SilverLight - MVVM绑定viewmodel属性到datagrid列 [英] SilverLight - MVVM binding viewmodel property to datagrid column

问题描述

在我的SilverLight应用程序中，我的ViewModel中有一个名为'vmProperty'的属性和名为'dgSource'的列表。

我将dgSource绑定到Datagrid作为ItemsSource，在此时每个datagrid行的datacontext将更改为dgSource中的每个项目。其中一列（称为复选框）需要绑定到vmProperty。但是由于ViewModel不再是行的数据文本，所以我无法访问此属性。

如何解决这个问题？如果问题不清楚，请让我知道，我会发一个样本。感谢提前。

解决方案

假设您的ViewModel被分配为LayoutRoot的DataContext，这应该工作：

  IsChecked ={Binding DataContext.vmProperty，ElementName = LayoutRoot}
  / pre> 
 
 
  
 
 
 
当然这不行，最终模板被复制因此，实际使用绑定的命名范围中不存在LayoutRoot。
 
 
 
最简单的解决方案是使用ViewModel来更改模型。在dgSource中可用的对象上显示所需的值，或者显示导航回ViewModel的 Parent 属性。 
 
In my SilverLight application, I have a property in my ViewModel called 'vmProperty' and a list called 'dgSource'. 

I bind my dgSource to the datagrid as ItemsSource at which point each datagrid row's datacontext changes to each item in dgSource. One of the columns, say a checkbox column, needs to bind to vmProperty. But since the ViewModel is no longer the row's datacontext, I cannot get access to this property. 

How do I get around this problem? If the question is not clear, please let me know and I will post a sample. Thanks in advance.
解决方案

Assuming your ViewModel is assigned as the LayoutRoot's DataContext this should work:- 
IsChecked="{Binding DataContext.vmProperty, ElementName=LayoutRoot}"


Of course this doesn't work, ultimately a template is replicated and therefore "LayoutRoot" does not exist in the namescope where the binding is actually used.

The simplest solution since this is a ViewModel is to change the model.  Expose the required value on the objects available in the dgSource or expose a Parent property that navigates back to the ViewModel.  

                        
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