I have been using a macro by Rodney Atkins called "ReinsertComments" that was posted by someone in the comments section over at CyberText (see macro below): https://cybertext.wordpress.com/2013/04/10/word-removing-reviewer-names/. It reinserts all the comments in a Word document. I would like to find a way, if at all possible, to reinsert only one user's comments, but not all other users' comments.
Is it possible to specify per author which comments will be
reinserted?
Could you run an If/Then statement to specify a user name for the comments to be reinserted? Perhaps something along the lines of
If myComment.Author = "Jane" Then
If a version of that is possible, where should I insert the If/Then
and End If portion in the macro below?
Thanks to all! :)
Sub CommentsReinsert()
Dim myComment As Comment
Dim myComText As String
Dim comStart
Dim comEnd
Dim i
On Error GoTo Done
Application.ScreenUpdating = False
If ActiveDocument.TrackRevisions = True Then
With ActiveDocument
.TrackRevisions = False
End With
End If
For i = ActiveDocument.Comments.Count To 1 Step -1
Set myComment = ActiveDocument.Comments(i)
myComText = myComment.Range.Text
comStart = myComment.Scope.Start
comEnd = myComment.Scope.End
myComment.Reference.Select
myComment.Delete
ActiveDocument.Range(comStart, comEnd).Select
ActiveDocument.Comments.Add _
Range:=Selection.Range, Text:=myComText
Next i
ActiveWindow.ActivePane.Close
Application.ScreenUpdating = True
Done:
End Sub
You could use code like:
Sub ReinsertComments()
Application.ScreenUpdating = False
Dim bRev As Boolean, i As Long, Rng As Range, Cmt As Comment
With ActiveDocument
bRev = .TrackRevisions
.TrackRevisions = False
For i = .Comments.Count To 1 Step -1
With .Comments(i)
Select Case .Author
Case "Jane"
Set Rng = .Range
Set Cmt = ActiveDocument.Comments.Add(Range:=.Scope, Text:="")
Cmt.Range.FormattedText = Rng.FormattedText ': Cmt.Author = "Anon."
Rng.Comments(1).Delete
End Select
End With
Next
.TrackRevisions = bRev
End With
Set Cmt = Nothing: Set Rng = Nothing
Application.ScreenUpdating = True
End Sub
Unlike the code you posted, the above preserves the original comment's formatting. Note the commented-out code for changing the inserted comment's author name.
An alternative approach would be to temporarily change the UserName & Initials in the Word Options, to that of the commentator, then simply change the Comment.Author property to whatever other name you want - without all the deletion/reinsertion circumlocution.
I imagine people would be concerned about the dates if there are replies to comments and the above macro was deleting then reinserting them with a different date...
As I said in my previous reply, you could use the alternative approach of:
Sub ChangeCommentAuthor()
Application.ScreenUpdating = False
Dim bRev As Boolean, i As Long, UsrNm As String, UsrIn As String, bUsr As Boolean
UsrNm = Application.UserName: UsrIn = Application.UserInitials: bUsr = Options.UseLocalUserInfo
'Nominate the UserName & Initials to change from
Application.UserName = "Figgie 10": Application.UserInitials = "F": Options.UseLocalUserInfo = True
With ActiveDocument
bRev = .TrackRevisions
.TrackRevisions = False
For i = .Comments.Count To 1 Step -1
With .Comments(i)
'Nominate the UserName & Initials to change to
If .Author = Application.UserName Then .Author = "John Doe": .Initial = "JD"
End With
Next
.TrackRevisions = bRev
End With
Application.UserName = UsrNm: Application.UserInitials = UsrIn: Options.UseLocalUserInfo = bUsr
Application.ScreenUpdating = True
End Sub
Related
I need to send an document as email body. The document contains several { FORMCHECKBOX } fields, which are either selected or not. Now, whenever I actually send the email, the checkboxes are not included at all. Because of this issue, I thougth about replacing the checkboxes with "Yes" or "No", depending on the selection.
