I need to proccess text by paragraphs. Next example shows taking a paragraph and removing its last character - the paragraph character. Then I proccess this text and try to replace old text with a new one. The problem is that it goes into infinite loop and freezes Word.
Sub Parser()
For Each para In ActiveDocument.Paragraphs
If Len(para.Range.Text) >= 150 Then
CleanedString = Left(para.Range.Text, Len(para.Range.Text) - 1)
'Some proccessing here
para.Range.Text = CleanedString & vbCr
End If
Next
End Sub
I also tried removing all the paragraphs and placing them back after processing, but it also failed.
Its an infinite loop because you are adding a new paragraph to the end of your text. Consequently every time you process a string the document gets 1 paragraph longer. The trick is to adjust the range before you process the text.
Sub Parser()
For Each para In ActiveDocument.Paragraphs
para.range.moveend unit:=wdcharacter, count =-1
If Len(para.Range.Text) >= 150 Then
CleanedString = para.Range.Text
'Some proccessing here
para.Range.Text = CleanedString
End If
Next
End Sub
Right solution was near, thanks #freeflow. I acutally needed to wrap it into With
For Each para In ActiveDocument.Paragraphs
If Len(para.Range.Text) >= 150 Then
With para.Range
.MoveEnd Unit:=wdCharacter, Count:=-1
NewText = .Text
.Text = NewText
End With
End If
Next
Related
I have the following code that searches for a certain point in a document and creates a search range until the end of the document. Then within that range it removes the paragraph following entirely bold paragraphs (subheadings), ignoring any styles that aren't Normal and aren't in a table. However, it seems to search the entire document (i.e. the beginning as well). How can I make it only search the range (i.e. from where I've positioned the cursor down to the end of the document)?
Dim aPara As Paragraph
Dim oSearchRange As Range
With Selection.Find
.Text = "Dear "
End With
Selection.MoveDown Unit:=wdParagraph, Count:=4
Set oSearchRange = Selection.Range
oSearchRange.End = ActiveDocument.Content.End
oSearchRange.MoveEnd wdParagraph, -1
For Each aPara In oSearchRange.Paragraphs
If aPara.Range.Font.Bold = True And aPara.Range.Next.Style = ActiveDocument.Styles("Normal") And Not aPara.Range.Next.Information(wdWithInTable) Then aPara.Range.Next.Delete
Next aPara
Thanks
I needed to add .Execute after the "Dear " search, thanks to Teamothy (:
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 code like so:
Sub MoveToBeginningSentence()
Application.ScreenUpdating = False
Dim selectedWords As Range
Dim selectedText As String
Const punctuation As String = " & Chr(145) & "
On Error GoTo ErrorReport
' Cancel macro when there's no text selected
Selection.Cut
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.MoveRight Unit:=wdCharacter, Count:=1, Extend:=wdExtend
Set selectedWords = Selection.Range
selectedText = selectedWords
If InStr(selectedText, punctuation) = 0 Then
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.Paste
Else
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.Paste
Selection.Paste
Selection.Paste
Selection.Paste
End If
ErrorReport:
End Sub
Basically, it help me move whatever text I have selected to the beginning of the sentence in Word. If there's no quotation mark, then paste once. If there is a quote mark, paste 4 times.
The problem is regardless of whether there's any quotation there or not, it will only paste once. If I set the macro to detect any other character, it will work fine. But every single time I try to force it to detect smart quotations, it will fail.
Is there any way to fix it?
Working with the Selection object is always a bit chancy; on the whole, it's better to work with a Range object. You can have only one Selection; you can have as many Ranges as you need.
Because your code uses the Selection object it's not 100% clear what the code does. Based on my best guess, I put together the following example which you can tweak if it's not exactly right.
At the beginning, I check whether there's something in the selection, or it's a blinking insertion point. If no text is selected, the macro ends. This is better than invoking Error handling, then not handling anything: If other problems crop up in your code, you wouldn't know about them.
A Range object is instantiated for the selection - there's no need to "cut" it, as you'll see further along. Based on this, the entire sentence is also assigned to a Range object. The text of the sentence is picked up, then the sentence's Range is "collapsed" to its starting point. (Think of this like pressing the left arrow on the keyboard.)
Now the sentence's text is checked for the character Chr(145). If it's not there, the original selection's text (including formatting) is added at the beginning of the sentence. If it's there, then it's added four times.
Finally, the original selection is deleted.
Sub MoveToBeginningSentence()
Application.ScreenUpdating = False
Dim selectedText As String
Dim punctuation As String
punctuation = Chr(145) ' ‘ "smart" apostrophe
Dim selRange As word.Range
Dim curSentence As word.Range
Dim i As Long
' Cancel macro when there's no text selected
If Selection.Type = wdSelectionIP Then Exit Sub
Set selRange = Selection.Range
Set curSentence = selRange.Sentences(1)
selectedText = curSentence.Text
curSentence.Collapse wdCollapseStart
If InStr(selectedText, punctuation) = 0 Then
curSentence.FormattedText = selRange.FormattedText
Else
For i = 1 To 4
curSentence.FormattedText = selRange.FormattedText
curSentence.Collapse wdCollapseEnd
Next
End If
selRange.Delete
End Sub
Please check out this code.
