how to pop-up the FILE OPEN dialog box ??? 1

[marco02]

Hi,

I'm trying to pop up the FILE OPEN common dialog box for my user. The code below doesn't work! (comes from the "Developer's Handbook: Vol 1)

Dim cdl as CommonDlg
Set cdl = New CommonDlg
cdl.ShowOpen

Bugs at the first line. Declaration not accepted !! ...

thanks for your help on this matter,

marco

 

[Kendel]

Put a common dialog on ur form and name it dlg1.

private sub CmdOpen_Click()
dlg.ShowOpen
end sub

 

[marco02]

i doesn't work Kendel,

moreover my calling this ShowOpen is not systematic in the procedure. It only happens in the specific case when the default path is not valid. So i don't want to attach it to a button.

Would my pbs have tsomething to do with a REFERENCE missing ? (Tools -> References ?)

thanks for your help!

marco

 

[marco02]

it doesn't work Kendel,

moreover my calling this ShowOpen is not systematic in the procedure. It only happens in the specific case when the default path is not valid. So i don't want to attach it to a button.

Would my pbs have tsomething to do with a REFERENCE missing ? (Tools -> References ?)

thanks for your help!

marco

 

[Kendel]

Doesn't work? Dialog doesn't popup? Did you put a dialog on your form? You need to add the Microsft Common Dialog Control 6.0 for the Components. To use the dialog, no need to add any refference.

With dlgCommon
.InitDir = App.Path
.DialogTitle = "Open File As ..."
.Filter = "All Files|*.*|Text Files|*.txt"
.ShowOpen
End With

 

[marco02]

IT WORKS !!!
sorry for the late answer guys i got sick ....

For the next guy, here is the complete detailed procedure:

1) In a Form (whatever form) Toolbox -> More control ->
put the "Microsoft Common Dialog Control 6.0" control
somewhere in the form.

2) Go in Access VB editor

3) Tools -> Reference -> check "Microsoft Common Dialog Control 6.0". Validate

4) Code as follow:

Dim commonDlg As New CommonDialog
commonDlg.ShowOpen

Thanks guys, marco.