Change the author of comments - vba

I have a list of words in a table. I have a VBA word macro that reads this table. When the macro identifies the words in the table, it adds comments or suggestions. For example: "don't use that word". Currently the author of those comments is defaulted to the name of whoever runs the macro. If user John Wayne runs the macro, each comment is displayed with the label "John Wayne" as the author. How can I change the author to "robot"? I added the comment.author line but it doesn't work.
Do While .Execute(Forward:=True) = True
suggestion = Documents(DocName).Tables(1).Rows(RowCounter).Cells(CellWithSuggestion).range.Text
ActiveDocument.Comment.Author = "robot"
ActiveDocument.Comments.Add findRange, Text:=suggestion
findRange.Collapse wdCollapseEnd 'to avoid getting into endless loop
Loop 'do while

You have to set the author per created comment.
Use this sub for it
Option Explicit
Private Const SystemAuthorForComments As String = "robot"
Sub addComment(rgComment As Range, strMessage As String, _
Optional ByVal strAuthor As String)
If LenB(strAuthor) = 0 Then strAuthor = SystemAuthorForComments
Dim doc As Word.Document: Set doc = rgComment.Parent
Dim c As Word.Comment
Set c = doc.Comments.Add(rgComment, strMessage)
c.Author = strAuthor
End Sub
You would have to change your code snippet to:
Do While .Execute(Forward:=True) = True
suggestion = Documents(DocName).Tables(1).Rows(RowCounter).Cells(CellWithSuggestion).Range.Text
addComment findrange, suggestion '--> this is how you call the sub
findrange.Collapse wdCollapseEnd 'to avoid getting into endless loop
Loop 'do while

Workaround to change the author of comments to "robot":
Add "_0" at the end of all comment's author name.
Process to add new comments.
Change comments to author = "robot". Comments with an author that ends in "_0" stay intact.
Dim brand As String
Dim Comment As Comment
brand = "robot"
If ActiveDocument.Comments.Count > 0 Then
For Each Comment In ActiveDocument.Comments
Comment.Author = Comment.Author & "_0"
Next
End If
If ActiveDocument.Comments.Count > 0 Then
For Each Comment In ActiveDocument.Comments
If InStr(1, Comment.Author, "_0", 1) = 0 Then
Comment.Author = brand
End If
Next
End If

Related

Sub to find text in a Word document by specified font and font size

Goal: Find headings in a document by their font and font size and put them into a spreadsheet.
All headings in my doc are formatted as Ariel, size 16. I want to do a find of the Word doc, select the matching range of text to the end of the line, then assign it to a variable so I can put it in a spreadsheet. I can do an advanced find and search for the font/size successfully, but can't get it to select the range of text or assign it to a variable.
Tried modifying the below from http://www.vbaexpress.com/forum/showthread.php?55726-find-replace-fonts-macro but couldn't figure out how to select and assign the found text to a variable. If I can get it assigned to the variable then I can take care of the rest to get it into a spreadsheet.
'A basic Word macro coded by Greg Maxey
Sub FindFont
Dim strHeading as string
Dim oChr As Range
For Each oChr In ActiveDocument.Range.Characters
If oChr.Font.Name = "Ariel" And oChr.Font.Size = "16" Then
strHeading = .selected
Next
lbl_Exit:
Exit Sub
End Sub
To get the current code working, you just need to amend strHeading = .selected to something like strHeading = strHeading & oChr & vbNewLine. You'll also need to add an End If statement after that line and probably amend "Ariel" to "Arial".
I think a better way to do this would be to use Word's Find method. Depending on how you are going to be inserting the data into the spreadsheet, you may also prefer to put each header that you find in a collection instead of a string, although you could easily delimit the string and then split it before transferring the data into the spreadsheet.
Just to give you some more ideas, I've put some sample code below.
Sub Demo()
Dim Find As Find
Dim Result As Collection
Set Find = ActiveDocument.Range.Find
With Find
.Font.Name = "Arial"
.Font.Size = 16
End With
Set Result = Execute(Find)
If Result.Count = 0 Then
MsgBox "No match found"
Exit Sub
Else
TransferToExcel Result
End If
End Sub
Function Execute(Find As Find) As Collection
Set Execute = New Collection
Do While Find.Execute
Execute.Add Find.Parent.Text
Loop
End Function
Sub TransferToExcel(Data As Collection)
Dim i As Long
With CreateObject("Excel.Application")
With .Workbooks.Add
With .Sheets(1)
For i = 1 To Data.Count
.Cells(i, 1) = Data(i)
Next
End With
End With
.Visible = True
End With
End Sub