Sub MoveToBeginningSentence()
' 19 Jan 2018
Dim Rng As Range
Dim SelText As String
Dim Repeats As Integer
Dim i As Integer
With Selection.Range
SelText = .Text ' copy the selected text
Set Rng = .Sentences(1) ' identify the current sentence
End With
If Len(SelText) Then ' Skip when no text is selected
With Rng
Application.ScreenUpdating = False
Selection.Range.Text = "" ' delete the selected text
Repeats = IIf(IsQuote(.Text), 4, 1)
If Repeats = 4 Then .MoveStart wdCharacter, 1
For i = 1 To Repeats
.Text = SelText & .Text
Next i
Application.ScreenUpdating = True
End With
Else
MsgBox "Please select some text.", _
vbExclamation, "Selection is empty"
End If
End Sub
Private Function IsQuote(Txt As String) As Boolean
' 19 Jan 2018
Dim Quotes
Dim Ch As Long
Dim i As Long
Quotes = Array(34, 147, 148, -24143, -24144)
Ch = Asc(Txt)
' Debug.Print Ch ' read ASCII code of first character
For i = 0 To UBound(Quotes)
If Ch = Quotes(i) Then Exit For
Next i
IsQuote = (i <= UBound(Quotes))
End Function
The approach taken is to identify the first character of the selected sentence using the ASC() function. For a normal quotation mark that would be 34. In my test I came up with -24143 and -24144 (opening and closing). I couldn't identify Chr(145) but found MS stating that curly quotation marks are Chr(147) and Chr(148) respectively. Therefore I added a function that checks all of them. If you enable the line Debug.Print Ch in the function the character code actually found will be printed to the immediate window. You might add more character codes to the array Quotes.
The code itself doesn't consider spaces between words. Perhaps Word will take care of that, and perhaps you don't need it.
You need to supply InStr with the starting position as a first parameter:
If InStr(1, selectedText, punctuation) = 0 Then
Also
Const punctuation As String = " & Chr(145) & "
is going to search for space-ampersand-space-Chr(145)-space-ampersand-space. If you want to search for the smart quote character then use
Const punctuation As String = Chr(145)
Hope that helps.
I wrote a macro to delete all the empty paragraphs in my document, but it exhibits weird behavior: If there are a number of empty paragraphs at the very end of the document, about half of them are deleted. Repeatedly running the macro gradually eliminates the empty paragraphs until only one empty paragraph remains. Even if there is a boundary condition so that I need a line of code to delete the last paragraph, but I still don't understand why only half of the empty paragraphs at the end are deleted. Can anyone explain why this is happening and how to correct this behavior? As an aside, I searched online and saw numerous posts about detecting paragraph markers (^p, ^13, and others, but only searching vbCr worked, which is another minor puzzle.)
Sub Delete_Empty__Paras_2() 'This macro looks for empty paragraphs and deletes them.
Dim original_num_of_paras_in_doc As Integer
Dim num_of_deleted_paras As Integer
original_num_of_paras_in_doc = ActiveDocument.Paragraphs.Count 'Count the number of paragraphs in the document to start
num_of_deleted_paras = 0 'In the beginning, no paragraphs have been deleted
Selection.HomeKey Unit:=wdStory 'Go to the beginning of the document.
For current_para_number = 1 To original_num_of_paras_in_doc 'Process each paragraph in the document, one by one.
If current_para_number + num_of_deleted_paras > original_num_of_paras_in_doc Then 'Stop processing paragraphs when the loop has processed every paragraph.
Exit For
Else 'If the system just deleted the 3rd paragraph of the document because
' it's empty, the next paragraph processed is the 3rd one again,
'so when we iterate the counter, we have to subtract the number of deleted paragraphs to account for this.
Set paraRange = ActiveDocument.Paragraphs(current_para_number - num_of_deleted_paras).Range
paratext = paraRange.Text
If paratext = vbCr Then 'Is the paragraph empty? (By the way, checking for vbCr is the only method that worked for checking for empty paras.)
paratext = "" 'Delete the paragraph.
ActiveDocument.Paragraphs(current_para_number - num_of_deleted_paras).Range.Text = paratext
num_of_deleted_paras = num_of_deleted_paras + 1 'Iterate the count of deleted paras.
End If
End If
Next current_para_number
End Sub
This code will delete all blank paragraphs...
Sub RemoveBlankParas()
Dim oDoc As Word.Document
Dim i As Long
Dim oRng As Range
Dim lParas As Long
Set oDoc = ActiveDocument
lParas = oDoc.Paragraphs.Count ' Total paragraph count
Set oRng = ActiveDocument.Range
For i = lParas To 1 Step -1
oRng.Select
lEnd = lEnd + oRng.Paragraphs.Count ' Keep track of how many processed
If Len(ActiveDocument.Paragraphs(i).Range.Text) = 1 Then
ActiveDocument.Paragraphs(i).Range.Delete
End If
Next i
Set para = Nothing
Set oDoc = Nothing
Exit Sub
End Sub
You can replace the paragraph marks:
ActiveDocument.Range.Find.Execute FindText:="^p^p", ReplaceWith:="^p", Replace:=wdReplaceAll
ActiveDocument.Range.Find.Execute "^p^p", , , , , , , , , "^p", wdReplaceAll ' might be needed more than once
I want to make a macro that will do the following:
Highlight every nth selection.
