VB 6公共控件打开对话框出现问题，请帮助 [英] Weard problem with VB 6 common control open dialog please help

问题描述

有人可以帮助我理解为什么VB对话框控件正在执行此操作吗.

我有2个表格和1个模块

Form1
Form2
Module1

Module1代码为

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

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

Form1代码是

Form1 code is

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

表格2的代码是

Form 2 code is

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

当我以启动代码的形式运行项目Form 2时，stCheckSign函数的返回值为0

但是，如果我从表单1运行项目并从路径中选择文件，则返回值将更改为-22669

两种代码都在做相同的事情，唯一的区别是Form1从公共对话框show open命令获取文件路径参数，而Form 2中的路径是硬编码的.

每当您单击显示打开对话框上的打开按钮时，我都无法理解返回值会发生变化.同样，一旦将值更改为-22669，Form2也不起作用，我需要关闭该项目并重新打开它，Form 2才能再次起作用.

谁能告诉我为什么常见的对话框打开按钮弄乱了strCheckSign函数的返回值吗?


[edit]已添加代码块-OriginalGriff [/edit]

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]

推荐答案

尝试实验:
将代码从Form 1 Command1_click函数添加到Form 2 Comand_Click.

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

现在看看您得到的值:每对应该相同

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

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