I found this code, to replace all checkboxes, but it doesn't care for the actual selection:
Sub ChkBxClr()
Dim i As Long, Rng As Range
With ActiveDocument
For i = .FormFields.Count To 1 Step -1
With .FormFields(i)
If .Type = wdFieldFormCheckBox Then
Set Rng = .Range
.Delete
Rng.Text = "[__]"
End If
End With
Next
Set Rng = Nothing
End With
End Sub
Also this code changes the whole document and not just the email body, which I would like to prevent.
The overall code:
Sub SendEmail()
Dim Subject As String
Dim From As String
On Error Resume Next
Selection.GoTo What:=wdGoToBookmark, Name:="From"
From = Selection.Text
Subject = "Title"
Call ChkBxClr
ActiveWindow.EnvelopeVisible = True
With ActiveDocument.MailEnvelope.Item
.To = "<some mail address>"
.Subject = Subject
.Send
End With
End Sub
Sub ChkBxClr()
Dim i As Long, Rng As Range
With ActiveDocument
For i = .FormFields.Count To 1 Step -1
With .FormFields(i)
If .Type = wdFieldFormCheckBox Then
Set Rng = .Range
.Delete
Rng.Text = "[__]"
End If
End With
Next
Set Rng = Nothing
End With
End Sub
So the question is, how can I either display the checkboxes within the email body or replace the checkboxes (preferably only in message body) with a text, depending on the selection?
I have the following code that I found when googleing on the problem. The problem with this code is that it overwrites the next-to-last section header (and footer though I only need the header preserved) to that of the last section, which is the default (strange) behavior of Word.
Is there a workaround to this in VBA?
Here is the code that has the inherent fault:
Sub DeleteLastSection()
'Deletes last section of a document including
'the section break
Dim doc As Document
Dim rng As Range
Dim ctr As Integer
Set doc = ActiveDocument
ctr = doc.Sections.Count
Set rng = doc.Sections(ctr).Range
Dim myheader As HeaderFooter
If ctr > 1 Then
With rng
.Select
.MoveStart Unit:=wdCharacter, Count:=-1
.Delete
End With
End If
End Sub
Note: The entire range of the last section is being deleted by the code and that is the required behavior. The inherent problem in the default behavior of Word is what I needed a workaround for in VBA code. One can found complex manual procedures to avoid it, but I needed a simple approach in code.
The problem here lies in the fact that the section break carries the section information. If you delete it, the last section becomes part of the section before. The trick I use below is to create a continuous section break instead of a page break, and then do all the rest:
Sub DeleteLastSection()
'Deletes last section of a document including
'the section break
Dim doc As Document
Dim rng As Range
Dim NewEndOfDocument As Range
Dim ctr As Integer
Set doc = ActiveDocument
ctr = doc.Sections.Count
Set rng = doc.Sections(ctr).Range
If ctr > 1 Then
' Create a section break at the end of the second to last section
Set NewEndOfDocument = doc.Sections(ctr - 1).Range
NewEndOfDocument.EndOf wdSection, wdMove
doc.Sections.Add NewEndOfDocument, wdSectionContinuous
With rng
.Select
.MoveStart Unit:=wdCharacter, Count:=-1
.Delete
End With
End If
End Sub
Ordinarily, deleting a Section break causes the Section preceding the break to assume the page layout of the following Section. The following macro works the other way, across multiple (selected) Section breaks. All common page layout issues (margins, page orientation, text columns, headers & footers) are addressed. As you can see by studying the code, it's no trivial task to do all these things.
