Weard problem with VB 6 common control open dialog please help

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Can someone help me understand this why VB dialog control is doing this.

I have 2 Forms and 1 Module

Form1
Form2
Module1

Module1 code is

VB

Public Declare Function stCheckSign Lib "NEWSTSIGN.dll" (ByVal szpath As String) As Integer

Form1 code is

VB

Private Sub Command1_Click()

Dim strFileToOpen As String
On Error GoTo Command1_Click_Exit
    With CommonDialog1
    .CancelError = True
    .InitDir = mstrLastDir
    .Flags = cdlOFNHideReadOnly & &H8
    .FileName = ""
    .Filter = "Text Files(*.mll)|*.txt|All Files(*.*)|*.*"
    .ShowOpen
    strFileToOpen = .FileName
    End With
On Error GoTo Command1_Click_Error

MsgBox Module1.stCheckSign(strFileToOpen)

Exit Sub
Command1_Click_Error:
MsgBox "The following error has occurred:" & vbNewLine _
& "Err # " & Err.Number & " - " & Err.Description, _
vbCritical, _
"Open Error"
Command1_Click_Exit:

End Sub

Form 2 code is

VB

Public spath As String

Private Sub Command1_Click()
MsgBox Module1.stCheckSign(spath)
End Sub

Private Sub Form_Load()
spath = C:\Users\anand.sharma\Desktop\Sign\1011200218.mll
End Sub

When I run the project Form 2 as start-up code works and the return value come 0 from the stCheckSign function

But if I run the project from form 1 and selects the file from the path the return value changes to -22669

Both code are doing the same thing only the difference is Form1 is getting the file path arrgument from common dialog show open command and in Form 2 the path is hardcoded.

I can't able to understand the return value gets changed whenever you click on open button on the show open dialog. Also once the value is changed to -22669 Form2 also does not work and I need to close the project and reopen it for Form 2 to work again.

Can anyone show me some light why the common dialog open button is messing the return value of strCheckSign function?

[edit]Code block added - OriginalGriff[/edit]

Posted 21-Apr-11 21:17pm

asmanu

Updated 3-Jul-11 18:37pm

Аslam Iqbal

v4

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Christian Graus 3-Jul-11 22:46pm

Can someone explain to me why anyone would use VB6 in this day and age ?

1 solution

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OriginalGriff · Answer 1 · 2011-04-21T22:04:00

Solution 1

Try an experiment:
Add the code from your Form 1 Command1_click function to your Form 2 Comand_Click.

Private Sub Command1_Click()

Dim strFileToOpen As String
On Error GoTo Command1_Click_Exit
    With CommonDialog1
    .CancelError = True
    .InitDir = mstrLastDir
    .Flags = cdlOFNHideReadOnly & &H8
    .FileName = ""
    .Filter = "Text Files(*.mll)|*.txt|All Files(*.*)|*.*"
    .ShowOpen
    strFileToOpen = .FileName
    End With
On Error GoTo Command1_Click_Error
MsgBox strFileToOpen
MsgBox spath
MsgBox Module1.stCheckSign(strFileToOpen)
MsgBox Module1.stCheckSign(spath)
End Sub

Now look at the values you are getting: each pair should be identical

Posted 21-Apr-11 22:04pm

OriginalGriff

Comments

asmanu 22-Apr-11 5:30am

Out put of above is

C:\Users\anand.sharma\Desktop\Sign\1011200218.mll
C:\Users\anand.sharma\Desktop\Sign\1011200218.mll
-20669
-20669

but if I reverse the function call

like this

Public spath As String

Private Sub Command1_Click()
Debug.Print spath
Debug.Print Module1.stCheckSign(spath)

Dim strFileToOpen As String
With CommonDialog1
.CancelError = True
.InitDir = mstrLastDir
.Flags = cdlOFNHideReadOnly & &H8
.FileName = ""
.Filter = "Text Files(*.mll)|*.txt|All Files(*.*)|*.*"
.ShowOpen
strFileToOpen = .FileName
End With

Debug.Print strFileToOpen
Debug.Print Module1.stCheckSign(strFileToOpen)
End Sub

Then the output is

C:\Users\anand.sharma\Desktop\Sign\1011200218.mll
0
C:\Users\anand.sharma\Desktop\Sign\1011200218.mll
-20669

OriginalGriff 22-Apr-11 5:48am

Interesting. Do you have the source for the stCheckSign function in NEWSTSIGN.DLL?