I looked up some examples, but I cannot quite understand how the Range object works. I am trying to loop through each of my headings (of level 4) and have a nested loop that looks through all the tables in between the headings. I cannot figure out how to set that specific range, so any help will be greatly appreciated.
Dim myHeadings As Variant
myHeadings = ActiveDocument.GetCrossReferenceItems(wdRefTypeHeading)
For iCount = LBound(myHeadings) To UBound(myHeadings)
level = getLevel(CStr(myHeadings(iCount)))
If level = 4 Then
'This is where I want to set a range between myHeadings(iCount) to myHeadings(iCount+1)
set aRange = ??
End If
Next iCount
You are on the right track here. The myHeadings variable you have simply gives a list of the strings of the Level 4 Headings in the document. What you need to do is then search the document for those strings to get the range of the Level 4 Headings.
Once you have the range of each of the headings you can check for the tables in the range between these headings. I've modified your code slightly to do this. Also note its good practice to put Option Explicit at the top of your module to ensure all variables are declared.
My code will tell you how many tables are between each of the Level 4 headings. NOTE: It does not check between the last heading and the end of the document, I'll leave that up to you ;)
Sub DoMyHeadings()
Dim iCount As Integer, iL4Count As Integer, Level As Integer, itabCount As Integer
Dim myHeadings As Variant, tbl As Table
Dim Level4Heading() As Range, rTableRange As Range
myHeadings = ActiveDocument.GetCrossReferenceItems(wdRefTypeHeading)
'We want to move to the start of the document so we can loop through the headings
Selection.HomeKey Unit:=wdStory
For iCount = LBound(myHeadings) To UBound(myHeadings)
Level = getLevel(CStr(myHeadings(iCount)))
If Level = 4 Then
'We can now search the document to find the ranges of the level 4 headings
With Selection.Find
.ClearFormatting 'Always clear find formatting
.Style = ActiveDocument.Styles("Heading 4") 'Set the heading style
.Text = VBA.Trim$(myHeadings(iCount)) 'This is the heading text (trim to remove spaces)
.Replacement.Text = "" 'We are not replacing the text
.Forward = True 'Move forward so we can each consecutive heading
.Wrap = wdFindContinue 'Continue to the next find
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
.Execute
End With
'Just make sure the text matches (it should be I have a habit of double checking
If Selection.Text = VBA.Trim$(myHeadings(iCount)) Then
iL4Count = iL4Count + 1 'Keep a counter for the L4 headings for redim
ReDim Preserve Level4Heading(1 To iL4Count) 'Redim the array keeping existing values
Set Level4Heading(iL4Count) = Selection.Range 'Set the range you've just picked up to the array
End If
End If
Next iCount
'Now we want to loop through all the Level4 Heading Ranges
For iCount = LBound(Level4Heading) To UBound(Level4Heading) - 1
'Reset the table counter
itabCount = 0
'Use the start of the current heading and next heading to get the range in between which will contain the tables
Set rTableRange = ActiveDocument.Range(Level4Heading(iCount).Start, Level4Heading(iCount + 1).Start)
'Now you have set the range in the document between the headings you can loop through
For Each tbl In rTableRange.Tables
'This is where you can work your table magic
itabCount = itabCount + 1
Next tbl
'Display the number of tables
MsgBox "You have " & itabCount & " table(s) between heading " & Level4Heading(iCount).Text & " And " & Level4Heading(iCount + 1).Text
Next iCount
End Sub
You could jump from one heading to the next using Goto. See below how to loop through level 4 headings.
Dim heading As Range
Set heading = ActiveDocument.Range(start:=0, End:=0)
Do ' Loop through headings
Dim current As Long
current = heading.start
Set heading = heading.GoTo(What:=wdGoToHeading, Which:=wdGoToNext)
If heading.start = current Then
' We haven't moved because there are no more headings
Exit Do
End If
If heading.Paragraphs(1).OutlineLevel = wdOutlineLevel4 Then
' Now this is a level 4 heading. Let's do something with it.
