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Adding Resources Server Side

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Scheduler
This is a migrated thread and some comments may be shown as answers.
Chad
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Chad asked on 24 Sep 2008, 03:29 PM
I need to be able to add a varying number of resources that utilize varying types of controls (checkbox, dropdownlist, etc).

The problem is that I cannot create an AdvancedEdit/Insert template using an aspx page, and need to do it on the server side.

So far I can add a resource with the following...

aResource =

new ResourceType("Calendar Color");

aResource.DataSource = GetColors();  // a DataTable with 7 colors and 1 column - "Color"

aResource.KeyField =

"Color";

aResource.TextField =

"Color";

aResource.ForeignKeyField =

"RadSchedulerColor"// the field in the Scheduler datasource containing 1 of  the 7 color values

theScheduler.ResourceTypes.Add(aResource);


The problem is that I want to add new resources with different controls, like a checkbox or a radiobutton. Then I want to be able to set the value based on the Scheduler's datasource. Currently this does not link to the Scheduler so the resource has no default value. I tried to override the OnFormCreated method:

void theScheduler_FormCreated(object sender, SchedulerFormCreatedEventArgs e)

{

if (e.Container.Mode == SchedulerFormMode.AdvancedInsert || e.Container.Mode == SchedulerFormMode.AdvancedEdit)

{

DropDownList colors = (DropDownList)e.Container.FindControl("Calendar Color");

colors.SelectedValue =

"Red";

colors.Width =

new Unit(100, UnitType.Percentage);

}

}


This only increased the width, it doesn't set the SelectedValue. Am I doing this right? Is what I'm trying to do possible without creating a custom UserControl with aspx elements?

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Peter
Telerik team
answered on 26 Sep 2008, 03:22 PM
Hello Chad,

If you want to use controls other than the default dropdownlist control to display resources, you should use the Customizing the advanced template example. I understand that you would rather not resort to using user controls, but unfortunately this is the only way to achieve the disired functionality.


Cheers,
Peter
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