[Help] Open a file automatically [VB6] [solved]

Hello, I am trying to find a way that when I push a button in Visual Basic, a file will open.
I have tried this code ( Dim RetVal
RetVal = Shell("C:\Users\User\Desktop\Mp3cyprusmaniacs_cont est\ini.bat", vbWindowState) ) and it works fine, but it is with a path..and when the file is not there, a message pops up with "error 53 - file not found"..
I would like to know if there is a way to open that file without using any path..
Thanks

uh just add a openfiledialog and add a button and double click on the button and type:

OpenFileDialog1.showdialog

its that simple lol

can you tell me what to do in more details because it is my first day in vb6? :P @justiman

vb6? the code i have dont work for vb6... get vb 2010 its ALOT better. then do what i said and there you go... or i could make it for you and send it to you

So you want to select a file and open it?
Its easy! You need a CommonDialogControl

Code:

With CommonDialog1
    .FileName = ""
    .ShowOpen
    
    If .FileName <> "" Then
        RetVal = Shell("CommonDialog1.FileName", vbWindowState)
    End If
End With

Originally Posted by Vanoq

So you want to select a file and open it?
Its easy! You need a CommonDialogControl

Code:

With CommonDialog1
    .FileName = ""
    .ShowOpen
    
    If .FileName <> "" Then
        RetVal = Shell("CommonDialog1.FileName", vbWindowState)
    End If
End With

my way is easier...

will this work for vb6? @Vanoq

Originally Posted by lelelololele

will this work for vb6? @Vanoq

i can just make one for you in like 30 seconds and send it to you

do you mean make a form in VB6 with a button which opens a file?

Originally Posted by justiman

i can just make one for you in like 30 seconds and send it to you

He wants to learn himself, he doesn't want to just take something he didn't make

Originally Posted by lelelololele

will this work for vb6? @Vanoq

Of course. My snippet is VB6

Originally Posted by lelelololele

do you mean make a form in VB6 with a button which opens a file?

yea thats what you do o.0

hmmm...I don't know..something is wrong here

-.-

Code:

Private Sub Command1_Click()
With CommonDialog1
    .FileName = "ini.bat"
    .ShowOpen
    
    If .FileName <> "ini.bat" Then
        RetVal = Shell("CommonDialog1.FileName", vbWindowState)
    End If
End With
End Sub

what did I do wrong?

Dont change the filename. You can use the code as i posted it.

Code:

.Filename = ""

sry for being so annoying, but, I am a newbie in vb6

so, i just create a button and put this code?

Code:

rivate Sub Command1_Click()
With CommonDialog1
    .FileName = ""
    .ShowOpen
    
    If .FileName <> "" Then
        RetVal = Shell("CommonDialog1.FileName", vbWindowState)
    End If
End With
End Sub

if yes, there is an error..

[Help] Open a file automatically [VB6] [solved]

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