' heading.Expand Unit:=wdParagraph
' Debug.Print heading.Text
End If
Loop
Don't look specifically for "Heading 4" because,
one may use non built-in styles,
it would not work with international versions of Word.
Check the wdOutlineLevel4 instead.
Now, to get the range for the whole level 4, here is a little known trick:
Dim rTableRange as Range
' rTableRange will encompass the region under the current/preceding heading
Set rTableRange = heading.GoTo(What:=wdGoToBookmark, Name:="\HeadingLevel")
This will work better for the last heading 4 in the document or the last one below a heading 3.
Related
I'm using a VBA script to try to find the starting number of a paragraph (they are list items not formatted as such - not trying to format, just find the numbers).
1. First Item
2. Second Item
No number - don't include despite 61.5 in paragraph.
25 elephants should not be included
12. Item Twelve, but don't duplicate because of Susie's 35 items
Is there any way to say in VBA "If start of paragraph has 1-2 numbers, return those numbers". In regex, what I'm looking for is ^(\d\+)\.
Here is a working bit of VBA code - haven't figured out how to CREATE the excel file yet, so if you go to test create a blank test.xslx in your temp folder. Of course this may be simple enough that testing isn't necessary.
Sub FindWordCopySentence()
On Error Resume Next
Dim appExcel As Object
Dim objSheet As Object
Dim aRange As Range
Dim intRowCount As Integer
intRowCount = 1
' Open Excel File
If objSheet Is Nothing Then
Set appExcel = CreateObject("Excel.Application")
'Change the file path to match the location of your test.xls
Set objSheet = appExcel.workbooks.Open("C:\temp\test.xlsx").Sheets("Sheet1")
intRowCount = 1
End If
' Word Document Find
Set aRange = ActiveDocument.Range
With aRange.Find
Do
.ClearFormatting
' Find 1-2 digit number
.Text = "[0-9]{1,2}"
.MatchWildcards = True
.Execute
If .Found Then
' Copy to Excel file
aRange.Expand Unit:=wdSentence
aRange.Copy
aRange.Collapse wdCollapseEnd
objSheet.Cells(intRowCount, 1).Select
objSheet.Paste
intRowCount = intRowCount + 1
End If
Loop While .Found
End With
Set aRange = Nothing
If Not objSheet Is Nothing Then
appExcel.workbooks(1).Close True
appExcel.Quit
Set objSheet = Nothing
Set appExcel = Nothing
End If
End Sub
Thanks!
I would go quite a bit simpler and just check the first few characters of the paragraph:
Option Explicit
Sub test()
Dim para As Paragraph
For Each para In ThisDocument.Paragraphs
With para.Range
If (.Characters(2) = ".") Or (.Characters(3) = ".") Then
If IsNumeric(para.Range.Words(1)) Then
Debug.Print "Do something with paragraph number " & _
para.Range.Words(1) & "."
End If
End If
End With
Next para
End Sub
A more efficient approach, which obviates the need to test every paragraph:
Sub Demo()
Application.ScreenUpdating = False
Dim StrOut As String
With ActiveDocument.Range
With .Find
.ClearFormatting
.Replacement.ClearFormatting
.Text = "^13[0-9.]{1,}" ' or: .Text = "^13[0-9]{1,}
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindStop
.MatchWildcards = True
.Execute
End With
Do While .Find.Found
StrOut = StrOut & .Text
' or: MsgBox Split(.Text, vbCr)(1)
.Collapse wdCollapseEnd
.Find.Execute
Loop
End With
Application.ScreenUpdating = True
MsgBox StrOut
End Sub
As coded, the macro returns the entire list strings where there may be multiple levels (e.g. 1.2). Comments show how to find just the first number where there may be multiple levels and how to extract that number for testing (the Find expression includes the preceding paragraph break).