Sub MergeSections()
Application.ScreenUpdating = False
Dim sPageHght As Single, sPageWdth As Single
Dim sHeaderDist As Single, sFooterDist As Single
Dim sTMargin As Single, sBMargin As Single
Dim sLMargin As Single, sRMargin As Single
Dim sGutter As Single, sGutterPos As Single
Dim lPaperSize As Long, lGutterStyle As Long
Dim lMirrorMargins As Long, lVerticalAlignment As Long
Dim lScnStart As Long, lScnDir As Long
Dim lOddEvenHdFt As Long, lDiffFirstHdFt As Long
Dim bTwoPagesOnOne As Boolean, bBkFldPrnt As Boolean
Dim bBkFldPrnShts As Boolean, bBkFldRevPrnt As Boolean
Dim lOrientation As Long, oHdFt As HeaderFooter
Dim Sctn1 As Section, Sctn2 As Section
With Selection
If .Sections.Count = 1 Then
MsgBox "Selection does not span a Section break", vbExclamation
Exit Sub
End If
Set Sctn1 = .Sections.First: Set Sctn2 = .Sections.Last
With Sctn1.PageSetup
lPaperSize = .PaperSize
lGutterStyle = .GutterStyle
lOrientation = .Orientation
lMirrorMargins = .MirrorMargins
lScnStart = .SectionStart
lScnDir = .SectionDirection
lOddEvenHdFt = .OddAndEvenPagesHeaderFooter
lDiffFirstHdFt = .DifferentFirstPageHeaderFooter
lVerticalAlignment = .VerticalAlignment
sPageHght = .PageHeight
sPageWdth = .PageWidth
sTMargin = .TopMargin
sBMargin = .BottomMargin
sLMargin = .LeftMargin
sRMargin = .RightMargin
sGutter = .Gutter
sGutterPos = .GutterPos
sHeaderDist = .HeaderDistance
sFooterDist = .FooterDistance
bTwoPagesOnOne = .TwoPagesOnOne
bBkFldPrnt = .BookFoldPrinting
bBkFldPrnShts = .BookFoldPrintingSheets
bBkFldRevPrnt = .BookFoldRevPrinting
End With
With Sctn2.PageSetup
.GutterStyle = lGutterStyle
.MirrorMargins = lMirrorMargins
.SectionStart = lScnStart
.SectionDirection = lScnDir
.OddAndEvenPagesHeaderFooter = lOddEvenHdFt
.DifferentFirstPageHeaderFooter = lDiffFirstHdFt
.VerticalAlignment = lVerticalAlignment
.PageHeight = sPageHght
.PageWidth = sPageWdth
.TopMargin = sTMargin
.BottomMargin = sBMargin
.LeftMargin = sLMargin
.RightMargin = sRMargin
.Gutter = sGutter
.GutterPos = sGutterPos
.HeaderDistance = sHeaderDist
.FooterDistance = sFooterDist
.TwoPagesOnOne = bTwoPagesOnOne
.BookFoldPrinting = bBkFldPrnt
.BookFoldPrintingSheets = bBkFldPrnShts
.BookFoldRevPrinting = bBkFldRevPrnt
.PaperSize = lPaperSize
.Orientation = lOrientation
End With
With Sctn2
For Each oHdFt In .Footers
oHdFt.LinkToPrevious = Sctn1.Footers(oHdFt.Index).LinkToPrevious
If oHdFt.LinkToPrevious = False Then
Sctn1.Headers(oHdFt.Index).Range.Copy
oHdFt.Range.Paste
End If
Next
For Each oHdFt In .Headers
oHdFt.LinkToPrevious = Sctn1.Headers(oHdFt.Index).LinkToPrevious
If oHdFt.LinkToPrevious = False Then
Sctn1.Headers(oHdFt.Index).Range.Copy
oHdFt.Range.Paste
End If
Next
End With
While .Sections.Count > 1
.Sections.First.Range.Characters.Last.Delete
Wend
Set Sctn1 = Nothing: Set Sctn2 = Nothing
End With
Application.ScreenUpdating = True
End Sub
Looking more into this on my own (I had to solve the issue in short order and could not wait), I came to the same conclusion as was noted in the comment by #CindyMeister that when deleting the last "section break" in actual fact the next-to-last section is being deleted, and what data and formatting heretofore belonged to the last section is apparently inherited by the new last section (i.e. the earlier next-to-last section). But in reality the last section remained and only the section break was deleted, so what was deleted was the next-to-last section (and the actual pages from the last section).
