VCALinkTool::OutputBaseFileName Property

Visual Studio 2015
 

Gets the name (but not the location) of the generated satellite resource DLL or DLL.

Namespace:   Microsoft.VisualStudio.VCProjectEngine
Assembly:  Microsoft.VisualStudio.VCProjectEngine (in Microsoft.VisualStudio.VCProjectEngine.dll)

C#
C++
F#
VB
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property String^ OutputBaseFileName {
	String^ get();
}

Property Value

Type: System::String^

The name of the generated satellite resource DLL.

The common language runtime, which requires that satellite resources have the same name, are distinguished by their directories. For example, French resources would have the name specified here and be in a subdirectory named "fr" under the directory in which the primary output resides. As a result, the name returned by OutputBaseFileName is always a single name, even when there are multiple satellite DLLs being generated.

See How to: Compile Example Code for Project Model Extensibility for information about how to compile and run this sample.

VB
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' Add a reference to Microsoft.VisualStudio.VCProjectEngine and have a 
' Visual C++ project loaded before running this example.
Imports EnvDTE
Imports Microsoft.VisualStudio.VCProjectEngine
Public Module Module1
    Sub Test()
        Dim prj As VCProject
        Dim cfgs, tools As IVCCollection
        Dim cfg As VCConfiguration
        Dim tool As VCALinkTool
        prj = DTE.Solution.Projects.Item(1).Object
        cfgs = prj.Configurations
        cfg = cfgs.Item(1)
        tool = cfg.Tools("VCALinkTool")
        MsgBox("Output base file name: " & tool.OutputBaseFileName)
    End Sub
End Module
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