I need to create a macros which removes whitespaces and indent before all paragraphs in the active MS Word document. I've tried following:
For Each p In ActiveDocument.Paragraphs
p.Range.Text = Trim(p.range.Text)
Next p
which sets macros into eternal loop. If I try to assign string literal to the paragraphs, vba always creates only 1 paragraph:
For Each p In ActiveDocument.Paragraphs
p.Range.Text = "test"
Next p
I think I have a general misconception about paragraph object. I would appreciate any enlightment on the subject.
The reason the code in the question is looping is because replacing one paragraph with the processed (trimmed) text is changing the paragraphs collection. So the code will continually process the same paragraph at some point.
This is normal behavior with objects that are getting deleted and recreated "behind the scenes". The way to work around it is to loop the collection from the end to the front:
For i = ActiveDocument.Paragraphs.Count To 1 Step -1
Set p = ActiveDocument.Paragraphs(i)
p.Range.Text = Trim(p.Range.Text)
Next
That said, if the paragraphs in the document contain any formatting this will be lost. String processing does not retain formatting.
An alternative would be to check the first character of each paragraph for the kinds of characters you consider to be "white space". If present, extend the range until no more of these characters are detected, and delete. That will leave the formatting intact. (Since this does not change the entire paragraph a "normal" loop works.)
Sub TestTrimParas()
Dim p As Word.Paragraph
Dim i As Long
Dim rng As Word.Range
For Each p In ActiveDocument.Paragraphs
Set rng = p.Range.Characters.First
'Test for a space or TAB character
If rng.Text = " " Or rng.Text = Chr(9) Then
i = rng.MoveEndWhile(" " + Chr(9))
Debug.Print i
rng.Delete
End If
Next p
End Sub
You could, of course, do this in a fraction of the time without a loop, using nothing fancier than Find/Replace. For example:
Find = ^p^w
Replace = ^p
and
Find = ^w^p
Replace = ^p
As a macro this becomes:
Sub Demo()
Application.ScreenUpdating = False
With ActiveDocument.Range
.InsertBefore vbCr
With .Find
.ClearFormatting
.Replacement.ClearFormatting
.Forward = True
.Wrap = wdFindContinue
.Format = False
.MatchWildcards = False
.Text = "^p^w"
.Replacement.Text = "^p"
.Execute Replace:=wdReplaceAll
.Text = "^w^p"
.Execute Replace:=wdReplaceAll
End With
.Characters.First.Text = vbNullString
End With
Application.ScreenUpdating = True
End Sub
Note also that trimming text the way you're doing is liable to destroy all intra-paragraph formatting, cross-reference fields, and the like; it also won't change indents. Indents can be removed by selecting the entire document and changing the paragraph format; better still, modify the underlying Styles (assuming they've been used correctly).
Entering "eternal" loop is a bit unpleasant. Only Chuck Norris can exit one. Anyway, try to make a check before trimming and it will not enter:
Sub TestMe()
Dim p As Paragraph
For Each p In ThisDocument.Paragraphs
If p.Range <> Trim(p.Range) Then p.Range = Trim(p.Range)
Next p
End Sub
As has been said by #Cindy Meister, I need to prevent endless creation of another paragraphs by trimming them. I bear in mind that paragraph range contains at least 1 character, so processing range - 1 character would be safe. Following has worked for me
Sub ProcessParagraphs()
Set docContent = ActiveDocument.Content
' replace TAB symbols throughout the document to single space (trim does not remove TAB)
docContent.Find.Execute FindText:=vbTab, ReplaceWith:=" ", Replace:=wdReplaceAll
For Each p In ActiveDocument.Paragraphs
' delete empty paragraph (delete operation is safe, we cannot enter enternal loop here)
If Len(p.range.Text) = 1 Then
p.range.Delete
' remove whitespaces
Else
Set thisRg = p.range
' shrink range by 1 character
thisRg.MoveEnd wdCharacter, -1
thisRg.Text = Trim(thisRg.Text)
End If
p.LeftIndent = 0
p.FirstLineIndent = 0
p.Reset
p.range.Font.Reset
Next
With Selection
.ClearFormatting
End With
End Sub
I saw a number of solutions here are what worked for me. Note I turn off track changes and then revert back to original document tracking status.