Word VBA Getting page number of a specific footer in section

Couldn't find the answer I was looking for.
I want to get the current page number String including its format.
For example: Some sections may have chapter identifier (1-1), some are in Roman style, etc..
My hope was to get the selection of the specific footer, then loop through the fields and get the Page field data (Output is the String I want).
So far as I can see, there is no option to loop through the footers of a given section, just get the general template and try working with it.
I'm aware of wdActiveEndAdjustedPageNumber from Selection.Range.Information, but it just gives me partial information.
Am I wrong? Is there a way to work with a specific footer I choose?
If not, can you guide me how to get the following data:
Closest chapter number value
Getting the page number value of a special format such as Roman, Alphabetical font (Meaning applying the page format on the wdActiveEndAdjustedPageNumber)
Thanks.
Edit for clarification:
In my word template, the Heading 1 style creates the following header: Chapter 1, followed by Chapter 2 and so on.
In page number format, there is an option to include the current Chapter value to the page number.
For example: Assuming the following setup
will result with these pages in the { PAGE } field: 1-1, 1-2, 1-3, ...
My goal is to somehow get this entire "value" for a specific page footer.
Here is a code snippet which won't work properly:
Sub getPageFieldInFooter()
' get current section number
Dim sectionNum As Integer
sectionNum = Selection.Range.Information(wdActiveEndSectionNumber)
'select first page footer, loop through its fields and find Page field
ActiveDocument.Sections(sectionNum).Footers(wdHeaderFooterPrimary).Range.Select
Dim f As Field
For Each f In Selection.Fields
If f.Type = wdFieldPage Then
' do something with the page data
MsgBox f.Data
End If
Next f
End Sub
The output of such a method is '1-1'
The reason it won't work is because it can retrieve the first page only (or the second using wdHeaderFooterEvenPages).
Same goes for Roman number format, or any other from that list.
For the following page number settings, I wish to get the "value" in a specific footer.
The code above will return the values for first or second page, and that's it.
Is there a way to access any footer in the document and perform my code example?
If not, how can I get the page number "value" for any footer I choose?
Hope this is clearer.
The following is working for me, although I'm not certain how reliable it is. Apparently, if I query the Footer (or Header) of the current selection in the document it will return the information for the Footer (or Header) of that page.
Things get very complicated as soon as you start working with multiple sections and Different First Page. I've done some testing for that in the code below, but I wouldn't swear it's "production code". However, it should give you a starting place.
Sub GetFormattedPageNumberFromSelection()
Dim sel As word.Selection
Dim sec As word.Section
Dim r As word.Range, rOriginal As word.Range
Dim fld As word.Field
Dim secCurrIndex As Long
Dim sNoPageNumber As String
Set sel = Selection
If Not sel.InRange(sel.Document.content) Then Exit Sub
Set sec = sel.Sections(1)
If Not sec.Footers(wdHeaderFooterFirstPage).exists Then
Set r = sec.Footers(wdHeaderFooterPrimary).Range
Else
Set r = sel.Range
Set rOriginal = r.Duplicate
secCurrIndex = sec.index
If secCurrIndex <> 1 Then
sel.GoToPrevious wdGoToPage
If sel.Sections(1).index = secCurrIndex Then
Set r = sec.Footers(wdHeaderFooterPrimary).Range
Else
Set r = sec.Footers(wdHeaderFooterFirstPage).Range
End If
rOriginal.Select 'return to original selection
ElseIf r.Information(wdActiveEndPageNumber) = 1 Then
Set r = sec.Footers(wdHeaderFooterFirstPage).Range
Else
Set r = sec.Footers(wdHeaderFooterPrimary).Range
End If
End If
For Each fld In r.Fields
sNoPageNumber = "No page number"
If fld.Type = wdFieldPage Then
Debug.Print fld.result
sNoPageNumber = ""
Exit For
End If
Next
If Len(sNoPageNumber) > 0 Then Debug.Print sNoPageNumber
End Sub
...and sometimes we don't see the simplest way.
Insert a Page field at the current selection, read the result, then delete it again:
Sub GetFormattedPageNumberFromSelection2()
Dim rng As word.Range
Dim fld As word.Field
Set rng = Selection.Range
Set fld = rng.Fields.Add(rng, wdFieldPage)
Debug.Print fld.result
fld.Delete
End Sub
What you haven't told us is how you're 'choosing' the page you want the reference for. Assuming it's based in whatever page is selected/displayed, you could use something like the following for a page header:
Sub Demo()
Application.ScreenUpdating = False
Dim Rng As Range, Fld As Field
ActiveWindow.ActivePane.View.SeekView = wdSeekCurrentPageHeader
For Each Fld In Selection.HeaderFooter.Range.Fields
If Fld.Type = wdFieldPage Then
MsgBox Fld.Result
Exit For
End If
Next
ActiveWindow.ActivePane.View.SeekView = wdSeekMainDocument
Application.ScreenUpdating = True
End Sub
Unfortunately, wdSeekCurrentPageFooter returns the next page's footer!, so you can't use that for the current footer. The following, however, should work wherever the PAGE # field is located:
Sub Demo()
Application.ScreenUpdating = False
Dim i As Long, Fld As Field, bExit As Boolean: bExit = False
With ActiveWindow.ActivePane.Pages(Selection.Information(wdActiveEndAdjustedPageNumber))
For i = 1 To .Rectangles.Count
With .Rectangles(i).Range
For Each Fld In .Fields
If Fld.Type = wdFieldPage Then
MsgBox Fld.Result
bExit = True: Exit For
End If
Next
End With
If bExit = True Then Exit For
Next
End With
Application.ScreenUpdating = True
End Sub

