Visual Basic

Luc Pattyn24-May-11 8:38

Luc Pattyn

24-May-11 8:38

That clearly is not the entire story. Are you waiting for the process to exit somewhere? what is in the "Run clicked" handler? Is your app responsive (can you move its main window, will it repaint) while the process runs?
one handler can not execute while another one is still running!

Smile | :)

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Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.

Andraw Tang24-May-11 9:01

24-May-11 9:01

That's the whole codes when "Run" is clicked.

Yes, I need to waiting for the process to exit, I need to get the return string and display them in the form. That's why I use
p.WaitForExit(5400000). But use p.WaitForExit(5400000) will hang progressbar form, if don;t use p.WaitForExit(5400000), when should I close the progressbar form?

Private Sub OnProcesited(ByVal sender As Object, ByVal e As System.EventArgs)
   Dim p As System.Diagnostics.Process
   p = sender

   Dim rtnStr As String = String.Empty
   rtnStr = p.StandardOutput.ReadToEnd()

   If rtnStr.Contains(" FAST terminated normally.") = True Then
       rtnStr = rtnStr & vbCrLf & "Please click Postprocess -> View Output File in menu to view the file."
       updateViewMenu()
   End If
   '''progBarFrm.Close()
   '''progBarFrm.Dispose()
   
   Dim frm As New FormShowFASTResult
   frm.txtFASTResult.Text = rtnStr
   frm.Button1.Select()
   frm.ShowDialog()
 End Sub

Dave Kreskowiak24-May-11 8:41

24-May-11 8:41

You still have absolutely nothing updating the progress bar. You called Show on the form to show the form with the progress bar on it, but you've got nothing at all that updates the progress bar, so it's not going to do anything.

Without seeing the code behind the progressbar form, this is about all we can say.

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Andraw Tang24-May-11 9:04

24-May-11 9:04

Progressbar value should be updated using timer inside the form, after call SHOW of the form, the timer will be started, sure you cannot see these codes from inside the function which create the form.

Andraw Tang24-May-11 8:45

24-May-11 8:45

I found only if I clearly call p.Kill(), the Exited event will be triggered, but when I should call the .Kill() for a process?

if I use p.WaitForExit(5400000), the program will be hanged to wait for process to exit, the progressbar form will be hanged.

Dave Kreskowiak24-May-11 8:51

24-May-11 8:51

The process will drop by itself. You shouldn't have to call Kill on it at all. Once the process you launched quits on its own, the Exited event will be raised.

5400000?? Really? 90 minutes??

You do realize WaitForExit is a blocking call, right? No other code will execute so long as WaitForExit is waiting. Remove this line and just let it run. When the target app is finished, your event handler will get called.

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Andraw Tang24-May-11 9:27

24-May-11 9:27

Sorry calculation error, I already comment out that line. Now I modify the source codes, also remove the event.

When "Run" menu is clicked, it will call Run_Tool() inside which it start a precess and open a form, I also show the codes for progressbar form below, but the progressbar color is grayed, no value is shown.

Would you please take a look where is wrong? thanks!

  Private Sub Run_Tool()

    Dim sAppPath As String = ".\xxxx.exe"
    Dim fileName As String = "XXXXXX.TXT"
    Dim p As New System.Diagnostics.Process()

    Dim rtnStr As String = String.Empty

    p.StartInfo.FileName = Application.StartupPath & "\" & sAppPath
    p.StartInfo.Arguments = fileName

    'Do not use the system shell to start the program this is so we can hide the command dos window
    p.StartInfo.UseShellExecute = False
    p.StartInfo.WorkingDirectory = strOutputFilePath
    ' Show no dos window if false
    p.StartInfo.CreateNoWindow = True
    p.StartInfo.RedirectStandardOutput = True

    Dim progBarFrm As New FormProgressBar
    progBarFrm.Show()


    Try
        p.Start()
    Catch ex As Exception
        ShowValidationMessage("Unexception error is caught, please close and try again.", "Error")
        'p.Kill()
        progBarFrm.Close()
        Return
    End Try


    rtnStr = p.StandardOutput.ReadToEnd()

    ''p.WaitForExit(120000)   '= 1.5 min: 1000 millisecond = 1 second
    ' if the process doesn't complete within 5400 seconds, kill it
    ''If Not p.HasExited Then
    ''p.Kill()
    ''End If