I hope this helps some.
Option Explicit
Public Function TrimParagraphSpaces()
Dim TrackChangeStatus: TrackChangeStatus = ActiveDocument.TrackRevisions
ActiveDocument.TrackRevisions = False
Dim oPara As Paragraph
For Each oPara In ActiveDocument.StoryRanges(wdMainTextStory).Paragraphs
Dim oRange As Range: Set oRange = oPara.Range
Dim endRange, startRange As Range
Set startRange = oRange.Characters.First
Do While (startRange = Space(1))
startRange.Delete 'Remove last space in each paragraphs
Set startRange = oRange.Characters.First
Loop
Set endRange = oRange
' NOTE: for end range must select the before last characted. endRange.characters.Last returns the chr(13) return
endRange.SetRange Start:=oRange.End - 2, End:=oRange.End - 1
Do While (endRange = Space(1))
'endRange.Delete 'NOTE delete somehow does not work for the last paragraph
endRange.Text = "" 'Remove last space in each paragraphs
Set endRange = oPara.Range
endRange.SetRange Start:=oRange.End - 1, End:=oRange.End
Loop
Next
ActiveDocument.TrackRevisions = TrackChangeStatus
End Function
I have a Word macro that allows to put his/her cursor anywhere in a Word document and it finds and saves the Heading 1, Heading 2 and Heading 3 text that is above the text selected by the user in order capture the chapter, section and sub-section that is associated with any sentence in the document.
I am currently using the code below which moves up the document line-by-line until it finds a style that contains "Heading x". When I have completed this task I move down the number of lines that I moved up to get to Heading 1, which may be many pages.
As you can imagine this is awkward, takes a long time (sometimes 60+ seconds) and is visually disturbing.
The code below is that subroutine that identifies the heading.
Dim str_heading_txt, hdgn_STYLE As String
Dim SELECTION_PG_NO as Integer
hdng_STYLE = Selection.Style
Do Until Left(hdng_STYLE, 7) = "Heading"
LINESUP = LINESUP + 1
Selection.MoveUp Unit:=wdLine, COUNT:=1
Selection.HomeKey Unit:=wdLine
Selection.EndKey Unit:=wdLine, Extend:=wdExtend
hdng_STYLE = Selection.Style
'reached first page without finding heading
SELECTION_PG_NO = Selection.Information(wdActiveEndPageNumber)
If SELECTION_PG_NO = 1 Then 'exit if on first page
a_stop = True
Exit Sub
End If
Loop
str_heading_txt = Selection.Sentences(1)
I tried another approach below in order to eliminate the scrolling and performance issues using the Range.Find command below.
I am having trouble getting the selection range to move to the text with the "Heading 1" style. The code selects the sentence at the initial selection, not the text with the "Heading 1" style.
Ideally the Find command would take me to any style that contained "Heading" but, if required, I can code separately for "Heading 1", "Heading 2" and "Heading 3".
What changes to the code are required so that "Heading 1" is selected or, alternatively, that "Heading" is selected?
Dim str_heading_txt, hdgn_STYLE As String
Dim Rng As Range
Dim Fnd As Boolean
Set Rng = Selection.Range
With Rng.Find
.ClearFormatting
.Style = "Heading 1"
.Forward = False
.Execute
Fnd = .Found
End With
If Fnd = True Then
With Rng
hdng_STYLE = Selection.Style
str_heading_txt = Selection.Sentences(1)
End With
End If
Any assistance is sincerely appreciated.
You can use the range.GoTo() method.
