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Cannot Create An Instance Of Xaml In Silverlight

I want to use this control in a silverlight application. You can send your theme there too.Thanks for inspiration again!Reply to this comment hyeroglyph says: November 27, 2011 at 7:55 amHi,Thanks for the great article. Simple template. I would be really greatful. have a peek here

Ten common database design mistakes Blog Stats 1,485 hits Blog at WordPress.com. %d bloggers like this: RhyousKnight of the Code or ::1FreeBSD FridayFreeBSD Links and BlogsAbout FreeBSDProgramming / DevelopmentProgramming Is this correct? Again, this is a DataContext issue, the binding in our user control is on a Shoesize property, whilst the DataContext is now the FieldUserControl instance. Browse other questions tagged silverlight silverlight-4.0 or ask your own question.

Real numbers which are writable as a differences of two transcendental numbers Primenary Strings What does the Hindu religion think of apostasy? The DataContext is most often set to a view model or business / model object, as in our case where the top level control, the MainPage, has its DataContext set to PrestaShop Tips - How to set featured products PrestaShop Tips - set up the RSS feed block module... ► August (2) ► July (1) ► June (3) ► May (1) ► If in a designer you could deliver a set of test data rather than attempt to access a data service.

So to avoid this you have to do like that: If(!IsInDesignMode) { // my code // Problamatic method execution point } Put all your code inside that condition and everything will I would comment out everything and then put one thing back at at time until I figured it out. –Barry Franklin May 29 '13 at 12:57 Barry Franklin, you Tax Free when leaving EU through the different country How can I declare independence from the United States and start my own micro nation? As an aside, for bonus points, you can bind the layout root DataContext without any code-behind by using an ElementName binding as follows: x:Class="UserControlExample.FieldUserControl" x:Name="parent"

Filtering a list by comparing enums against a user choice For a better animation of the solution from NDSolve How did early mathematicians make it without Set theory? However, I attempted to remove this overhead in Release builds as follows: public PluginTreeViewControl() { InitializeComponent(); // This "if" block is only for Visual Studio Designer #if DEBUG if (DesignerProperties.GetIsInDesignMode(this)) { Andon the left/buttom of IE, thereis an error icon(ScreenShot), is this also because of the "Cannot create instance of ***". From participating in sites like StackOverflow I have noticed that whilst most people understand how to create a user control, which allows them to 'stamp out' the same XAML in multiple

I'd discovered that some controls inside these two usercontrols were using the same Style (Style="{StaticResource GenericHyperlinkButtonStyle}") Once I'd removed the style, they started working as usual. The first step is to create a new user control, FieldUserControl, and move our XAML into there: x:Class="UserControlExample.FieldUserControl" ...> Orientation="Horizontal"> Thanks.Attachments1.JPG26.2 KBLike • Show 0 Likes0 Actions JNery-esristaff Feb 6, 2012 10:23 AMUnmark CorrectCorrect AnswerPlease check... Ideally this property should support binding, just like any other property of the framework UI controls.

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thanks Matt Tuesday, August 12, 2008 9:04 AM Reply | Quote 0 Sign in to vote I also have met this problem, and hope someone can solve it The navigate here How small could an animal be before it is consciously aware of the effects of quantum mechanics? UI for Silverlight Resources Buy Try Feed for this thread 2 posts, 0 answers Prasad 2 posts Member since: Apr 2012 Posted 16 Apr 2012 Link to this post Hi , You can use the static proeprty DesignerProperties.IsInDesignTool to determine if your code is currently being used in a designer tool or not.

more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed ahura Proposed as answer by thinksomid Sunday, January 26, 2014 8:55 PM Thursday, October 30, 2008 8:54 AM Reply | Quote 0 Sign in to vote thanks ahura. it's really [email protected] Sunday, January 26, 2014 8:56 PM Reply | Quote Microsoft is conducting an online survey to understand your opinion of the Msdn Web site. Check This Out So we add another dependency property to our user control.

asked 6 years ago viewed 4915 times active 3 years ago Related 3How to use binding to bind to the grand-parent element in Silverlight?3What is a good pattern for binding a Thanks.Like • Show 0 Likes0 Actions JNery-esristaff Feb 2, 2012 2:02 PMMark CorrectCorrect AnswerYou can check that when you type "So I researched and found this: DesignerProperties.GetIsInDesignMode MethodAfter reading about this, I changed my code in my Constructor to this: public PluginTreeViewControl() { InitializeComponent(); // This "if" block is only for

After all, users like to be presented with a consistent interface, so re-use makes sense. There was a null reference exception happening. This is a tip for a beginner PrestaShop owner or developer. My manager said I spend too much time on Stack Exchange.

Two of my usercontrols did not want to render in their containing page and was I getting the same 'cannot create an instance of [x]' exception. Most people's first reaction is to set the DataContext of the user control to itself (I distinctly recall doing this myself the first time I encountered this problem!). More recently I have diversified to include HTML5, JavaScript and iOS development. @ColinEberhardt MORE BY COLIN Recursive Pattern Matching and Transformation of JavaScript AST 22nd Jun · 6 min read Six http://skimwp.org/cannot-create/cannot-create-an-instance-of-viewmodel-xaml.php C:\...\Views\CompareCycles.xaml 270 13The detail shows that Cannot find a Resource with the Name/Key CommonBorder [Line: 10 Position: 44] at System.Windows.Application.LoadComponent(Object component, Uri resourceLocator) at xxx.UserControls.GraphComparison.InitializeComponent() in C:\...\obj\Debug\UserControls\GraphComparison.g.i.cs:line 60 at xxx.UserControls.GraphComparison..ctor() in

I have a viewmodel and I would like to add it to the application's resources in the xaml, but when I try it says "Cannot create an instance of 'AppViewModel'. Fill in your details below or click an icon to log in: Email (required) (Address never made public) Name (required) Website You are commenting using your WordPress.com account. (LogOut/Change) You are Another way to check is to put a break point on GraphicSource class, to see if it comes through there. This problem can be fixed by setting the DataContext of the FieldUserControl's root element to itself.

Custom controls are rather special, with the logic being de-coupled from the XAML in order to support templating. PrestaShop Tips - How to set currency correctly for default currency and Paypal restriction PrestaShop supports multiple currencies. Is adding the ‘tbl’ prefix to table names really a problem? Click="myClickHandler") chances are they don't line up with the associated methods in the code behind.

Join them; it only takes a minute: Sign up Cannot create instance of viewmodel in xaml up vote 5 down vote favorite 2 I've been searching for an answer to this I fixed it, thanks. –Barry Franklin Nov 2 '11 at 17:13 add a comment| 1 Answer 1 active oldest votes up vote 4 down vote accepted I just figured out that This workaround makes me super happy!Like this:Like Loading...Related Category: csharp, WPF |Comment (RSS) |Trackback3 Comments hyeroglyph says: November 27, 2011 at 1:15 pmThank you very much. If the technique of binding the layout root of the user control to itself is a bit confusing - the following diagram, which shows the visual tree of our simple application,

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silverlight silverlight-4.0 share|improve this question edited Jul 9 '11 at 16:36 Gone Coding 64k11108136 asked Jul 9 '11 at 11:52 happysmile 2,3102782160 solution: InitializeComponent(); if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode‌(this)) { //Code that Most functionality can be developed separately and independently. ...