How to use a file name to start an application by using Visual Basic .NET or Visual Basic 2005 (305201)

SUMMARY

This article demonstrates how to start the application that is associated with a given document extension or file type without needing to know the name or location of the associated application. For example, you can start Arcade.bmp with the application that is associated with the .bmp file extension, which is MSPaint.exe in most cases.

Requirements

• Microsoft Visual Basic .NET or Microsoft Visual Basic 2005

Specify The ProcessStartInfo information

You can use the ProcessStartInfo structure of the .NET Framework Process class to specify options when you start a process. This article outlines how to use the file name option. Another member, UseShellExecute, specifies that the process be started based on a file extension or file type instead of the name of the executable (.exe). This property is set to true by default. It is set explicitly in this code for illustrative purposes.

Dim fileName as String = "\eula.txt"
Dim sysFolder As String = _
  Environment.GetFolderPath(Environment.SpecialFolder.System)
Dim pInfo As New ProcessStartInfo()
pInfo.FileName = sysFolder & fileName
pInfo.UseShellExecute = True
				

Start the application

This example opens a file named Eula.txt. The file is opened with the application that is associated with the .txt file extension, which is normally Notepad.exe. You can substitute any file name or type that has an associated application.

Dim p As Process = Process.Start(pInfo)
				

Shortcut to start the application

Because UseShellExecute is true by default for a process, it is not required that you use ProcessStartInfo when you start a process. You can start the associated application with a single line of code, as follows:

Dim p As Process = Process.Start("c:\winnt\system32\eula.txt")
				

Complete code sample

    'How to Start a Process with a File Name
    'Specify a file.
    Dim fileName as String = "\eula.txt"
    'Get the name of the system folder.

    Dim sysFolder As String = _
      Environment.GetFolderPath(Environment.SpecialFolder.System)
    'Create a new ProcessStartInfo structure.
    Dim pInfo As New ProcessStartInfo()
    'Set the file name member of pinfo to Eula.txt in the system folder.
    pInfo.FileName = sysFolder & fileName
    'UseShellExecute is true by default. It is set here for illustration.
    pInfo.UseShellExecute = True
    'Start the process as specified in the process info structure.
    Dim p As Process = Process.Start(pInfo)
				

Troubleshooting

It is possible that an individual computer may not have the associated application installed, or the associations in the registry may not be correct. It is best to wrap this code in a try...catch block so that your application is alerted in the event of a failure.