I found that the LinkToPrevious property of the HeaderFooter object allows a simplistic approach to "inherit" the settings from the previous section.
So by adding a few lines to set this property to true in each instance and then change it back to false, I can get the required behavior of the next-to-last section remaining the same as before.
(Please note that it worked for me because I simply had different text in the primary header, and did not have special formatting and else. But I suspect that based on the workings of the LinkToPrevious property this is a panacea. Please comment if otherwise.)
These are the lines to set the property:
for each hf in .Sections(1).Headers
hf.LinkToPrevious = True
hf.LinkToPrevious = False
next
for each hf in .Sections(1).Footers
hf.LinkToPrevious = True
hf.LinkToPrevious = False
next
The full working code for progeny:
Sub DeleteLastSection()
'Deletes last section of a document including
'the section break
Dim doc As Document
Dim rng As Range
Dim ctr As Integer
Set doc = ActiveDocument
ctr = doc.Sections.Count
Set rng = doc.Sections(ctr).Range
Dim myheader As HeaderFooter
If ctr > 1 Then
With rng
'Added lines to "inherit" the settings from the next-to-last section
for each hf in .Sections(1).Headers
hf.LinkToPrevious = True
hf.LinkToPrevious = False
next
for each hf in .Sections(1).Footers
hf.LinkToPrevious = True
hf.LinkToPrevious = False
next
.Select
.MoveStart Unit:=wdCharacter, Count:=-1
.Delete
End With
End If
End Sub
Deleting the last section of a word document is not a trivial task.
Things you might have to do if items are different between the 'next to last' and 'last' section of a document.
Ensure that in the last section any 'linktoprevious' in a header or footer is set to to false
Copy all headers and footers from the next to last section to the last section
Copy the relevant page format items of the next to last section to the last section (paper size, orientation, margins etc)
Get the range for the last section in the document. Move the end of the range backward until the ascii value is >=32.
Then you can safely delete the adjusted range from your document without any nasty side effects
This is the code I just created that works well:
Sub DeleteLastPage()
Dim pgSetUp As PageSetup
Dim iSect As Integer
iSect = ActiveDocument.Sections.Count - 1
Set pgSetUp = ActiveDocument.Sections(iSect).PageSetup
With ActiveDocument.Sections.Last.PageSetup
.LineNumbering.Active = pgSetUp.LineNumbering.Active
.Orientation = pgSetUp.Orientation
.TopMargin = pgSetUp.TopMargin
.BottomMargin = pgSetUp.BottomMargin
.LeftMargin = pgSetUp.LeftMargin
.RightMargin = pgSetUp.RightMargin
.Gutter = pgSetUp.Gutter
.HeaderDistance = pgSetUp.HeaderDistance
.FooterDistance = pgSetUp.FooterDistance
.PageWidth = pgSetUp.PageWidth
.PageHeight = pgSetUp.PageHeight
.SectionStart = wdSectionNewPage
.OddAndEvenPagesHeaderFooter = pgSetUp.OddAndEvenPagesHeaderFooter
.DifferentFirstPageHeaderFooter = pgSetUp.DifferentFirstPageHeaderFooter
.VerticalAlignment = wdAlignVerticalTop
End With
With ActiveDocument.Sections.Last.Headers(wdHeaderFooterPrimary)
.LinkToPrevious = true
End With
With ActiveDocument.Sections.Last.Footers(wdHeaderFooterPrimary)
.LinkToPrevious = true
End With
ActiveDocument.Sections.Last.Range.Characters.Delete
Selection.EndKey Unit:=wdStory
Selection.MoveLeft Unit:=wdCharacter, Count:=1, Extend:=wdExtend
Selection.Delete Unit:=wdCharacter,Count:=1
End Sub
Copy pasting 1 line of text from word to excel using VBA.
When the code reaches the below line I am getting the below error.
ActiveSheet.Paste
Run Time Error '1004': Paste Method Of worksheet Class Failed error
But if I click Debug button and press F8 then it's pasting the data in excel without any error.
