Accessing Data from two different Databases in the same dataset

Hello Experts,

I am using Visual Studio 2012 (C#) and SQL Server 2008. I would like to create a single DataSet in Visual Studio (C#) that accesses and binds tables from both separate databases.

I tried doing this by ...

1. Creating a new DataSet in C# (VS 2012)
2. Added a connection to each Database in Server Explorer (VS2012)
3. Added the desired objects from each database on the DataSet palette.

VS didn't would only allow me to choose one of the Database objects at a time. In other words I could only choose one of the databases at a time.

Does anyone know how to do this?


would it be better to use the Linked server function in SQL Server?

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Koen Van WielinkBusiness Intelligence SpecialistCommented:
Just to clarify, are you getting data from different databases on the SAME server, or different servers? If it's the same server you wouldn't need the linked server function in SQL at all, just use fully qualified names when referring to the respective database I think.
Either way, coming from a DB background rather than a .NET background, I'd handle this on the database side through a view or a stored procedure, then just link your .NET application to that instead. Easier to maintain in SQL I think.

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Jacques Bourgeois (James Burger)PresidentCommented:
A DataSet is simply a container for DataTable objects, nothing else.

Because each DataTable needs to be built separately, simply use a different connection for each DataTable.

If you want to mix the data from two databases into a single table, then you do not need a DataSet. You might want to read my article on the subject.

You simply need a DataTable. And as suggested by Kown Van Wielink, this is better done at the database level, in a stored procedure. Call that stored procedure and use the result as your DataTable.
SaxitalisAuthor Commented:
Yes I am accessing all databases on the same server so can use fully qualified naming in my queries. I was able to get my data displayed using a ReportViewer object. Thanks
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