[Okta Webinar] Learn how to a build a cloud-first strategyRegister Now

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 2905
  • Last Modified:

System.Diagnostics.Process.Start

I don't know if I am doing this incorrect.  I am trying to connect to a network drive using this command in VB.NET and passing a username and password.  

The network drive is already connected but after a fresh restart of the computer, it wants a password.

I am using:
System.Diagnostics.Process.Start("X:", "username", "password")

Am I doing something wrong because I keep receiving a compile error:  Overload resolution failed because no accessible 'start' accepts this number of arguments.
0
chellert
Asked:
chellert
  • 8
  • 6
1 Solution
 
Fernando SotoCommented:
Hi chellert;

The reason why you are getting the error is that the System.Diagnostics.Process.Start take 0, 1 or 2 parameters. So to start it with arguments the command should be

System.Diagnostics.Process.Start("X:", "username password")

But I am not sure that the username and password are going to be passed to the login screen. If the login screen does not accept parameter input it will not work.

Fernando
0
 
chellertAuthor Commented:
Is there a shell command that will allow me to login into the network drive?

0
 
Fernando SotoCommented:
Hi chellert;

I do not know. I have did a little research and I think you may be able to do what you need using the net use command

    System.Diagnostics.Process.Start("net", "use X: \\ComputerName\ Other info from the Documentation")

Documentation for Net Use

    http://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/net_use.mspx?mfr=true

I hope that this helps;

Fernando
0
VIDEO: THE CONCERTO CLOUD FOR HEALTHCARE

Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.

 
chellertAuthor Commented:
I am still having my original problem, where I have to start the program twice to just connect properly.
0
 
Fernando SotoCommented:
Sorry but I am not sure how else you can get around the problem
0
 
chellertAuthor Commented:
OK, I have tried this:

    Dim myProcess As System.Diagnostics.Process = New System.Diagnostics.Process()
    Dim pw As New System.Security.SecureString
   
    For Each ch As Char In "password"
            pw.AppendChar(ch)
    Next

        myProcess.StartInfo.Arguments = "X:"
        myProcess.StartInfo.FileName = "\\toonvhcl10_vol5_server\vol5\BVTS"
        myProcess.StartInfo.UserName = "username"
        myProcess.StartInfo.Password = pw
        myProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden
        myProcess.Start()

Once I hit the myProcess.Start(), I receive this message  "The Process object must have the UseShellExecute property set to false in order to start a process as a user."

What does this mean???
0
 
Fernando SotoCommented:
Hi chellert;

Read the comment near the bottom.

        Dim myProcess As System.Diagnostics.Process = New System.Diagnostics.Process()
        Dim pw As New System.Security.SecureString

        For Each ch As Char In "password"
            pw.AppendChar(ch)
        Next

        myProcess.StartInfo.Arguments = "X:"
        'myProcess.StartInfo.FileName = "\\toonvhcl10_vol5_server\vol5\BVTS"
        myProcess.StartInfo.FileName = "notepad.exe"
        myProcess.StartInfo.UserName = "username"
        myProcess.StartInfo.Password = pw
        myProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden

        ' You need to set this to False in able to start the process as a user
        ' as defined in myProcess.StartInfo.UserName not being an empty string.
        myProcess.StartInfo.UseShellExecute = False ' <== Need to set to False

        myProcess.Start()


Fernando
0
 
chellertAuthor Commented:
Now I receive this error "The parameter is incorrect"
0
 
Fernando SotoCommented:
Does it say which The parameter is incorrect? which line?
0
 
chellertAuthor Commented:
       myProcess.StartInfo.Arguments = "X:"
        myProcess.StartInfo.FileName = "file.txt"
        myProcess.StartInfo.UserName = "cdncmo"
        myProcess.StartInfo.WorkingDirectory = "\\toonvhcl10_vol5_server\vol5\BVTS"
        myProcess.StartInfo.Password = pw
        myProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden
        myProcess.StartInfo.UseShellExecute = False
        myProcess.Start()
        myProcess.WaitForExit()

The error comes at myProcess.Start() again
0
 
Fernando SotoCommented:
The thing about having to have myProcess.StartInfo.UseShellExecute = False is that the variable myProcess.StartInfo.FileName needs to be a exe type file because you are not using the system shell.
So if you want to open a text file then you need to do it something like this

        myProcess.StartInfo.Arguments = "\\toonvhcl10_vol5_server\vol5\BVTS\file.txt"
        myProcess.StartInfo.FileName = "notepad.exe"
        myProcess.StartInfo.UserName = "cdncmo"
        myProcess.StartInfo.WorkingDirectory = "c:\Windows\system32"
        myProcess.StartInfo.Password = pw
        myProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden
        myProcess.StartInfo.UseShellExecute = False
        myProcess.Start()
        myProcess.WaitForExit()
0
 
Fernando SotoCommented:
I am glad that you were able to work through the problem. You have a good day. :=)
0
 
chellertAuthor Commented:
Thank you for your time and patience.
0
 
Fernando SotoCommented:
Only glad to help.
0

Featured Post

Vote for the Most Valuable Expert

It’s time to recognize experts that go above and beyond with helpful solutions and engagement on site. Choose from the top experts in the Hall of Fame or on the right rail of your favorite topic page. Look for the blue “Nominate” button on their profile to vote.

  • 8
  • 6
Tackle projects and never again get stuck behind a technical roadblock.
Join Now