This error occurs each time the loop goes on and pressing debug and F8 pasting the data nicely.
I did several testing and unable to find the root cause of this issue.
Also used DoEvents before pasting the data code but nothing worked.
Any suggestions?
EDIT:-
I am posting the code since both of you are saying the same. Here is the code for your review.
Sub FindAndReplace()
Dim vFR As Variant, r As Range, i As Long, rSource As Range
Dim sCurrRep() As String, sGlobalRep As Variant, y As Long, x As Long
Dim NumCharsBefore As Long, NumCharsAfter As Long
Dim StrFind As String, StrReplace As String, CountNoOfReplaces As Variant
'------------------------------------------------
Dim oWord As Object
Const wdReplaceAll = 2
Set oWord = CreateObject("Word.Application")
'------------------------------------------------
Application.ScreenUpdating = False
vFR = ThisWorkbook.Sheets("Sheet1").Range("A1").CurrentRegion.Value
On Error Resume Next
Set rSource = Cells.SpecialCells(xlCellTypeConstants)
On Error GoTo 0
If Not rSource Is Nothing Then
For Each r In rSource.Cells
For i = 2 To UBound(vFR)
If Trim(vFR(i, 1)) <> "" Then
With oWord
.Documents.Add
DoEvents
r.Copy
.ActiveDocument.Content.Paste
NumCharsBefore = .ActiveDocument.Characters.Count
With .ActiveDocument.Content.Find
.ClearFormatting
.Font.Bold = False
.Replacement.ClearFormatting
.Execute FindText:=vFR(i, 1), ReplaceWith:=vFR(i, 2), Format:=True, Replace:=wdReplaceAll
End With
.Selection.Paragraphs(1).Range.Select
.Selection.Copy
r.Select
ActiveSheet.Paste'Error occurs in this line pressing debug and F8 is pasting the data
StrFind = vFR(i, 1): StrReplace = vFR(i, 2)
NumCharsAfter = .ActiveDocument.Characters.Count
CountNoOfReplaces = (NumCharsBefore - NumCharsAfter) / (Len(StrFind) - Len(StrReplace))
.ActiveDocument.UndoClear
.ActiveDocument.Close SaveChanges:=False
If CountNoOfReplaces Then
x = x + 1
ReDim Preserve sCurrRep(1 To 3, 1 To x)
sCurrRep(1, x) = vFR(i, 1)
sCurrRep(2, x) = vFR(i, 2)
sCurrRep(3, x) = CountNoOfReplaces
End If
CountNoOfReplaces = 0
End With
End If
Next i
Next r
End If
oWord.Quit
'Some more gode goes here... which is not needed since error occurs in the above loop
End Sub
If you want to know why I have chosen word for replacement then please go through the below link.
http://www.excelforum.com/excel-programming-vba-macros/1128898-vba-characters-function-fails-when-the-cell-content-exceeds-261-characters.html
Also used the code from the below link to get the number of replacements count.
http://word.mvps.org/faqs/macrosvba/GetNoOfReplacements.htm
Characters(start, length).Delete() method really seems not to work with longer strings in Excel :(. So a custom Delete() method could be written which will work with decoupled formating informations and texts. So the text of the cell can be modified without loosing the formating information. HTH.
