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  • Tree and List on the same page

    - by Jamie
    Hi All! Using Grails and the RichUI plugin to display a tree, and it works fine. When I click one of the Nodes in the tree I show a list(table) from a controller. I should be able to create new, edit and sort. My problem is that pagination doesn't work and also sorting!!! Are there anyone who has done this, or can it be done differently ? Best Regards Jamie

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  • Simple Fundamental Algorithm Question, binary tree traversal

    - by Ben
    I'm trying to explain to non-computer science major student with many questions. (1)What traverses tree? Is it just logic or actual on off switch generates 1s and 0s traveling the circuit board? where is this tree and node exists CPU/Memory in between? (2)If it is 1s and 0s HOW the circuits understand the line for example p=p.getLeft(); I said search the google or wiki.

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  • Recursive breadth first tree traversal

    - by dugogota
    I'm pulling my hair out trying to figure out how to implement breadth first tree traversal in scheme. I've done it in Java and C++. If I had code, I'd post it but I'm not sure how exactly to begin. Given the tree definition below, how to implement breadth first search using recursion? (define tree1 '( A ( B (C () ()) (D () ()) ) (E (F () ()) (G () ())) )) Any help, any, is greatly appreciated.

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  • 2 sided winform Tree - exist?

    - by Avi Harush
    Hi, I'm looking for a 2 sided winform Tree control. something like what you see in the math books. meaning that the tree can go both right and left in the control. Something like http://www.math.bas.bg/~nkirov/2010/NETB201/slides/ch06/pic3.jpg Thanks Avi

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  • C++ print out a binary search tree

    - by starcorn
    Hello, Got nothing better to do this Christmas holiday, so I decided to try out making a binary search tree. I'm stuck with the print function. How should the logic behind it work? Since the tree is already inserting it in a somewhat sorted order, and I want to print the tree from smallest values to the biggest. So I need to travel to the furthest left branch of the tree to print the first value. Right, so after that how do I remember the way back up, do I need to save the previous node? A search in wikipedia gave me an solution which they used stack. And other solutions I couldn't quite understand how they've made it, so I'm asking here instead hoping someone can enlight me. I also wonder my insert function is OK. I've seen other's solution being smaller. void treenode::insert(int i) { if(root == 0) { cout << "root" << endl; root = new node(i,root); } else { node* travel = root; node* prev; while(travel) { if(travel->value > i) { cout << "travel left" << endl; prev = travel; travel = travel->left; } else { cout << "travel right" << endl; prev = travel; travel = travel->right; } } //insert if(prev->value > i) { cout << "left" << endl; prev->left = new node(i); } else { cout << "right" << endl; prev->right = new node(i); } } } void treenode::print() { node* travel = root; while(travel) { cout << travel->value << endl; travel = travel->left; } }

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  • Javascript: getting element in dom tree when mouseover

    - by oimoim
    Hi, When using : document.onmouseover = function(e) {} Is there a property which gives me the element in the dom tree ? For example, I can set a style to e.srcElement But, how can I later access this element to (for example) reset its style ? And how can I know at which place in the dom tree it is ? I want to be able to situate it in the whole page dump. Many thanks.

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  • Algorithm for finding symmetries of a tree

    - by Paxinum
    I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them). The sectors live in the complex plane, and for n even, sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced. These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them). Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree. Now, given such a representation, how can I see if it is symmetric wrt the real axis? For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line, so it is symmetric wrt the real axis. If the distances between (015) and (1245) is equal to distance from (1245) to (234), this is also symmetric wrt the imaginary axis. The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal. Edit: This is actually for my research. I have posted the question at mathoverflow as well, but my days in competition programming tells me that this is more like an IOI task. Code in mathematica would be excellent, but java, python, or any other language readable by a human suffices. Here are some examples (pretend the double edges are single and we have a tree) http://www2.math.su.se/~per/files.php?file=contr_ex_1.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_2.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_5.pdf Example 1 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,3). Example 2 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,1). Example 5 is described as (0,1,4,5)(1,2,3,4) with distances (2). So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry. (These symmetries corresponds to special kinds of potential in the Schroedinger equation, which has nice properties in quantum mechanics.)

