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Column grouping in RIA service

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Maciej
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Maciej asked on 16 Nov 2010, 09:41 AM
Hello,

I'm developing an Silverlight application using RIA services (with DomainDataSource load size of 50), dataGrid, and DataPager (with page size of 10). I would like to group my data by selected column which has 12 different values. And here is the problem: insted of getting 10 first groups (two other are on second page of pager), I only get 10 first grouped results (with 5 pager pages available). Is there any way to change that behaviour and group whole 50 result, or even better whole data accesible in database table? It was working fine in webservice approach, unfortunately i have to use RIA.

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Rossen Hristov
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answered on 16 Nov 2010, 09:54 AM
Hello Maciej,

Unfortunately, WCF RIA Services do not support real grouping. When you group on the server, it simply sorts the data according to the grouping, but it does not perform real grouping, i.e. it does not return groups. Instead it returns sorted Entities.

Unless MS decide to implement grouping -- there is nothing that can be done as far as grouping on the server is concerned.

I hope this helps.

Sincerely yours,
Ross
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Maciej
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answered on 16 Nov 2010, 11:00 AM
Thank you for fast reply Ross.

Regards,
Maciej.
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Jonx
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answered on 20 Jan 2011, 01:23 AM
Helo Ross,
Does your RadDomainDatasource solve the problem?
It seems to me that it doesn't but maybe I missed something.

I have the same problem then Maciej, It does only group the items on the current page which is useless...

Any hints about how this may be done by "hand"?

Thanks,
John. 
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Vlad
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answered on 20 Jan 2011, 08:23 AM
Hello,

 RadDomainDatasource will work in this case exactly like regular DomainDatasource since our component is just a client for WCF RIA Services. The data source will just tell the service to return groups and it is up to the service how this grouping will be created and returned. 

Kind regards,
Vlad
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Jonx
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answered on 20 Jan 2011, 11:05 AM
I understand... Thank you for the explanation...

But know, woud you have any idea where I should start to make things rights by myself?
Where would you start?

Maciej, you say it's working with webservices? Would you be so kind and point me to an exemple somewhere working with webservices?
That yould be so helpful to me,
Thanks a lot in advance...

John.
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Jonx
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answered on 03 Feb 2011, 06:56 PM
Still... What alternatives do I have?
Could the VirtualQueryableCollectionView be a solution?
I absolutely need to find a way...
Any hints?
Thanks a lot in advance,
John.
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Jonx
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answered on 03 Feb 2011, 07:06 PM
What about the new DomainCollectionView added in SP1?
Does it also have the same limitation in terms of grouping?
Thank you for your help guys...
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Jonx
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answered on 03 Feb 2011, 09:07 PM
Ok after that post I wait for an answer before adding a new post ;)

I just stumbled on that post of yours Vlad:
http://blogs.telerik.com/vladimirenchev/posts/10-01-14/how_to_server_sorting_filtering_grouping_and_paging_with_radgridview_for_silverlight_without_wcf_ria_services.aspx

Is this what I was looking for ? A grid with grouping working correctly with a pager ?

Thank you for your help ;)

John.
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Vlad
Telerik team
answered on 04 Feb 2011, 07:21 AM
Hello,

 The approach in this blog post is not related to RIA services - I've used plain WCF service to achieve this. 

Greetings,
Vlad
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Jonx
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answered on 04 Feb 2011, 08:13 AM
If it's the only way to get working groups, I'll go that way.

Can you please confirm, the groups are working correclty before I try to go down that road?

Thanks...
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Vlad
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answered on 04 Feb 2011, 09:04 AM
Hello,

 You can download the group items in this way however currently the virtual collection will not virtualize groups - this is still not supported. We have plans to improve this however we don't have any clear release date for such feature so far. Once we have more info we will announce it publicly. 

Best wishes,
Vlad
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Jonx
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answered on 05 Feb 2011, 05:54 PM
Ok Vlad,

I guess that I will use standard filtering for now instead of grouping and wait for better solutions.

Thank you for your help.
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Maciej
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Rossen Hristov
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Maciej
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Jonx
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Vlad
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