您的代码将控件引用视为名称(字符串变量)，因此您将获得类型名称 (System.windows.datagridview) 而不是控件的名称.由于模板 DGV 恰好有名称，请使用它:

Your code is treating a control reference as if it were a name (string variable), so you get the Type name (System.windows.datagridview) rather than the name of the control. Since the template DGV happarently has the name, use it:

Dim myGridName As String         ' name is a String, not DGV
Dim myGridItemIndex As Integer
Dim myDGV As DataGridView        ' no NEW - not creating a new one
                                 ' just a reference var
'...
' this is now a For/Each loop
For Each row As DataGridViewRow in DsSaveTemplates.tblTemplates.Rows
    myGridName  = row.Cell("GridName")
    myGridItemIndex  = row.Cell("GridItemIndex")

    ' assuming this code is in a Form:
    ' get a reference to the control
    myDGV = CType(Me.Controls(myGridName), DataGridView)
    ' talk to it like a DGV:
    myDGV.Item(0, myGridItemIndex).Value = True

Next

注意:Option Strict 可能需要对名称和索引进行一些转换

Note: Option Strict would likely require some conversions for the name and index

如果 DGV 位于面板或选项卡等容器控件中，您必须找到"该控件，因为它们将位于那个控件的集合中，而不是表单的集合中.而不是 myDGV = CType(Me.Controls(myGridName), DataGridView):

If the DGV(s) reside in container controls like Panels or Tabs, you have to "find" the control because they will be in that control's collection, not the Form's. Instead of myDGV = CType(Me.Controls(myGridName), DataGridView):

' have the form search for the ctl by name
' the TRUE param tells it to search child controls like panels 
Dim tmpArry = Me.Controls.Find(myGridName, True)

' test for a return 
If tmpArry.Length > 0 Then
    ' tmpArry will be Type Control, so cast it 
    myDGV = CType(tmpArry(0), DataGridView)
End If

这通常从一开始就更好，因此您不必在编码时记住表单的布局.

This is usually better from the start so you do not have to remember how the form is laid out when coding.