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  • Efficient Array Storage for Binary Tree

    - by Sundararajan S
    We have to write the nodes of a binary tree to a file. What is the most space efficient way of writing a binary tree . We can store it in array format with parent in position 'i' and its childs in 2i,2i+1. But this will waste lot of space in case of sparse binary trees.

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  • question about tree

    - by davit-datuashvili
    i have question for example i want to implement binary tree with array i want understand what will indexes for left child and rigth child ?my array is 0 based i want implement searching in tree using array can anybody help me?

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  • What tool to use to draw file tree diagram

    - by Michael
    Given a file tree - a directory with directories in it etc, what software would you recommend to create a diagram of the file-tree as a graphic file that I can embed in a word processor document I prefer vector (SVG, EPS, EMF...) files. The tool must run on Windows, but preferably cross-platform. The tool may be commercial but preferably free.

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  • Looking for a good Python Tree data structure

    - by morpheous
    I am looking for a good Tree data structure class. I have come across this package, but since I am relatively new to Python (not programming), I dont know if there are any better ones out there. I'd like to hear from the Pythonistas on here - do you have a favorite tree script that you regularly use and would recommend?

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  • Mutating the expression tree of a predicate to target another type

    - by Jon
    Intro In the application I 'm currently working on, there are two kinds of each business object: the "ActiveRecord" type, and the "DataContract" type. So for example, we have: namespace ActiveRecord { class Widget { public int Id { get; set; } } } namespace DataContracts { class Widget { public int Id { get; set; } } } The database access layer takes care of "translating" between hierarchies: you can tell it to update a DataContracts.Widget, and it will magically create an ActiveRecord.Widget with the same property values and save that. The problem I have surfaced when attempting to refactor this database access layer. The Problem I want to add methods like the following to the database access layer: // Widget is DataContract.Widget interface DbAccessLayer { IEnumerable<Widget> GetMany(Expression<Func<Widget, bool>> predicate); } The above is a simple general-use "get" method with custom predicate. The only point of interest is that I 'm not passing in an anonymous function but rather an expression tree. This is done because inside DbAccessLayer we have to query ActiveRecord.Widget efficiently (LINQ to SQL) and not have the database return all ActiveRecord.Widget instances and then filter the enumerable collection. We need to pass in an expression tree, so we ask for one as the parameter for GetMany. The snag: the parameter we have needs to be magically transformed from an Expression<Func<DataContract.Widget, bool>> to an Expression<Func<ActiveRecord.Widget, bool>>. This is where I haven't managed to pull it off... Attempted Solution What we 'd like to do inside GetMany is: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( predicate.Body, predicate.Parameters); // use lambda to query ActiveRecord.Widget and return some value } This won't work because in a typical scenario, for example if: predicate == w => w.Id == 0; ...the expression tree contains a MemberAccessExpression instance which has a MemberInfo property (named Member) that point to members of DataContract.Widget. There are also ParameterExpression instances both in the expression tree and in its parameter expression collection (predicate.Parameters); After searching a bit, I found System.Linq.Expressions.ExpressionVisitor (its source can be found here in the context of a how-to, very helpful) which is a convenient way to modify an expression tree. Armed with this, I implemented a visitor. This simple visitor only takes care of changing the types in member access and parameter expressions. It may not be complete, but it's fine for the expression w => w.Id == 0. internal class Visitor : ExpressionVisitor { private readonly Func<Type, Type> dataContractToActiveRecordTypeConverter; public Visitor(Func<Type, Type> dataContractToActiveRecordTypeConverter) { this.dataContractToActiveRecordTypeConverter = dataContractToActiveRecordTypeConverter; } protected override Expression VisitMember(MemberExpression node) { var dataContractType = node.Member.ReflectedType; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); var converted = Expression.MakeMemberAccess( base.Visit(node.Expression), activeRecordType.GetProperty(node.Member.Name)); return converted; } protected override Expression VisitParameter(ParameterExpression node) { var dataContractType = node.Type; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); return Expression.Parameter(activeRecordType, node.Name); } } With this visitor, GetMany becomes: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var visitor = new Visitor(...); var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( visitor.Visit(predicate.Body), predicate.Parameters.Select(p => visitor.Visit(p)); var widgets = ActiveRecord.Widget.Repository().Where(lambda); // This is just for reference, see below Expression<Func<ActiveRecord.Widget, bool>> referenceLambda = w => w.Id == 0; // Here we 'd convert the widgets to instances of DataContract.Widget and // return them -- this has nothing to do with the question though. } Results The good news is that lambda is constructed just fine. The bad news is that it isn't working; it's blowing up on me when I try to use it (the exception messages are really not helpful at all). I have examined the lambda my code produces and a hardcoded lambda with the same expression; they look exactly the same. I spent hours in the debugger trying to find some difference, but I can't. When predicate is w => w.Id == 0, lambda looks exactly like referenceLambda. But the latter works with e.g. IQueryable<T>.Where, while the former does not (I have tried this in the immediate window of the debugger). I should also mention that when predicate is w => true, it all works just fine. Therefore I am assuming that I 'm not doing enough work in Visitor, but I can't find any more leads to follow on. Can someone point me in the right direction? Thanks in advance for your help!