Dim rngHead As Range, str_heading_txt As String, hdgn_STYLE As String
Set rngHead = Selection.GoTo(wdGoToHeading, wdGoToPrevious)
'Grab the entire text - headers are considered a paragraph
rngHead.Expand wdParagraph
' Read the text of your heading
str_heading_txt = rngHead.Text
' Read the style (name) of your heading
hdgn_STYLE = rngHead.Style
I noticed that you used Selection.Sentences(1) to grab the text, but headings are already essentially a paragraph by itself - so you can just use the range.Expand() method and expand using wdParagraph
Also, a bit of advice:
When declaring variables such as:
Dim str_heading_txt, hdgn_STYLE As String
Your intent was good, but str_heading_txt was actually declared as type Variant. Unfortunately with VBA, if you want your variables to have a specific data type, you much declare so individually:
Dim str_heading_txt As String, hdgn_STYLE As String
Or some data types even have "Shorthand" methods known as Type Characters:
Dim str_heading_txt$, hdgn_STYLE$
Notice how the $ was appended to the end of your variable? This just declared it as a String without requiring the As String.
Some Common Type-Characters:
$ String
& Long
% Integer
! Single
# Double
You can even append these to the actual value:
Dim a
a = 5
Debug.Print TypeName(a) 'Prints Integer (default)
a = 5!
Debug.Print TypeName(a) 'Prints Single
Try something based on:
Sub Demo()
Dim Rng As Range, StrHd As String, s As Long
s = 10
With Selection
Set Rng = .Range
Set Rng = Rng.GoTo(What:=wdGoToBookmark, Name:="\HeadingLevel")
StrHd = Rng.Paragraphs.First.Range.Text
Do While Right(Rng.Paragraphs.First.Style, 1) > 1
Rng.End = Rng.Start - 1
Set Rng = Rng.GoTo(What:=wdGoToBookmark, Name:="\HeadingLevel")
With Rng.Paragraphs.First
If Right(.Style, 1) < s Then
s = Right(.Style, 1)
StrHd = .Range.Text & StrHd
End If
End With
Loop
MsgBox StrHd
End With
End Sub
I'm new to VBA and would greatly appreciate some help on a problem.
I have long Word documents where I need to apply standard comments to the same set of keywords, but only in selected sections of the document. The following macro worked to find a keyword and apply a comment (from question here https://superuser.com/questions/547710/macro-to-insert-comment-bubbles-in-microsoft-word):
Sub label_items()
'
' label_items Macro
'
'
Do While Selection.Find.Execute("keyword1") = True
ActiveDocument.Comments.Add range:=Selection.range, Text:="comment for keyword 1"
Loop
End Sub
The two modifications are:
1) only apply the comments to user selected text, not the whole document. I tried a "With Selection.Range.Find" approach but I don't think comments can be added this way (??)
2) repeat this for 20+ keywords in the selected text. The keywords aren't totally standard and have names like P_1HAI10, P_1HAI20, P_2HAI60, P_HFS10, etc.
EDIT: I have tried to combine code from similar questions ( Word VBA: finding a set of words and inserting predefined comments and Word macro, storing the current selection (VBA)) but my current attempt (below) only runs for the first keyword and comment and runs over the entire document, not just the text I have highlighted/selected.
Sub label_items()
'
' label_items Macro
'
Dim selbkup As range
Set selbkup = ActiveDocument.range(Selection.range.Start, Selection.range.End)
Set range = selbkup
Do While range.Find.Execute("keyword 1") = True
ActiveDocument.Comments.Add range, "comment for keyword 1"
Loop
Set range = selbkup
Do While range.Find.Execute("keyword 2") = True
ActiveDocument.Comments.Add range, "comment for keyword 2"
Loop
'I would repeat this process for all of my keywords
End Sub
I've combed through previous questions and the Office Dev Center and am stuck. Any help/guidance is greatly appreciated!
It's a matter of adding a loop and a means of Finding the next keyword you're looking for. There are a few suggestions in the code example below, so please adjust it as necessary to fit your requirements.
