Search Results

Search found 10447 results on 418 pages for 'keyboard layout'.

Page 142/418 | < Previous Page | 138 139 140 141 142 143 144 145 146 147 148 149  | Next Page >

  • How can I recreate the Evernote 5 iOS tabbed layout using CSS3 an jQuery?

    - by Ismailp
    Looking for a tutorial on how I could make a web interface similar to the Evernote 5 iOS tabbed layout, using CSS3 and jQuery?! Here is a link showing the tabs: http://www.google.se/search?q=evernote+5+tabbed&hl=en&tbo=d&source=lnms&tbm=isch&sa=X&ei=wxWoUKPxF-TO4QTQk4DwDQ&ved=0CAgQ_AUoAQ&biw=320&bih=416#i=8 Do you guys know any good tutorials/resources for this? All help appreciated! Thanks in advance

    Read the article

  • Drupal: Can Book layout menu be added and expanded in the primary menu?

    - by kelly
    I'd like to take a book menu and just add it right to the primary links. Any way to do that? It can appear alreayd in the Navigation links but I'd like it to expand to deeper child levels. Also, I'm using a theme (Newswire) that creates a suckerfish menu from the primary links, so if I can automatically add my book pages to the primary links that would be ideal... Sample Book Layout

    Read the article

  • Is there any pros to duplicate browser/keyboard functionality?

    - by metal-gear-solid
    Is it good for user experience to duplicate browser/keyboard functionality? For example: to provide these links on a web-page. "Back to top" link "Print this page" link "Add to Favorite" link "Back" button/link "Text zoom" button Are they really create Site's usability and accessibility? How screen reader will behave these links, will these confuse to screen reader users?

    Read the article

  • how to center layout to vertical in android through java code?

    - by UMMA
    friends, i want to set android:layout_centerVertical="true" property of layout through java code of an image. can any one guide me how to achieve this. here is my code. RelativeLayout.LayoutParams params = new RelativeLayout.LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.WRAP_CONTENT); params.height = (int)totalHeight; img.setLayoutParams(params); i have tried using setScaleType(ScaleType.FIT_CENTER) but no use. any help would be appriciated.

    Read the article

  • Azure application working on emulator but not on azure cloud

    - by Hisham Riaz
    firstly i am developing my MVC3 application on visual web developer 2010 express, by migrating my MVC3 (cshtml) files on MVC2. it works great on local system using the emulator, but once i deploy the application on azure it gives runtime errors. example: The layout page "~/Views/Shared/test_page.cshtml" could not be found at the following path: "~/Views/Shared/test_page.cshtml". Source Error: Line 8: //Layout = "~/Views/Shared/upload.cshtml"; Line 9: //Layout = "~/Views/Shared/_Layout2.cshtml"; Line 10: Layout = "~/Views/Shared/test_page.cshtml"; Line 11: } Line 12: else CODE IS AS FOLLOWS: _ViewStart.cshtml file @{ string AccId = Request.QueryString["AccId"].ToString(); if (AccId=="0") { //Layout = "~/Views/Shared/upload.cshtml"; //Layout = "~/Views/Shared/_Layout2.cshtml"; Layout = "~/Views/Shared/test_page.cshtml"; } else { string LayOutPagePath = MVCTest.Models.ComponentClass.GetLayOutPagePath(AccId); Layout = LayOutPagePath; } } ......... how ever the page exist, and is working fine on azure emulator, but not in azure cloud. CODE FOR test_page.cshtml @{ var result = "1234567890"; var temp_xml = MVCTest.Models.ComponentClass.GetTemplateAndTheme("1");//returning xml string LayOutPagePath = MVCTest.Models.ComponentClass.GetLayOutPagePath("1");//returning string } @RenderBody() test_page @temp_xml @result @LayOutPagePath

