code path of a application

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hi all,

i kept my xml file in the following path
i need to get this path dynamically
where sample xml is my project file name
D:\OfficeWork\practice\SampleXml\SampleXml

i am getting executable file by using this path D:\OfficeWork\practice\SampleXml\SampleXml\bin\Debug

Posted 19-Sep-11 3:18am

s.chiranjeevi

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#realJSOP · Answer 1 · 2011-09-19T03:37:00

C#

Dim path As String = Application.StartupPath
Dim found As Boolean = False
Do
    found = File.Exists(System.IO.Path.Combine(path, "sample.xml"))
    If Not found And path.Length > 3 Then
        path = System.IO.Path.GetDirectory(path)
    End If
Loop While Not found OrElse String.IsNullOrEmpty(path)

If found Then
    path = System.IO.Path.Combine(path, "sample.xml")
    ' do something with the path
End If

fjdiewornncalwe · Answer 2 · 2011-09-19T03:39:00

I would suggest for simplicity that you add the SampleXml file to your project and then select it in your solution explorer. In the properties window select the "Copy to output directory" and make sure it is set to Copy always or Copy if newer. Then it will appear in the bin\Debug directory when you build and you can just reference the file name instead of having to calculate an absolute path for the file.

code path of a application

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