Option Explicit
Sub label_items()
Dim myDoc As Document
Dim targetRange As Range
Set myDoc = ActiveDocument
Set targetRange = Selection.Range
'--- drop a bookmark to return the cursor to it's original location
Const RETURN_BM = "OrigCursorLoc"
myDoc.Bookmarks.Add Name:=RETURN_BM, Range:=Selection.Range
'--- if nothing is selected, then search the whole document
If Selection.Start = Selection.End Then
Selection.Start = 0
targetRange.Start = 0
targetRange.End = myDoc.Range.End
End If
'--- build list of keywords to search
Dim keywords() As String
keywords = Split("SMS,HTTP,SMTP", ",", , vbTextCompare)
'--- search for all keywords within the user selected range
Dim i As Long
For i = 0 To UBound(keywords)
'--- set the cursor back to the beginning of the
' originally selected range
Selection.GoTo What:=wdGoToBookmark, Name:=RETURN_BM
Do
With Selection.Find
.Forward = True
.Wrap = wdFindStop
.Text = keywords(i)
.Execute
If .Found Then
If (Selection.Start < targetRange.End) Then
Selection.Comments.Add Selection.Range, _
Text:="Found the " & keywords(i) & " keyword"
Else
Exit Do
End If
Else
Exit Do
End If
End With
Loop
Next i
'--- set the cursor back to the beginning of the
' originally selected range
Selection.GoTo What:=wdGoToBookmark, Name:=RETURN_BM
End Sub
I need to count the number of occurrences of a specific style, within a column of a table. My program finds the number of occurrences within the whole document instead of only in the selection.
Sub Find()
Selection.Tables(1).Columns(1).Select
With Selection.Find
.Style = "Style2"
iCount = 0
While .Execute
iCount = iCount + 1
Wend
MsgBox (iCount)
End With
End Sub
Performing Find inside a table is a tricky proposition as Find has a nasty tendancy to "bounce" inside of a cell. When I tested your code, having no information on how the style is applied in the table cells, the macro went into a loop and didn't stop until I forced it to. So I was a bit surprised that your code worked at all...
The problem with doing find on a column is that, in the underlying structures of the document a column is not a contiguous set of characters, the way it appears on-screen. The Word table information runs top-to-bottom in the cell, left-to-right across a row, then to the next row and repeat. The column selection is an illusion maintained by the Word application. So macro code basing on Selection or Range can't follow the usual rules that apply.
The following worked for me. In essence, it searches inside the entire table, but when it hits a cell not in the specified column the target range is moved to the next cell in the column and the search is run again. Only the "hits" inside cells in the column are counted.
Sub FindStyleInstanceInTableColumn()
Dim iCount As Long, iCellCount As Long, iCounter As Long
Dim cel As word.Cell
Dim col As word.Column
Dim rngFind As word.Range, rngCel As word.Range
Dim bFound As Boolean
Set col = Selection.Tables(1).Columns(1)
iCount = 0
iCellCount = col.Cells.Count
iCounter = 1
Set rngCel = col.Cells(iCounter).Range
Set rngFind = rngCel.Duplicate
'Don't include end-of-cell marker
rngFind.MoveEnd wdCharacter, -1
rngFind.Select 'For debugging
With rngFind.Find
.Style = "Style2"
bFound = .Execute(wrap:=wdFindStop)
Do
rngFind.Select 'For debugging
If bFound Then
'If the found range is within a column cell
'then increase the counter
If rngFind.InRange(rngCel) Then
iCount = iCount + 1
'If the found range is not in a column cell
'then the style wasn't found in the cell so
'go to the next cell
ElseIf iCounter < iCellCount Then
iCounter = iCounter + 1
Set rngCel = col.Cells(iCounter).Range
rngFind.Start = rngCel.Start
rngFind.End = rngCel.Start
End If
rngFind.Collapse wdCollapseEnd
End If
bFound = .Execute(Format:=True, wrap:=wdFindStop)
Loop Until iCounter = iCellCount And Not bFound
End With
MsgBox (iCount)
End Sub
EDIT: Adjusted the code to take into account no hits in first cell and hits in the last cell of the column. The difference is to make sure the starting point for rngFind is in the same cell as rngCel.