    Read the article

  • Dojo DataGrid, programmatic creation

    - by djna
    I'm trying to create a DataGrid dynamically. Declarative creation is working fine: A source of data: <span dojoType="dojo.data.ItemFileReadStore" jsId="theStore" url="dataonly001.json"> </span> a simple layout: <script type="text/javascript" > var layout = [ { field: 'custId', name: 'Id', } // more items elided ... ]; </script> The grid: <body class="tundra" id="mainbody"> <div id="grid" dojoType="dojox.grid.DataGrid" store="theStore" structure="layout" autoWidth="true" ></div> </body> And I get the grid displayed nicely. Instead, I want to create the grid dynamically. For the sake of figuring out what's broken I've tried to use exactly the same layout and store, removing just teh grid declaration and adding this script: <script type="text/javascript" djConfig="parseOnLoad: true, debugAtAllCosts: true"> dojo.addOnLoad(function(){ // initialise and lay out home page console.log("Have a store:" + theStore); console.log("Have a structure:" + layout); var grid = new dojox.grid.DataGrid({ id:"grid", store : theStore, clientSort : "true", structure : layout }); grid.startup(); dojo.place(grid.domNode, dojo.body(), "first"); }); </script> The effect that I get is a completely empty rectangle is displayed. Using FireBug I can see that the DataGrid widget is created but that it has no rows or columns. So my guess is that the datastore or layout are not being passed correctly. However, it does appear that at the point of creation the values of theStore and layout are correct. Suggestions please, or indeed a working example of a programmic grid might solve the problem.

    Read the article

  • Dojo 1.8: Getting dest.appendChild is not a function while rendering a custom template

    - by Jim Pedid
    After adding in the WidgetsInTemplateMixin, I am receiving an error dest.appendChild is not a function In the documentation, it claims that there will be an error if this.containerNode is not able to contain any child objects. However, I have marked the containerNode attachment point for a div with dojo type "dijit/layout/ContentPane". Can anyone explain to me why this isn't working? Here is the Template file <div class="${baseClass}"> <div class="${baseClass}Container" data-dojo-type="dijit/layout/BorderContainer" data-dojo-props="design: 'headline'"> <div data-dojo-type="dijit/layout/ContentPane" data-dojo-props="region: 'top'"> Top </div> <div data-dojo-type="dijit/layout/ContentPane" data-dojo-props="region: 'center'" data-dojo-attach-point="containerNode"> </div> <div data-dojo-type="dijit/layout/ContentPane" data-dojo-props="region: 'leading', splitter: true"> Sidebar </div> </div> </div> Here is the javascript definition define([ "dojo/_base/declare", "dijit/_WidgetBase", "dijit/_OnDijitClickMixin", "dijit/layout/BorderContainer", "dijit/layout/ContentPane", "dijit/layout/TabContainer", "dijit/_TemplatedMixin", "dijit/_WidgetsInTemplateMixin", "dojo/text!./templates/MainContainer.html" ], function (declare, _WidgetBase, _OnDijitClickMixin, BorderContainer, ContentPane, TabContainer, _TemplatedMixin, _WidgetsInTemplateMixin, template) { return declare([_WidgetBase, _OnDijitClickMixin, _TemplatedMixin, _WidgetsInTemplateMixin], { templateString:template, baseClass:"main" }); }); The custom widget defined declaratively <div data-dojo-type="main/ui/MainContainer" data-dojo-props="title: 'Main Application'"> Hello Center! </div>

    Read the article

  • Qt support for VNC

    - by muchala123
    i want to test whether qt is supporting VNC or not. For that i have written a small layout program using Qt library. the source code for the layout program is as follows: layout.cpp #include <QApplication> #include <QHBoxLayout> #include <QSlider> #include <QSpinBox> int main(int argc, char *argv[]) { QApplication app(argc, argv); QWidget *window = new QWidget; window->setWindowTitle("Enter The Age of the person"); QSpinBox *spinBox = new QSpinBox; QSlider *slider = new QSlider(Qt::Horizontal); spinBox->setRange(0, 130); slider->setRange(0, 130); QObject::connect(spinBox, SIGNAL(valueChanged(int)), slider, SLOT(setValue(int))); QObject::connect(slider, SIGNAL(valueChanged(int)), spinBox, SLOT(setValue(int))); spinBox->setValue(35); QHBoxLayout *layout = new QHBoxLayout; layout->addWidget(spinBox); layout->addWidget(slider); window->setLayout(layout); window->show(); return app.exec(); } i want to run this as server application on my linux PC.For that what i configured Qt and installed like this. ./configure -qt-gfx-vnc make make install The program is working fine. But if i run the application as VNC server application like ./layout -qws -display VNC:0 i am encountering an error.it says that "_X11TransSocketINETConnect() can't get address for VNC:6000: Temporary failure in name resolution".. pls help me what i need to do. Thanks

    Read the article

< Previous Page | 138 139 140 141 142 143 144 145 146 147 148 149  | Next Page >