    If rtnStr.Contains(" FAST terminated normally.") = True Then
        rtnStr = rtnStr & vbCrLf & "Please click Postprocess -> View Output File in menu to view the file."
        updateViewMenu()
    End If
    progBarFrm.Close()

    Dim frm As New FormShowFASTResult
    frm.txtFASTResult.Text = rtnStr
    frm.Button1.Select()
    frm.ShowDialog()

End Sub

Source code for the progressbar form as the following

Public Class FormProgressBar

    Private Sub FormProgressBar_Disposed(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Disposed
        Timer1.Stop()
    End Sub



    Private Sub FormProgressBar_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
        Me.CenterToScreen()
        ProgressBar1.Minimum = 0
        ProgressBar1.Maximum = 100
        ProgressBar1.Value = 0
        Timer1.Interval = 1
        Timer1.Start()

    End Sub


    Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick
        If ProgressBar1.Value < 100 Then
            ProgressBar1.Value = ProgressBar1.Value + 1
        Else
            ProgressBar1.Value = 0
        End If
    End Sub
End Class

Dave Kreskowiak24-May-11 9:49

24-May-11 9:49

First, ReadToEnd isn't going to give you what you're looking for. It'll block until the console has to something to return, and even then, it'll return block of text, most likely NOT all of the text your app output to the console, and it will only read ONE block of text.

Now, if you want to gather all the text the app outputs, and still have a responsive UI (such as the progress bar redrawing itself), you'll have to read the console output asynchronously.

Something like:

Private Sub StartConsoleRead(ByRef st As Stream)
        Dim rdo As New ReadData
        'rdo.s = _conStream.BaseStream
        rdo.s = st

        Dim consoleReadAsyncCallback As New AsyncCallback(AddressOf ConsoleReadCallback)
        _conStream.BaseStream.BeginRead(rdo.buffer, 0, rdo.buffer.Length, consoleReadAsyncCallback, rdo)
    End Sub

    Private Sub ConsoleReadCallback(ByVal asyncResult As IAsyncResult)
        Dim rdo As ReadData = DirectCast(asyncResult.AsyncState, ReadData)
        Dim bytesRead As Integer = rdo.s.EndRead(asyncResult)
        Dim enc As Encoding = Encoding.ASCII
        Dim s As String = enc.GetString(rdo.buffer, 0, bytesRead)

        CmdOutput.Invoke(UpdateCmdOutputDelegate, New Object() {s})
        StartConsoleRead(rdo.s)
    End Sub
 
 

Public Class ReadData
    Public buffer(20480) As Byte
    Public s As Stream
End Class

Also, you're Timer.Interval on the progress bar form should not be 1. Make it something more reasonable, like 250 to 500 (quarter to half a second). Setting it to one will just make your app hog the CPU.

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Andraw Tang24-May-11 9:41

24-May-11 9:41

If I don't start the process, the progressbar in the form works fine. So the p.Start action hang the progressbar form.

 Private Sub Run_FAST_Tool()

    Dim sAppPath As String = ".\FAST.exe"
    Dim fileName As String = objFASTPrimaryData.FASTPrimaryFileName_FST
    Dim p As New System.Diagnostics.Process()

    Dim rtnStr As String = String.Empty

    p.StartInfo.FileName = Application.StartupPath & "\" & sAppPath
    p.StartInfo.Arguments = fileName

    'Do not use the system shell to start the program this is so we can hide the command dos window
    p.StartInfo.UseShellExecute = False
    p.StartInfo.WorkingDirectory = strOutputFilePath
    ' Show no dos window if false
    p.StartInfo.CreateNoWindow = True
    p.StartInfo.RedirectStandardOutput = True

    Dim progBarFrm As New FormProgressBar
    progBarFrm.Show()
end sub

Dave Kreskowiak24-May-11 9:51

24-May-11 9:51

The reason why your UI thread blocks is not because you started the process. It's blocking because of another line AFTER your call to Start. Your executing some statement that blocks, like WaitForExit, freezing your UI. You cannot do that and expect your UI to update itself.

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Andraw Tang24-May-11 12:13

24-May-11 12:13

Thanks, David.

