Use system open with dialogue box

[jobs_arghya]

Hi All,

Hope u all are in fine tune. Friends I am just trying to do some R & D with the system dialogue boxes. can anyone help me to know how I can use the system Open With Dialogue Box to open a file in the VB Exe.

For example:

Suppose I am having a form with a textbox or a RTFBox. Now I will click on a text file on the desktop and indicate it to open in my form using the system Open With Dialogue Box. Is this possible. If Yes the how I can do it. Please anyone give me some idea.

Thanx in advance

Arghya

[Siddharth Rout]

Places this in a module

Public Declare Function GetOpenFileName Lib "comdlg32.dll" _
Alias "GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long
 
Public Type OPENFILENAME
    lStructSize As Long
    hwndOwner As Long
    hInstance As Long
    lpstrFilter As String
    lpstrCustomFilter As String
    nMaxCustFilter As Long
    nFilterIndex As Long
    lpstrFile As String
    nMaxFile As Long
    lpstrFileTitle As String
    nMaxFileTitle As Long
    lpstrInitialDir As String
    lpstrTitle As String
    flags As Long
    nFileOffset As Integer
    nFileExtension As Integer
    lpstrDefExt As String
    lCustData As Long
    lpfnHook As Long
    lpTemplateName As String
End Type
 
Public Function SelectFileOpenDialog()
Dim strTemp, strTemp1, pathStr As String
Dim i, n, j As Long
Dim OpenFile As OPENFILENAME
Dim lReturn As Long
Dim sFilter As String
Dim Fname As String
 
OpenFile.lStructSize = Len(OpenFile)
sFilter = "Text Files (*.txt)" & Chr(0) & "*.TXT" & Chr(0)
OpenFile.lpstrFilter = sFilter
OpenFile.nFilterIndex = 1
OpenFile.lpstrFile = String(257, 0)
OpenFile.nMaxFile = Len(OpenFile.lpstrFile) - 1
OpenFile.lpstrFileTitle = OpenFile.lpstrFile
OpenFile.nMaxFileTitle = OpenFile.nMaxFile
OpenFile.lpstrInitialDir = "C:\"
OpenFile.lpstrTitle = "Select File"
OpenFile.flags = 0
 
lReturn = GetOpenFileName(OpenFile)
 
If lReturn = 0 Then
MsgBox "You didn't select any file"
Else
 
Fname = Trim$(OpenFile.lpstrFileTitle) ' copy the filename to "Fname"
 
n = FileLen(OpenFile.lpstrFile) 'length of the file
End If
 
End Function

Place this in the command button click

Private Sub Command1_Click()
    SelectFileOpenDialog
End Sub

Please amend the code as per your requirements...

[westconn1]

to display the open with dialog is the default action for shellexecute if the file type is unregistered, to force the dialog to display you need to use shellexecuteex like this

Const SEE_MASK_INVOKEIDLIST = &HC
Const SEE_MASK_NOCLOSEPROCESS = &H40
Const SEE_MASK_FLAG_NO_UI = &H400
Type SHELLEXECUTEINFO
    cbSize As Long
    fMask As Long
    hwnd As Long
    lpVerb As String
    lpFile As String
    lpParameters As String
    lpDirectory As String
    nShow As Long
    hInstApp As Long
    lpIDList As Long
    lpClass As String
    hkeyClass As Long
    dwHotKey As Long
    hIcon As Long
    hProcess As Long
End Type
Private Declare Function ShellExecuteEx Lib "shell32.dll" (SEI As SHELLEXECUTEINFO) As Long
 
Sub openwith()
    Dim SEI As SHELLEXECUTEINFO
    Dim r As Long, filename As String
    filename = "c:\test\123.txt"
    
    With SEI
        'Set the structure's size
        .cbSize = Len(SEI)
        'Seet the mask
        .fMask = SEE_MASK_NOCLOSEPROCESS Or _
         SEE_MASK_INVOKEIDLIST Or SEE_MASK_FLAG_NO_UI
        'Set the owner window
        .hwnd = 0 ' or someform.hwnd
        'Show the properties
        .lpVerb = "openas"
        'Set the filename
        .lpFile = filename
        .lpParameters = vbNullChar
        .lpDirectory = vbNullChar
        .nShow = 0
        .hInstApp = 0
        .lpIDList = 0
    End With
    r = ShellExecuteEx(SEI)
End Sub