Check that selection to ensure it is a word (and not numerical or punctuation).
Cut the word and paste it into another document.
Replace the word with a blank space.
Repeat until the end of the document.
The hard part is checking a selection to validate that it is indeed a word and not something else.
I found some code written by someone else that might work, but I don't understand how to implement it in my macro with the rest of the commands:
Function IsLetter(strValue As String) As Boolean
Dim intPos As Integer
For intPos = 1 To Len(strValue)
Select Case Asc(Mid(strValue, intPos, 1))
Case 65 To 90, 97 To 122
IsLetter = True
Case Else
IsLetter = False
Exit For
End Select
Next
End Function
Sub Blank()
Dim OriginalStory As Document
Set OriginalStory = ActiveDocument
Dim WordListDoc As Document
Set WordListDoc = Application.Documents.Add
Windows(OriginalStory).Activate
sPrompt = "How many spaces would you like between each removed word?"
sTitle = "Choose Blank Interval"
sDefault = "8"
sInterval = InputBox(sPrompt, sTitle, sDefault)
Selection.HomeKey Unit:=wdStory
Do Until Selection.Bookmarks.Exists("\EndOfDoc") = True
Selection.MoveRight Unit:=wdWord, Count:=sInterval, Extend:=wdMove
Selection.MoveRight Unit:=wdWord, Count:=1, Extend:=wdExtend
If IsLetter = True Then
Selection.Cut
Selection.TypeText Text:="__________ "
Windows(WordListDoc).Activate
Selection.PasteAndFormat (wdFormatOriginalFormatting)
Selection.TypeParagraph
Windows(OriginalStory).Activate
Else
Selection.MoveRight Unit:=wdWord, Count:=1, Extend:=wdMove
Selection.MoveRight Unit:=wdWord, Count:=1, Extend:=wdExtend
Loop
Loop
End Sub
The function should sit 'above' the rest of the code right? But I get an error 'argument not optional' when I run it.
Any ideas or tips much appreciated.
I think the code below will do most of what you want. Note that some of the comments relate to the reasons for which I discarded some of your code while others may prove helpful in understanding the present version.
Sub InsertBlanks()
' 02 May 2017
Dim Doc As Document
Dim WordList As Document
Dim Rng As Range
Dim Interval As String, Inter As Integer
Dim Wd As String
' you shouldn't care which Window is active,
' though it probably is the one you want, anyway.
' The important thing is which document you work on.
' Windows(OriginalStory).Activate
Set Doc = ActiveDocument
Application.ScreenUpdating = False
Set WordList = Application.Documents.Add
' If you want to use all these variables you should also declare them.
' However, except for the input itself, they are hardly necessary.
' sPrompt = "How many spaces would you like between each removed word?"
' sTitle = "Choose Blank Interval"
' sDefault = "8"
Do
Interval = InputBox("How many retained words would you like between removed words?", _
"Choose Blank Interval", CStr(8))
If Interval = "" Then Exit Sub
Loop While Val(Interval) < 4 Or Val(Interval) > 25
Inter = CInt(Interval)
' you can modify min and max. Exit by entering a blank or 'Cancel'.
' You don't need to select anything.
' Selection.HomeKey Unit:=wdStory
Set Rng = Doc.Range(1, 1) ' that's the start of the document
' Set Rng = Doc.Bookmarks("James").Range ' I used another start for my testing
Do Until Rng.Bookmarks.Exists("\EndOfDoc") = True
Rng.Move wdWord, Inter
Wd = Rng.Words(1)
If Asc(Wd) < 65 Then
Inter = 1
Else
Set Rng = Rng.Words(1)
With Rng
' replace Len(Wd) with a fixed number of repeats,
' if you don't want to give a hint about the removed word.
.Text = String(Len(Wd) - 1, "_") & " "
.Collapse wdCollapseEnd
End With
With WordList.Range
If .Words.Count > 1 Then .InsertAfter Chr(11)
.InsertAfter Wd
End With
Inter = CInt(Interval)
End If
Loop
Application.ScreenUpdating = True
End Sub
In order to avoid processing non-words my above code tests, roughly, if the first character is a letter (ASCII > 64). This will preclude numbers and it will allow a lot of symbols. For example "€100" would be accepted for replacement but not "100". You may wish to refine this test, perhaps creating a function like you originally did. Another way I thought of would be to exclude "words" of less than 3 characters length. That would eliminate CrLf (if Word considers that one word) but it would also eliminate a lot of prepositions which you perhaps like while doing nothing about "€100". It's either very simple, the way I did it, or it can be quite complicated.
Variatus - thank you so much for this. It works absolutely perfectly and will be really useful for me.
And your comments are helpful for me to understand some of the commands you use that I am not familiar with.
I'm very grateful for your patience and help.