After I remove calling WaitForExit() and ReadToEnd(), the progressbar works now, but the Exited event function OnProcesited is not called, so the progressbar is always shown. How can I make OnProcesited() be called?

Private Sub OnProcesited(ByVal sender As Object, ByVal e As System.EventArgs)
        Dim p As System.Diagnostics.Process
        p = sender

        Dim rtnStr As String = String.Empty
        rtnStr = p.StandardOutput.ReadToEnd()

        If rtnStr.Contains(" FAST terminated normally.") = True Then
            rtnStr = rtnStr & vbCrLf & "Please click Postprocess -> View Output File in menu to view the file."
            updateViewMenu()
        End If
        progBarFrm.Close()
        progBarFrm.Dispose()

        Dim frm As New FormShowFASTResult
        frm.txtFASTResult.Text = rtnStr
        frm.Button1.Select()
        frm.ShowDialog()

    End Sub


    Private Sub Run_Tool()

        Dim sAppPath As String = ".\XXXXX.exe"
        Dim fileName As String = "XXXXX.txt"
        Dim p As New System.Diagnostics.Process()



        p.StartInfo.FileName = Application.StartupPath & "\" & sAppPath
        p.StartInfo.Arguments = fileName

        'Do not use the system shell to start the program this is so we can hide the command dos window
        p.StartInfo.UseShellExecute = False
        p.StartInfo.WorkingDirectory = strOutputFilePath
        ' Show no dos window if false
        p.StartInfo.CreateNoWindow = True
        p.StartInfo.RedirectStandardOutput = True
        p.EnableRaisingEvents = True
        AddHandler p.Exited, AddressOf OnProcesited


        progBarFrm = New FormProgressBar

        Try
            p.Start()

            progBarFrm.Show()
            progBarFrm.Refresh()            
        Catch ex As Exception
            ShowValidationMessage("Unexception error is caught, please close and try again.", "Error")
            
            Return
        End Try    
   End Sub

Dave Kreskowiak24-May-11 12:26

VB.NET Handle SQL DB in Multi user application

24-May-11 12:26

You can't force it. Your external process just hasn't stopped yet for some reason. Does your external process wait for the user to "Press any key to exit"?? I sure hope not!

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Dave Kreskowiak

All Time Programming23-May-11 20:44

23-May-11 20:44

I have an vb.net application that is being used by multi-users. I execute all sql statements in a seperate block all at one time one after another. Many times it so happens that 2 records (referrenced by 2 persons) are stored in a single id which should be stored in different id.

What should I do to avoid this type of conflict ?

Thanks & Regards,

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AspDotNetDev23-May-11 22:01

AspDotNetDev

23-May-11 22:01

Use identity columns so that the ID's are generated automatically.

All Time Programming23-May-11 22:34

23-May-11 22:34

Thanks AspDotNetDev, Its a high-end reqd to generate ID programmatically and not auto.

Thanks & Regards,

Thomas Krojer23-May-11 23:13

Thomas Krojer

23-May-11 23:13

tell your customer that he made an request which is VERY EXPANSIVE to handle - he will agree with identity columns.

I cannot remember: What did I before google?

Dave Kreskowiak24-May-11 1:21

24-May-11 1:21

Well, then you have a serious design flaw in your database. One which will take a redesign to fix and a major overhaul of your code.

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Dave Kreskowiak

All Time Programming24-May-11 1:36

24-May-11 1:36

Thanks AspDotNet, Thomas, Dave.

>>One which will take a redesign to fix and a major overhaul of your code.
Dave, can you explain a bit on what are you trying to say and how and what factors to look for in the code.

Thanks & Regards,

Dave Kreskowiak24-May-11 1:55

24-May-11 1:55

Simple. Either you don't have a valid primary key (for example, the VIN is unique to each car, but you should never use it as a primary key) or your code is generating the key values (REALLY bad idea!).

In order to fix this you are either going to have to add proper primary keys to your tables, and update your code accordingly or you're going to have to add a layer to your database where every instance of your app is going to have to go through a singleton running on a server to have a primary key generated. Only one instance of your app is going to be able to call this singleton at a time which is going to kill your scalability. Oh, and you're still going to have to update your code to support this.

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Dave Kreskowiak

All Time Programming24-May-11 1:58