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我正在創建一個.vbs文件,該文件應該打開訪問權限,並在內部訪問表單調用「問題詳細信息」,但傳遞參數,這意味着如果我在我的「問題」表中有10個問題爲每個文件創建一個vbs文件,單擊時應打開正確的記錄(對於表中的每個記錄將是一個ID)。到目前爲止,這是開放的訪問,它是打開表格(問題詳情),但它是空白的。我錯過了什麼?幫助,在這裏變得瘋狂...檢查下面的代碼。這裏奇怪的是,如果我再次雙擊它將刷新並打開正確的記錄,而無需打開任何窗口..我該如何解決這個問題?我不想做兩次:)訪問VBA傳遞參數vbs文件
Public Sub sendMRBmail(mrbid)
DoCmd.OpenForm "Issue Details", , , "[ID] = " & mrbid
End Sub
Private Sub Create_Click()
On Error GoTo Err_Command48_Click
Dim snid As Integer
snid = Me.ID
Dim filename As String
filename = "S:\Quality Control\vbs\QC" & snid & ".vbs"
Dim proc As String
proc = Chr(34) & "sendMRBmail" & Chr(34)
Dim strList As String
strList = "On Error Resume Next" & vbNewLine
strList = strList & "dim accessApp" & vbNewLine
strList = strList & "set accessApp = createObject(" & Chr(34) & "Access.Application" & Chr (34)")" & vbNewLine
strList = strList & "accessApp.OpenCurrentDataBase(" & Chr(34) & "S:\Quality Control\Quality DB\Quality Database.accdb" & Chr(34) & ")" & vbNewLine
strList = strList & "accessApp.Run " & proc & "," & Chr(34) & snid & Chr(34) & vbNewLine
strList = strList & "set accessApp = nothing" & vbNewLine
Open filename For Output As #1
Print #1, strList
Close #1
Err_Command48_Click:
If Err.Number <> 0 Then
MsgBox "Email Error #: " & Err.Number & ", " & "Description: " & Err.Description
Exit Sub
End If
End Sub
這裏面是什麼一個創建VBS文件
On Error Resume Next
dim accessApp
set accessApp = GetObject("S:\Quality Control\Quality DB\Quality Database.accdb")
accessApp.Run"sendMRBmail","231"
set accessApp = nothing
可能重複http://stackoverflow.com/questions/22646849/vbs-passing-parameters - 怪異行爲) – HansUp
它是一個dup,但我假設它是爲了添加正確的標籤而創建的。上面的代碼看起來像是在創建VBS文件。發佈其中一個VBS文件的內容以供審閱。 – Sorceri
是的,它是重複的,我在這裏移動它來添加正確的標籤,但沒有從那裏刪除。 – Shakka