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access vba close msgbox

geduo zhao answered on November 11, 2022 Popularity 3/10 Helpfulness 1/10

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Declare Function MessageBoxTimeout Lib "user32.dll" Alias "MessageBoxTimeoutA" ( _
    ByVal hwnd As Long, _
    ByVal lpText As String, _
    ByVal lpCaption As String, _
    ByVal uType As Long, _
    ByVal wLanguageID As Long, _
    ByVal lngMilliseconds As Long) As Long

Private Declare Function FindWindow Lib "user32" Alias "FindWindowA" ( _
    ByVal lpClassName As String, _
    ByVal lpWindowName As String) As Long

Public Function PopUpBox(Optional stMessage As String _
        = "Yes or No? leaving this window for 1 min is the same as clicking Yes.", _
        Optional stTitle As String = "PopUp Window", _
        Optional HalfSecTimer As Long = 120, Optional lgVBmsgType As Long = vbYesNo) As Long

    Dim RetVal As Long

    HalfSecTimer = HalfSecTimer * 500
    RetVal = MessageBoxTimeout(FindWindow(vbNullString, Title), stMessage, stTitle, lgVBmsgType, _ 
        0, HalfSecTimer)

    PopUpBox = RetVal
End Function

Popularity 3/10 Helpfulness 1/10 Language vb

Source: stackoverflow.com

Tags: msgbox vba

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Contributed on Nov 11 2022

geduo zhao

0 Answers Avg Quality 2/10

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With CreateObject("Scripting.WsShell")
    .Popup "Watch me disappear in 5 seconds", 5, Application.Name & ": test", vbInformation + vbOkCancel
End With

Popularity 1/10 Helpfulness 1/10 Language vb

Source: stackoverflow.com

Tags: msgbox vba

Link to this answer
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Contributed on Nov 11 2022

geduo zhao

0 Answers Avg Quality 2/10

access vba close msgbox

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