Word VBA: how to select found text rather than where the cursor is positioned

This is probably simple but I can't get it to work.
I need to search through my document, find words that contain the string 'alog' and add 'ue'. For example, 'catalogs' --> 'catalogues'.
The above works fine but I can't get the next bit to work: if a found string already has 'ue' after the 'log' I don't want to add another 'ue'.
The subroutine accessed from the macro is below. I've tried adding the following lines into the 'while execute' part, but 'selection' always turns out to be the word where the cursor happens to be.
With Selection
.Expand unit:=wdWord
End With
How do I i) select the content of the found range and ii) expand that new selection by two characters to see if those two characters are 'ue' ?
Many thanks.
Sub do_replace2(old_text As String, new_text As String, Count_changes As Integer)
' Replaces 'log' with 'logue'
' Ignores paragraphs in styles beginning with 'Question'
Dim rg As Range
Set rg = ActiveDocument.Range
With rg.Find
.Text = old_text
While .Execute
If Left(rg.Paragraphs(1).Style, 8) <> "Question" Then
rg.Text = new_text
With ActiveDocument.Comments.Add(rg, "Changed from '" & old_text & "'")
.Initial = "-logs"
.Author = "-logs"
End With
Count_changes = Count_changes + 1
End If
rg.Collapse wdCollapseEnd
Wend
End With
End Sub
I'm not quite sure I follow the first part of your question "How do I select the content of the found range". The rg variable already contains the search result. If you want to select it, just use rg.Select. This might be useful in debugging (so you can see where the Range is when you're stepping through the code), but there isn't really any other reason to use the Selection object in the code from your question. You can just use the Range object instead.
As to part 2 of your question "How do I ... expand that new selection by two characters", all you need to do is add 2 to the .End property of the Range. Since you're only using this for a test (and because the .Find method can be dodgy), test this on a copy of rg:
With rg.Find
.Text = old_text
While .Execute
If Left(rg.Paragraphs(1).Style, 8) <> "Question" Then
Dim test As Range
Set test = rg.Duplicate 'copy the found Range.
test.Collapse wdCollapseEnd 'move to the end of it.
test.End = test.End + 2 'expand to the next 2 characters.
If test.Text <> "ue" Then 'see if it's "ue".
rg.Text = new_text
With ActiveDocument.Comments.Add(rg, "Changed from '" & old_text & "'")
.Initial = "-logs"
.Author = "-logs"
End With
Count_changes = Count_changes + 1
End If
End If
rg.Collapse wdCollapseEnd
Wend
End With