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  • How to create a new widget for dojox.grid.cells.dijit?

    - by the_drow
    I am trying to create a button widget for dojox.grid. My problems are: 1) The button is only shown when I double click the grid. 2) I can't figure out how to set attributes through declarative markup. It seems that the markupFactory function is responsible for it but it doesn't set the widget's label. The following code demonstrates what I've got so far: dojo.require("dojox.grid.DataGrid"); dojo.require("dojo.data.ItemFileWriteStore"); dojo.require("dijit.form.Button"); dojo.require("dojox.grid.cells.dijit"); dojo.require("dojo.parser"); dojo.declare("dojox.grid.cells.Button", dojox.grid.cells._Widget, { widgetClass: dijit.form.Button, alwaysEditing: true, constructor: function(inCell) { this.inherited(arguments); this.widget = new dijit.form.Button; }, setValue: function(inRowIndex, inValue){ if (this.widget) { this.widget.attr('value', inValue); } else { this.inherited(arguments); } } }); dojox.grid.cells.Button.markupFactory = function(node, cell) { dojox.grid.cells._Widget.markupFactory(node, cell); }

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  • Is there a programming language with not a tree but tags idea behind OOP?

    - by kolupaev
    I'm thinking about tree structures, and I feel that I don't like them. It's like when you have a shop, then you try to put all products to tree-like catalog, and then you need to place one product to multiple categories, now you have multiple routing, bla-bla. I don't feel like everything in the world could be put to a tree. Instead, I like idea of tags. I would like to store everything with tags. With tags I could do much more. I can even simulate trees if I want. I want to have tag-based filesystem! But hey - modern OOP paradigm with inheritance is based on tree. I want to see how it is when you don't have such basement. Closest thing I found is mixins in some languages. Do you know what else is also about this ideas?

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  • Why create a Huffman tree per character instead of a Node?

    - by Omega
    For a school assignment we're supposed to make a Java implementation of a compressor/decompresser using Huffman's algorithm. I've been reading a bit about it, specially this C++ tutorial: http://www.cprogramming.com/tutorial/computersciencetheory/huffman.html In my program, I've been thinking about having Nodes that have the following properties: Total Frequency Character (if a leaf) Right child (if any) Left child (if any) Parent (if any) So when building the Huffman tree, it is just a matter of linking a node to others, etc. However, I'm a bit confused with the following quote (emphasis mine): First, every letter starts off as part of its own tree and the trees are ordered by the frequency of the letters in the original string. Then the two least-frequently used letters are combined into a single tree, and the frequency of that tree is set to be the combined frequency of the two trees that it links together. My question: why should I create a tree per letter, instead of just a node per letter and then do the linking later? I have not begun coding, I'm just studying the algorithm first, so I guess I'm missing an important detail. What is it?

