如何打开指定其路径的工作簿

srt*_*srt 5 excel vba excel-vba

Sub openwb()  
    ChDir "E:\sarath\PTMetrics\20131004\D8 L538-L550 16MY"
    Workbooks("D8 L538-L550_16MY_Powertrain Metrics_20131002.xlsm").Open    
End sub
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在这里,我Subscript out of range在第3行收到错误.如何打开指定其路径的工作簿?

Sid*_*out 19

Workbooks.open("E:\sarath\PTMetrics\20131004\D8 L538-L550 16MY\D8 L538-L550_16MY_Powertrain Metrics_20131002.xlsm")

或者,以更有条理的方式......

Sub openwb()
    Dim sPath As String, sFile As String
    Dim wb As Workbook

    sPath = "E:\sarath\PTMetrics\20131004\D8 L538-L550 16MY\"
    sFile = sPath & "D8 L538-L550_16MY_Powertrain Metrics_20131002.xlsm"

    Set wb = Workbooks.Open(sFile)
End Sub
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Pun*_* GP 5

您还可以通过提示打开所需的文件,这有助于从不同路径和不同文件中选择文件。

Sub openwb()
Dim wkbk As Workbook
Dim NewFile As Variant

NewFile = Application.GetOpenFilename("microsoft excel files (*.xlsm*), *.xlsm*")

If NewFile <> False Then
Set wkbk = Workbooks.Open(NewFile)
End If
End Sub
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