Using word wildcards to find unaccepted changes

I have some word documents with unaccepted, tracked changes. I want to accept them but still have them shown in red in my documents. I think a good way to do this would be doing a wildcard search for unaccepted changes and replacing them with the same text in red, however I dont know if this is possible.
I am also happy with other ways of achieving my goal, without wildcards.
Applying formatting to revisions cannot be done using Word's standard find & replace operation. However, you can write a macro that enumerates all revisions and then applies formatting to each of them.
There is a bloc post by Chris Rae who provides a macro that converts revisions to standard formatting:
Enumerating edits on large documents (AKA converting tracked changes to conventional formatting)
The macro may not yet do exactly what you need, but it should get you started.
For reference, here is a copy of the macro:
Sub EnumerateChanges()
Dim rAll As Revision
Dim dReport As Document
Dim dBigDoc As Document
Set dBigDoc = ActiveDocument
If dBigDoc.Revisions.Count = 0 Then
MsgBox "There are no revisions in the active document.", vbCritical
ElseIf MsgBox(“This will enumerate the changes in '" + dBigDoc.Name + "' in a new document and close the original WITHOUT saving changes. Continue?", vbYesNo) <> vbNo Then
Set dReport = Documents.Add
dBigDoc.Activate ' really just so we can show progress by selecting the revisions
dBigDoc.TrackRevisions = False ' Leaving this on results in a disaster
For Each rAll In dBigDoc.Revisions
' Now find the nearest section heading downwards
Dim rFindFirst As Range, rFindLast As Range
Set rFindLast = rAll.Range.Paragraphs(1).Range
While Not IsNumberedPara(rFindLast.Next(wdParagraph))
Set rFindLast = rFindLast.Next(wdParagraph)
Wend
' Now head back up to the next numbered section header
Set rFindFirst = rFindLast
Do
Set rFindFirst = rFindFirst.Previous(wdParagraph)
Loop Until IsNumberedPara(rFindFirst) Or (rFindFirst.Previous(wdParagraph) Is Nothing)
ConvertNumberedToText rFindFirst
Dim rChangedSection As Range
Set rChangedSection = dBigDoc.Range(rFindFirst.Start, rFindLast.End)
' Properly tag all the revisions in this whole section
Dim rOnesInThisSection As Revision
For Each rOnesInThisSection In rChangedSection.Revisions
rOnesInThisSection.Range.Select ' just for visual update
DoEvents ' update the screen so we can see how far we are through
If rOnesInThisSection.Type = wdRevisionDelete Then
rOnesInThisSection.Reject
With Selection.Range
.Font.ColorIndex = wdRed
.Font.StrikeThrough = True
End With
dBigDoc.Comments.Add Selection.Range, “deleted”
Else
If rOnesInThisSection.Type = wdRevisionInsert Then
rOnesInThisSection.Accept
With Selection.Range
.Font.ColorIndex = wdBlue
End With
dBigDoc.Comments.Add Selection.Range, “inserted”
End If
End If
Next
' Now copy the whole thing into our new document
rChangedSection.Copy
Dim rOut As Range
Set rOut = dReport.Range
rOut.EndOf wdStory, False
rOut.Paste
Next rAll
' There should end up being no numbered paragraphs at all in the
' new doc (they were converted to text), so delete them
Dim pFinal As Paragraph
For Each pFinal In dReport.Paragraphs
If IsNumberedPara(pFinal.Range) Then
pFinal.Range.ListFormat.RemoveNumbers
End If
Next
dBigDoc.Close False
End If
End Sub
Sub ConvertNumberedToText(rOf As Range)
If InStr(rOf.ListFormat.ListString, “.”) > 0 Then
rOf.InsertBefore "Changes to section " + rOf.ListFormat.ListString + " "
End If
End Sub
Function IsNumberedPara(rOf As Range) As Boolean
If rOf Is Nothing Then ‘ if the document doesn’t have numbered sections, this will cause changes to be enumerated in the whole thing
IsNumberedPara = True
ElseIf rOf.ListFormat.ListString <> "" Then
If Asc(rOf.ListFormat.ListString) <> 63 Then
IsNumberedPara = True
End If
End If
End Function

