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@finsterthecat
Last active January 16, 2018 20:59
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' Alt-F11 in Word to get into VBA mode
' Then Insert>>Module
' Paste this in and Run
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 and footer.
Set rng = docOutline.Content
astrHeadings = _
docSource.GetCrossReferenceItems(wdRefTypeHeading)
Set regex = CreateObject("vbscript.regexp")
regex.Global = True
regex.pattern = "\d+\.\d+\s"
For intItem = LBound(astrHeadings) To UBound(astrHeadings)
' Get the text and the level.
strText = Trim$(astrHeadings(intItem))
intLevel = GetLevel(CStr(astrHeadings(intItem)))
If (intLevel = 2) Then
' Add the text to the document.
rng.InsertAfter regex.Replace(strText, "") & vbNewLine
' Set the style of the selected range and
' then collapse the range for the next entry.
' Another option for Style "Heading " & intLevel
rng.Style = "Normal"
rng.Collapse wdCollapseEnd
End If
Next intItem
End Sub
Private Function GetLevel(strItem As String) As Integer
' Return the heading level of a header from the
' array returned by Word.
' The number of leading spaces indicates the
' outline level (2 spaces per level: H1 has
' 0 spaces, H2 has 2 spaces, H3 has 4 spaces.
Dim strTemp As String
Dim strOriginal As String
Dim intDiff As Integer
' Get rid of all trailing spaces.
strOriginal = RTrim$(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
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