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Accepted
Charles Kenyon
Asked: 2024-12-16 12:51:05 +0800 CST

循环通过样式来更改 NextParagraphStyle 时出现问题 - 运行时错误 91 - Word vba

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循环通过样式来更改 NextParagraphStyle 时出现问题 - 运行时错误 91 - Word vba

编辑：Tim Williams 发现了这个问题。我没有将 Set 用于对象变量。

我的宏是：

Sub StyleFollowingBodyText()
    ' Charles Kenyon 15 December 2024
    ' Set the following style for most QuickStyle paragraph styles to be Body Text
    Dim StyleCount As Long
    Dim thisStyle As Style
    Dim iCount As Long
    '
    With ActiveDocument
        Let StyleCount = .Styles.Count
        For iCount = 1 To StyleCount
            Let thisStyle = .Styles(iCount)
            If thisStyle.QuickStyle = True Then
                If thisStyle.Type = wdStyleTypeParagraph Or wdStyleTypeLinked Then
                    If thisStyle.NameLocal <> "Normal" Then
                        thisStyle.NextParagraphStyle = "Body Text"
                    End If
                End If
            End If
        Next iCount
    End With
End Sub

运行它时出现以下错误消息：

错误 91 的屏幕截图

此错误消息似乎有点“包罗万象”，并没有多大帮助。正如 Tim Williams 指出的那样，问题不在于对对象变量使用 Set 命令。

它正在 停止Let thisStyle = .Styles(iCount)。

宏的目的是将大部分段落 QuickStyles 更改为具有以下样式的正文样式。我试图更改 50 多个 [快速] 样式集，因为我不想让我的大部分文档使用普通样式。

vba

2 个回答

  1. Best Answer

    例如：

    Sub StyleFollowingBodyText()
Dim s As Long
With ActiveDocument
  For s = 1 To .Styles.Count
    With .Styles(s)
      If .QuickStyle = True Then
        If .NameLocal <> "Normal" Then
          If .Type = wdStyleTypeParagraph Or .Type = wdStyleTypeParagraph Then .NextParagraphStyle = "Body Text"
        End If
      End If
    End With
  Next
End With
End Sub
    • 1

  2. 我尝试丢弃问题陈述并且不为样式分配变量，但这样做却有效。

    我只是使用.Styles(iCount)而不是变量StyleCount。

    Sub StyleFollowingBodyText2()
    ' Charles Kenyon 15 December 2024
    ' Set the following style for most QuickStyle paragraph styles to be Body Text
    Dim StyleCount As Long
    Dim thisStyle As Style
    Dim iCount As Long
    '
    With ActiveDocument
        Let StyleCount = .Styles.Count
        For iCount = 1 To StyleCount
'            Let thisStyle = .Styles(iCount)
            If .Styles(iCount).QuickStyle = True Then
                If .Styles(iCount).Type = wdStyleTypeParagraph Or wdStyleTypeLinked Then
                    If .Styles(iCount).NameLocal <> "Normal" Then
                        .Styles(iCount).NextParagraphStyle = "Body Text"
                    End If
                End If
            End If
        Next iCount
    End With
End Sub

    Tim Williams 对我的问题的评论非常准确，解释了我为什么会收到错误。我使用“Let”而不是“Set”作为对象变量。

    我还发现了 Jay Freedman 编写的以下宏，用于处理 Normal 模板的样式。

    Sub StylesToBodyText()
    ' Jay Freedman 2023-12-18
    ' Change next style to Body Text, Based on style to Body Text in Normal
    ' You can use the macro below to change all of the "based on Normal" and "following paragraph Normal" styles
    '    in the Normal.dotm template to be based on and followed by Body Text.
    ' Before you run the macro, save a copy of the Normal.dotm template in some folder other
    '    than the Templates folder, as a backup in case something goes wrong.
    ' I deliberately did not include a Save in the macro,
    '    as you should examine the results in the Styles pane before you manually save and close the Normal.dotm window.
    '    If anything is amiss, close the window without saving and see where the problem occurred.
    ' https://answers.microsoft.com/en-us/msoffice/forum/all/how-to-change-the-style-for-following-paragraph/64644dfc-92a7-44b7-baa8-53fe622344b8
    '
    Dim sty As Style
    NormalTemplate.OpenAsDocument ' Comment this line out to apply active document or template and not normal template
    For Each sty In ActiveDocument.Styles
        If sty.Type = wdStyleTypeParagraph Then
            If sty.BaseStyle = "Normal" Then
                sty.BaseStyle = "Body Text"
            End If
'
            If sty.NextParagraphStyle = "Normal" Then
                sty.NextParagraphStyle = "Body Text"
            End If
        End If
    Next sty
    '   After checking styles, save and close normal template
End Sub

    我可能会采用这种结构，但我认为它也可以帮助其他人。他使用的是 For...Each 结构，而不是计数。

    • 0

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