How do you remove hyperlinks from a Microsoft Word document?

I'm writing a VB Macro to do some processing of documents for my work.
The lines of text are searched and the bracketed text is put in a list(box).
The problem comes when I want to remove all hyperlinks in the document and then generate new ones (not necessarily in the location of the original hyperlinks)
So the problem is How do I remove the existing hyperlinks?
My current issue is that every time a link gets added, the hyperlinks count goes up one, but when you delete it, the count does NOT reduce. (as a result I now have a document with 32 links - all empty except for 3 I put in myself - they do not show up in the document)
At the end of the code are my attempts at removing the hyperlinks.
Private Sub FindLinksV3_Click()
ListOfLinks.Clear
ListOfLinks.AddItem Now
ListOfLinks.AddItem ("Test String 1")
ListOfLinks.AddItem ActiveDocument.FullName
SentenceCount = ActiveDocument.Sentences.Count
ListOfLinks.AddItem ("Sentence Count:" & SentenceCount)
counter = 0
For Each myobject In ActiveDocument.Sentences ' Iterate through each element.
ListOfLinks.AddItem myobject
counter = counter + 1
BracketStart = (InStr(1, myobject, "("))
If BracketStart > 0 Then
BracketStop = (InStr(1, myobject, ")"))
If BracketStop > 0 Then
ListOfLinks.AddItem Mid$(myobject, BracketStart + 1, BracketStop - BracketStart - 1)
ActiveDocument.Sentences(counter).Select
ActiveDocument.Hyperlinks.Add Anchor:=Selection.Range, Address:= _
"http://testnolink/" & counter, ScreenTip:="" 'TextToDisplay:=""
End If
End If
Next
'ActiveDocument.Sentences(1).Select
'
'Selection.Range.Hyperlinks(1).Delete
ActiveDocument.Hyperlinks.Item(1).Delete
Debug.Print ActiveDocument.Hyperlinks.Count
End Sub
This is an old post, so am adding this VBA code in case it is useful to someone.
Hyperlinks (Collections) need to be deleted in reverse order:
Sub RemoveHyperlinksInDoc()
' You need to delete collection members starting from the end going backwards
With ActiveDocument
For i = .Hyperlinks.Count To 1 Step -1
.Hyperlinks(i).Delete
Next
End With
End Sub
Sub RemoveHyperlinksInRange()
' You need to delete collection members starting from the end going backwards
With Selection.Range
For i = .Hyperlinks.Count To 1 Step -1
.Hyperlinks(i).Delete
Next
End With
End Sub
The line removing the hyperlink is commented out. The following line will remove the first hyperlink within the selected range:
Selection.Range.Hyperlinks(1).Delete
This will also decrement Selection.Range.Hyperlinks.Count by 1.
To see how the count of links is changing you can run the following method on a document:
Sub AddAndRemoveHyperlink()
Dim oRange As Range
Set oRange = ActiveDocument.Range
oRange.Collapse wdCollapseStart
oRange.MoveEnd wdCharacter
Debug.Print ActiveDocument.Range.Hyperlinks.Count
ActiveDocument.Hyperlinks.Add oRange, "http://www.example.com"
Debug.Print ActiveDocument.Range.Hyperlinks.Count
ActiveDocument.Hyperlinks.Item(1).Delete
Debug.Print ActiveDocument.Range.Hyperlinks.Count
End Sub