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OutputBaseFileName Property

Visual Studio .NET 2003

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

[Visual Basic .NET]
Public ReadOnly Property OutputBaseFileName() As String
[Visual Basic 6]
Property Get OutputBaseFileName() As String
[C++]
HRESULT __stdcall get_OutputBaseFileName(
   /* [out, retval] */ BSTR* retVal
);
[C#]
public string OutputBaseFileName {get;}
[JScript .NET]
public function get OutputBaseFileName() : String

Return Value

A string representing the name of the generated satellite resource DLL.

Remarks

The common language runtime requires that satellite resources have the same name but be differentiated based on directory. For example, French resources would have the name specified here and be in a subdirectory called "fr" under the directory the primary output is found. As a result, the name returned by OutputBaseFileName is always a single name, even when there are multiple satellite DLLs being generated.

Example

' Add a reference to Microsoft.VisualStudio.VCProjectEngine and have a 
' Visual C++ .NET 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

See Samples for Project Model Extensibility for information on how to compile and run this sample.

See Also

Applies To: VCALinkTool Object

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