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Word文書から見出しを取得する

VBAを使用してWord文書のすべての見出しの一覧を取得するにはどうすればよいですか?

22
user35762

あなたはこれを意味します createOutline 関数(実際にはすべての見出しをソースWord文書から新しいWord文書にコピーします):

(私はastrHeadings = _docSource.GetCrossReferenceItems(wdRefTypeHeading)関数がこのプログラムのキーであり、あなたが求めているものを取得できるようにするはずです)

_Public Sub CreateOutline()
    Dim docOutline As Word.Document
    Dim docSource As Word.Document
    Dim rng As Word.Range

    Dim astrHeadings As Variant
    Dim strText As String
    Dim intLevel As Integer
    Dim intItem As Integer

    Set docSource = ActiveDocument
    Set docOutline = Documents.Add

    ' Content returns only the main body of the document, not the headers/footer.        
    Set rng = docOutline.Content
    ' GetCrossReferenceItems(wdRefTypeHeading) returns an array with references to all headings in the document
    astrHeadings = docSource.GetCrossReferenceItems(wdRefTypeHeading)

    For intItem = LBound(astrHeadings) To UBound(astrHeadings)
        ' Get the text and the level.
        strText = Trim$(astrHeadings(intItem))
        intLevel = GetLevel(CStr(astrHeadings(intItem)))

        ' Add the text to the document.
        rng.InsertAfter strText & vbNewLine

        ' Set the style of the selected range and
        ' then collapse the range for the next entry.
        rng.Style = "Heading " & intLevel
        rng.Collapse wdCollapseEnd
    Next intItem
End Sub

Private Function GetLevel(strItem As String) As Integer
    ' Return the heading level of a header from the
    ' array returned by Word.

    ' The number of leading spaces indicates the
    ' outline level (2 spaces per level: H1 has
    ' 0 spaces, H2 has 2 spaces, H3 has 4 spaces.

    Dim strTemp As String
    Dim strOriginal As String
    Dim intDiff As Integer

    ' Get rid of all trailing spaces.
    strOriginal = RTrim$(strItem)

    ' Trim leading spaces, and then compare with
    ' the original.
    strTemp = LTrim$(strOriginal)

    ' Subtract to find the number of
    ' leading spaces in the original string.
    intDiff = Len(strOriginal) - Len(strTemp)
    GetLevel = (intDiff / 2) + 1
End Function
_

2018年3月6日@kolによる更新

astrHeadingsは配列です(IsArrayTrueを返し、TypeNameString()を返します)_type mismatch_エラーが発生しますVBScriptでその要素にアクセスしようとすると(Windows 10 Pro 1709 16299.248のv5.8.16384)。 WordのVBAエディターで同じコードを実行すると要素にアクセスできるため、これはVBScript固有の問題である必要があります。 TOCの行を反復することになりました。VBScriptからでも機能するためです。

_For Each Paragraph In Doc.TablesOfContents(1).Range.Paragraphs
  WScript.Echo Paragraph.Range.Text
Next
_
18
VonC

見出しのリストを取得する最も簡単な方法は、ドキュメントの段落をループすることです。次に例を示します。

 Sub ReadPara()

    Dim DocPara As Paragraph

    For Each DocPara In ActiveDocument.Paragraphs

     If Left(DocPara.Range.Style, Len("Heading")) = "Heading" Then

       Debug.Print DocPara.Range.Text

     End If

    Next


End Sub

ところで、段落範囲の最後の文字を削除することをお勧めします。それ以外の場合、文字列をメッセージボックスまたはドキュメントに送信すると、追加の制御文字が表示されます。例えば:

Left(DocPara.Range.Text, len(DocPara.Range.Text)-1)
14
JonnyGold

このマクロは私にとってはうまく機能しました(Word 2010)。機能を少し拡張しました。これで、ユーザーに最小レベルの入力を求め、そのレベルより下の小見出しを抑制します。

Public Sub CreateOutline()
' from http://stackoverflow.com/questions/274814/getting-the-headings-from-a-Word-document
    Dim docOutline As Word.Document
    Dim docSource As Word.Document
    Dim rng As Word.Range

    Dim astrHeadings As Variant
    Dim strText As String
    Dim intLevel As Integer
    Dim intItem As Integer
    Dim minLevel As Integer

    Set docSource = ActiveDocument
    Set docOutline = Documents.Add

    minLevel = 1  'levels above this value won't be copied.
    minLevel = CInt(InputBox("This macro will generate a new document that contains only the headers from the existing document. What is the lowest level heading you want?", "2"))

    ' Content returns only the
    ' main body of the document, not
    ' the headers and footer.
    Set rng = docOutline.Content
    astrHeadings = _
     docSource.GetCrossReferenceItems(wdRefTypeHeading)

    For intItem = LBound(astrHeadings) To UBound(astrHeadings)
        ' Get the text and the level.
        strText = Trim$(astrHeadings(intItem))
        intLevel = GetLevel(CStr(astrHeadings(intItem)))

        If intLevel <= minLevel Then

            ' Add the text to the document.
            rng.InsertAfter strText & vbNewLine

            ' Set the style of the selected range and
            ' then collapse the range for the next entry.
            rng.Style = "Heading " & intLevel
            rng.Collapse wdCollapseEnd
        End If
    Next intItem
End Sub

Private Function GetLevel(strItem As String) As Integer
    ' from http://stackoverflow.com/questions/274814/getting-the-headings-from-a-Word-document
    ' Return the heading level of a header from the
    ' array returned by Word.

