would one move a node (and its children etc) to another list without leaving the previous list dangling? For example, if Node* myNode exists in ListOne and I want to append it to listTwo. Using pointers, listOne is left with a hole in its list since the next and previous pointers are changed. One solution is pass by value of the appended Node. So our insertAfter method would become: void insertAfter(Node* after, Node newNode); This seems like an awkward syntax. Another option is doing the copying internally, so you'd have: void insertAfter(Node* after, const Node* newNode) { Node *new_node = new Node(*newNode); Node* next = after->next; after->next = new_node; new_node->previous = after; next->previous = new_node; new_node->next = next; new_node->parent = after->parent; } Finally, you might create a moveNode method for moving and prevent raw insertion or appending of a node that already has been assigned siblings and parents. // default pointer value is 0 in constructor and a operator bool(..) // is defined for the Node bool isInList(const Node* node) const { return (node->previous || node->next || node->parent); } // then in insertAfter and friends if(isInList(newNode) // throw some error and bail I thought I'd toss this out there and see what folks came up with.

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