SilverLight - MVVM binding viewmodel property to datagrid column

Question

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.

Do really intend for each checkbox instance for each row to bind back to the same property on the parent object, doesn't seem to make a lot of sense to me? — AnthonyWJones
– AnthonyWJones, Commented Dec 1, 2009 at 12:11

AnthonyWJones · Accepted Answer · 2009-12-01 19:02:34Z

0

~~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.

edited Dec 1, 2009 at 19:02

answered Dec 1, 2009 at 12:27

AnthonyWJones

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3 Comments

etrast81 Over a year ago

Thanks for your responses. I used the ElementName method you have mentioned. But I run into the same issue as in 'stackoverflow.com/questions/1089650/…'. I put this outside the datagrid and this piece code works great. But within the datagrid it doesnt. Anyways, I used the link provided in that thread 'blogs.msdn.com/jaimer/archive/2008/11/22/…', but that seems to be relevant only for WPF. I am unable to access AddOwner or OverrideMetadata. Thoughts?

etrast81 Over a year ago

Thanks Anthony, Managed to expose it as part of dgSource. Not happy about it as it is replication. But it is working. So... :)

AnthonyWJones Over a year ago

Thats the advantage of a view model. It would be most unsatisfactory if it were simply the model but the whole point of a view model is that its munged form of data that supports a specific view or task.

bniwredyc · Accepted Answer · 2009-12-01 12:39:10Z

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Try to set checked property of your chechbox column to that:

{Binding RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type YourViewClassName}}, Path=DataContext.vmProperty}

That's mean that parent element which type is your view type will be found and than vmProperty of it's DataContext will be retrieved as value.

UPDATE:

It was solution for WPF.

For silverlight I think you can try to use construction like this:

{Binding Path=DataContext.vmProperty, ElementName=YourElement}

Where DataContext of YourElement is instance of your view model. I think it can be a grid for example.

<Grid x:Name="YourElement" DataContext={Binding}>
<!-- DataGrid here -->
</Grid>

edited Dec 1, 2009 at 12:39

answered Dec 1, 2009 at 11:19

bniwredyc

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2 Comments

etrast81 Over a year ago

I think that is WPF, isnt it? SilverLight does not have RelativeSource or AncestorType, I believe.

AnthonyWJones Over a year ago

It has RelativeSource but only in very limited capacity

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