Question

出于某种原因，我在第一行收到错误。一切都正确命名。我使用按钮创建器向导打开我想要的报告。

Private Sub cmdReport2013_Click()
On Error GoTo Err_cmdReport2013_Click


    stDocName = "rptAnalyst_Comp_2013"


Exit_cmdReport2013_Click:
    Exit Sub

Err_cmdReport2013_Click:
    MsgBox Err.Description
    Resume Exit_cmdReport2013_Click

End Sub

编辑：编译错误：未定义变量

EDIT2：此报告基于查询。这有关系吗？

EDIT3：

Private Sub cmdReport2013_Click()
Dim stDocName As String

On Error GoTo Err_cmdReport2013_Click

    stDocName = "rptAnalyst_Comp_2013"
    DoCmd.OpenReport (stDocName, acViewNormal)


Exit_cmdReport2013_Click:
    Exit Sub

Err_cmdReport2013_Click:
    MsgBox Err.Description
    Resume Exit_cmdReport2013_Click

End Sub

它说DoCmd.OperReport部分的语法错误。

编辑4：解决了语法的工作代码。

Private Sub cmdReport2013_Click()
Dim stDocName As String

On Error GoTo Err_cmdReport2013_Click

    stDocName = "rptAnalyst_Comp_2013"
    DoCmd.OpenReport stDocName, acViewReport


Exit_cmdReport2013_Click:
    Exit Sub


Err_cmdReport2013_Click:
    MsgBox Err.Description
    Resume Exit_cmdReport2013_Click

End Sub

Answer 1

Private Sub cmdReport2013_Click()
On Error GoTo Err_cmdReport2013_Click

docmd.openReport "rptAnalyst_Comp_2013"


Exit_cmdReport2013_Click:
Exit Sub

Err_cmdReport2013_Click:
MsgBox Err.Description
Resume Exit_cmdReport2013_Click

End Sub

Answer 2

Private Sub cmdReport2013_Click()
Dim stDocName as String

On Error GoTo Err_cmdReport2013_Click

    stDocName = "rptAnalyst_Comp_2013"
    DoCmd.OpenReport (stDocName)


Exit_cmdReport2013_Click:
    Exit Sub

Err_cmdReport2013_Click:
    MsgBox Err.Description
    Resume Exit_cmdReport2013_Click

End Sub

Access 2007，VBA：单击以打开报表时出错

2 个答案: