RDLC - Adding a Data Source in VS2010
Greetings.
I have an RDLC file and am wanting to add a data source to it, although without any luck so far. The data source is a custom class written by myself (just to add, we do this all the time). We recently converted over to the VS2010 RDLC format which caused some problems, but we've made some changes to our implementation that workaround the more major issues.
So, back to the issue at hand, when I attempt to add my data source to the DummyDataSource list in the RDLC view in VS2010 it just does nothing, however it does add the data source to the list of data sources, but you can't select it from the drop-down list in the RDLC view which means I can't add the data source at all.
Has anyone come across this problem? Is there anything I need to check? I've searched with fervour and had no luck.
Answer
There seems to be a bit of black magic going on here. Or at least I haven't figured out all of the incantation to make this happen reliably.
I think I was having a similar problem. Not sure if this will help you, but here's how I got around it.
In the VS2010 report designer, use view->Report Data to show the Report Data pane.
Click the New button and choose Dataset... to get the Dataset Properties dialog.
Name your Dataset, if you've done this before, you probably know that the dataset name here needs to match the name provided in code when you bind your ReportDataSource.
The new part that I just tried, is to click the New... button next to the Data source drop down list. The resulting wizard walks you through selecting your assembly and CLR class (use the checkbox to select your class).
When the wizard finished, my new dataset appeared.
One thing to note: The first time I tried to reproduce this, the wizard completed without adding my dataset. I went back to my class definition and decorated it with [DataContract] and [DataMember] attributes and then re-ran the wizard and it seemed to work great.
Perhaps someone with deeper knowledge can comment on why those attributes make this work, or why the wizard fails silently without them.