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  • Why don't we store the syntax tree instead of the source code?

    - by Calmarius
    We have a lot of programming languages. Every language is parsed and syntax checked before translated into code so an abstract syntax tree is built. We have this abstract syntax tree, why don't we store this syntax tree instead of the source code (or next to the source code)? By using an AST instead of the source code. Every programmer in a team can serialize this tree to any language, they want (with the appropriate context free grammar) and parse back to AST when they finished. So this would eliminate the debate about the coding style questions (where to put the { and }, where to put whitespace, indentation, etc.) What are the pros and cons of this approach?

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  • dojox.enhancedGrid get Selected Row

    - by user256007
    How can I get the Selected Row Object of dojox.enhancedGrid ? I am using selectionMode: 'single' e.g. with Radio Buttons. dijit.byId("gridViewWidget").selection.selectedIndex Returns the rowIndex. But how to get the rowObject of that Index ? I can get the rowNode()But What I need is value of the id column of that Row. Its possible to travarse the HTML DOM returned by rowNode() But Is theer any straight forward way ?

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  • Removing duplicates from a Dojo FilteringSelect

    - by chrismacp
    Hi, I'm trying to remove duplicates from a Dojo FilteringSelect without changing the contents of the attached itemFileReadStore data store. I can't seem to find any information on how this is done, if it is indeed possible. I'm thinking I may have to extend the FilteringSelect Dijit and provide the functionality myself but I'm hoping to not have to. The reason I'm trying to remove duplicates with the FilteringSelect and not the data store is because I'm using the same data store with three instances of the FitleringSelect, each displaying different values from each row of the store.

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  • Dojo Select widget validation style

    - by jamstooks
    Many dojo form widgets, like DateTextBox will style themselves in red (claro theme) with an "!" to indicate that the field isn't valid after focus. I can't seem to get this type of styling to work with the Select widget. I have the value set to required, but if the user leaves it in the default state, blank, then it remains unstyled. I tried to test this programmatically: wigs = dijit.findWidgets(node); for( i = 0; i < wigs.length; i++ ) { wigs[i].focus(); } This triggers validation on several other widgets and they are styled appropriately, but not he Select widgets. Anyone know how to make it obvious that those select widgets are not validating? Oh, and wigs[i].isValid() is definitely false for those select widgets.

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  • Java+DOM: How do I convert a DOM tree without namespaces to a namespace-aware DOM tree?

    - by java.is.for.desktop
    Hello, everyone! I receive a Document (DOM tree) from a certain API (not in JDK). Sadly, this Document is not namespace-aware. As far as I know DOM, once generated, namespace-awareness can't be "added" afterwards. When converting this Document using a Transformer to a string, the XML is correct. Elements have xmlns:... attributes and name prefixes. But from the DOM point of view, there are no namespaces and no prefixes. I need to be able to convert this Document into a new Document which is namespace-aware. Yes, I could do this by just converting it to a string and back to DOM with namespaces enabled. But: nodes of the original tree have user-objects set. Converting to string and back would make a mapping of these user-objects to the new Document very complicated, if not impossible. So I really need a way to convert non-namespace DOM to namespace DOM. Are there any more-or-less straightforward solutions for this? Worst case (what I'm hoping to avoid) would be to manually iterate through old Document tree and create new namespace-aware Node for each old Node. Doing so, one had to manually "parse" namespace prefixes, watch out for xmlns-attributes, and maintain a mapping between prefixes and namespace-URIs. Lots of things to go wrong.

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