    ' The number of leading spaces indicates the
    ' outline level (2 spaces per level: H1 has
    ' 0 spaces, H2 has 2 spaces, H3 has 4 spaces.

    Dim strTemp As String
    Dim strOriginal As String
    Dim intDiff As Integer

    ' Get rid of all trailing spaces.
    strOriginal = RTrim$(strItem)

    ' Trim leading spaces, and then compare with
    ' the original.
    strTemp = LTrim$(strOriginal)

    ' Subtract to find the number of
    ' leading spaces in the original string.
    intDiff = Len(strOriginal) - Len(strTemp)
    GetLevel = (intDiff / 2) + 1
End Function
2
joshoff

すべての見出しを抽出する最速の方法(LEVEL5まで)。

Sub EXTRACT_HDNGS()
Dim WDApp As Word.Application    'Word APP
Dim WDDoc As Word.Document       'Word DOC

Set WDApp = Word.Application
Set WDDoc = WDApp.ActiveDocument

For Head_n = 1 To 5
Head = ("Heading " & Head_n)
WDApp.Selection.HomeKey wdStory, wdMove

    Do
       With WDApp.selection
      .MoveStart Unit:=wdLine, Count:=1    
      .Collapse Direction:=wdCollapseEnd
       End with
        With WDApp.Selection.Find
          .ClearFormatting:          .text = "":     
          .MatchWildcards = False:   .Forward = True
          .Style = WDDoc.Styles(Head)
         If .Execute = False Then GoTo Level_exit
            .ClearFormatting
        End With

       Heading_txt = RemoveSpecialChar(WDApp.Selection.Range.text, 1):              Debug.Print Heading_txt
       Heading_lvl = WDApp.Selection.Range.ListFormat.ListLevelNumber:              Debug.Print Heading_lvl
       Heading_lne = WDDoc.Range(0, WDApp.Selection.Range.End).Paragraphs.Count:    Debug.Print Heading_lne
       Heading_pge = WDApp.Selection.Information(wdActiveEndPageNumber):            Debug.Print Heading_pge

       If Wdapp.Selection.Style = "Heading 1" Then GoTo Level_exit
       Wdapp.Selection.Collapse Direction:=wdCollapseStart
   Loop
Level_exit:
Next Head_n

End Sub
1
dxc

WikiがVonCの回答についてコメントした後、これが私のために働いたコードです。機能が速くなります。

Public Sub CopyHeadingsInNewDoc()
    Dim docOutline As Word.Document
    Dim docSource As Word.Document
    Dim rng As Word.Range

    Dim astrHeadings As Variant
    Dim strText As String
    Dim longLevel As Integer
    Dim longItem As Integer

    Set docSource = ActiveDocument
    Set docOutline = Documents.Add

    ' Content returns only the
    ' main body of the document, not
    ' the headers and footer.
    Set rng = docOutline.Content
    astrHeadings = _
     docSource.GetCrossReferenceItems(wdRefTypeHeading)

    For intItem = LBound(astrHeadings) To UBound(astrHeadings)
        ' Get the text and the level.
        strText = Trim$(astrHeadings(intItem))
        intLevel = GetLevel(CStr(astrHeadings(intItem)))

        ' Add the text to the document.
        rng.InsertAfter strText & vbNewLine

        ' Set the style of the selected range and
        ' then collapse the range for the next entry.
        rng.Style = "Heading " & intLevel
        rng.Collapse wdCollapseEnd
    Next intItem
End Sub

Private Function GetLevel(strItem As String) As Integer
    ' Return the heading level of a header from the
    ' array returned by Word.

    ' The number of leading spaces indicates the
    ' outline level (2 spaces per level: H1 has
    ' 0 spaces, H2 has 2 spaces, H3 has 4 spaces.

    Dim strTemp As String
    Dim strOriginal As String
    Dim longDiff As Integer

    ' Get rid of all trailing spaces.
    strOriginal = RTrim$(strItem)

    ' Trim leading spaces, and then compare with
    ' the original.
    strTemp = LTrim$(strOriginal)

    ' Subtract to find the number of
    ' leading spaces in the original string.
    longDiff = Len(strOriginal) - Len(strTemp)
    GetLevel = (longDiff / 2) + 1
End Function
1
MagTun

ホイールを何度も再発明するのはなぜですか?

「すべての見出しのリスト」は、ドキュメントの標準的なWordインデックスです。

これは、ドキュメントにインデックスを追加しながらマクロを記録することで得られたものです。

Sub Macro1()
    ActiveDocument.TablesOfContents.Add Range:=Selection.Range, _
        RightAlignPageNumbers:=True, _
        UseHeadingStyles:=True, _
        UpperHeadingLevel:=1, _
        LowerHeadingLevel:=5, _
        IncludePageNumbers:=True, _
        AddedStyles:="", _
        UseHyperlinks:=True, _
        HidePageNumbersInWeb:=True, _
        UseOutlineLevels:=True
End Sub
1
jumpjack

ドキュメントに目次を作成してコピーすることもできます。これにより、para refとタイトルが区別されます。これは、別のコンテキストで表示する必要がある場合に便利です。ドキュメントに目次が不要な場合は、コピーと貼り付けの後に削除してください。 JK。

0
JoeK