create X.xaml and X.xaml.cs files vs should understand they are xaml and code behind
Date : March 29 2020, 07:55 AM
will be helpful for those in need This is because Visual Studio doesn't know they are associated. It's just a naming convention to have .xaml and .xaml.cs. IN the project file you need to enforce the relationship via: <Compile Include="NodeControl.xaml.cs">
<DependentUpon>NodeControl.xaml</DependentUpon>
</Compile>
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Merge XAML file into Xaml code
Date : March 29 2020, 07:55 AM
this will help I want to merge one XAML file, that only contains styles into the Window.Resource of my application. I tried using MergedDictionaries as described in WPF Reference custom resource defined in another xaml file . , Here is a demo wpf project: -WpfApplication
|--App.xaml
|--MyStyle[here is a floder]
|---MyStyleDocument1.xaml
|---MyStyleDocument2.xaml
|---MyStyleDocument3.xaml
<Application.Resources>
<ResourceDictionary>
<ResourceDictionary.MergedDictionaries>
<!--add style file here like this-->
<ResourceDictionary Source="/MyStyle/MyStyleDocument1.xaml" />
</ResourceDictionary.MergedDictionaries>
<ResourceDictionary>
<Application.Resources>
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How are XAML elements accessed from code behind in a different XAML file?
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , What are you trying to do here? The code you posted has got nothing to do with accessing "XAML objects in another file". The Application.Properties property is just a dictionary for sharing data, in a thread-safe fashion between different parts of your application. It is very rarely used in my experience - in fact I have never seen it used. You are getting a null exception because you probably haven't added the 'favoritesItem' to the dictionary first. If you want to access a named XAML element from another class you will need to expose it via a public property. (Named XAML elements create private member fields in their defining class).
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Loading a grid in xaml code from a file, then making it active in main code
Tag : chash , By : Thierry Brunet
Date : March 29 2020, 07:55 AM
I wish this helpful for you A really simple solution would just be to house each of your rooms inside a user control and then place the control into your container grid when you need to change rooms. Here's the rough idea: public partial class MainWindow : Window
{
UserControl _currentRoom;
public MainWindow()
{
InitializeComponent();
}
protected override void OnPreviewMouseDown(MouseButtonEventArgs e)
{
this.areaContainer.Children.Clear();
_currentRoom = null;
if (e.LeftButton == MouseButtonState.Pressed)
_currentRoom = new Room1();
if(e.RightButton == MouseButtonState.Pressed)
_currentRoom = new Room2();
this.areaContainer.Children.Add(_currentRoom);
base.OnPreviewMouseDown(e);
}
}
<UserControl x:Class="Test.Room1"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
mc:Ignorable="d"
d:DesignHeight="300" d:DesignWidth="300">
<Button Height="100" Width="100">
Hello world!
</Button>
</UserControl>
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Use UIElement from XAML file in a window's XAML code
Tag : chash , By : gorbiz
Date : March 29 2020, 07:55 AM
Hope this helps You may perhaps bind the Content property of a ContentControl, with a Binding Converter that returns the result of XamlReader.Load.
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