Add new class named MyCharacter. It will contain information about text and
formating of one character:
Public Text As String
Public Index As Integer
Public Name As Variant
Public FontStyle As Variant
Public Size As Variant
Public Strikethrough As Variant
Public Superscript As Variant
Public Subscript As Variant
Public OutlineFont As Variant
Public Shadow As Variant
Public Underline As Variant
Public Color As Variant
Public TintAndShade As Variant
Public ThemeFont As Variant
Add next new class named MyCharcters and wrap the code of the new
Delete method in it. With Filter method a new collection of MyCharacter is created. This collection contains only the characters which should remain. Finally in method Rewrite the text is re-written from this collection back to target range along with formating info:
Private m_targetRange As Range
Private m_start As Integer
Private m_length As Integer
Private m_endPosition As Integer
Public Sub Delete(targetRange As Range, start As Integer, length As Integer)
Set m_targetRange = targetRange
m_start = start
m_length = length
m_endPosition = m_start + m_length - 1
Dim filterdChars As Collection
Set filterdChars = Filter
Rewrite filterdChars
End Sub
Private Function Filter() As Collection
Dim i As Integer
Dim newIndex As Integer
Dim newChar As MyCharacter
Set Filter = New Collection
newIndex = 1
For i = 1 To m_targetRange.Characters.Count
If i < m_start Or i > m_endPosition Then
Set newChar = New MyCharacter
With newChar
.Text = m_targetRange.Characters(i, 1).Text
.Index = newIndex
.Name = m_targetRange.Characters(i, 1).Font.Name
.FontStyle = m_targetRange.Characters(i, 1).Font.FontStyle
.Size = m_targetRange.Characters(i, 1).Font.Size
.Strikethrough = m_targetRange.Characters(i, 1).Font.Strikethrough
.Superscript = m_targetRange.Characters(i, 1).Font.Superscript
.Subscript = m_targetRange.Characters(i, 1).Font.Subscript
.OutlineFont = m_targetRange.Characters(i, 1).Font.OutlineFont
.Shadow = m_targetRange.Characters(i, 1).Font.Shadow
.Underline = m_targetRange.Characters(i, 1).Font.Underline
.Color = m_targetRange.Characters(i, 1).Font.Color
.TintAndShade = m_targetRange.Characters(i, 1).Font.TintAndShade
.ThemeFont = m_targetRange.Characters(i, 1).Font.ThemeFont
End With
Filter.Add newChar, CStr(newIndex)
newIndex = newIndex + 1
End If
Next i
End Function
Private Sub Rewrite(chars As Collection)
m_targetRange.Value = ""
Dim i As Integer
For i = 1 To chars.Count
If IsEmpty(m_targetRange.Value) Then
m_targetRange.Value = chars(i).Text
Else
m_targetRange.Value = m_targetRange.Value & chars(i).Text
End If
Next i
For i = 1 To chars.Count
With m_targetRange.Characters(i, 1).Font
.Name = chars(i).Name
.FontStyle = chars(i).FontStyle
.Size = chars(i).Size
.Strikethrough = chars(i).Strikethrough
.Superscript = chars(i).Superscript
.Subscript = chars(i).Subscript
.OutlineFont = chars(i).OutlineFont
.Shadow = chars(i).Shadow
.Underline = chars(i).Underline
.Color = chars(i).Color
.TintAndShade = chars(i).TintAndShade
.ThemeFont = chars(i).ThemeFont
End With
Next i
End Sub
How to use it:
Sub test()
Dim target As Range
Dim myChars As MyCharacters
Application.ScreenUpdating = False
Set target = Worksheets("Demo").Range("A1")
Set myChars = New MyCharacters
myChars.Delete targetRange:=target, start:=300, length:=27
Application.ScreenUpdating = True
End Sub
Before:
After:
To make it more stable, you should:
Disable all events while operating
Never call .Activate or .Select
Paste directly in the targeted cell with WorkSheet.Paste
Cancel the Copy operation with Application.CutCopyMode = False
Reuse the same document and not create one for each iteration
Do as less operations as possible in an iteration
Use early binding [New Word.Application] instead of late binding [CreateObject("Word.Application")]
Your example refactored :
Sub FindAndReplace()
Dim dictionary(), target As Range, ws As Worksheet, cell As Range, i As Long
Dim strFind As String, strReplace As String, diffCount As Long, replaceCount As Long
Dim appWord As Word.Application, content As Word.Range, find As Word.find
dictionary = [Sheet1!A1].CurrentRegion.Value
Set target = Cells.SpecialCells(xlCellTypeConstants)
' launch and setup word
Set appWord = New Word.Application
Set content = appWord.Documents.Add().content
Set find = content.find
find.ClearFormatting
find.Font.Bold = False
find.replacement.ClearFormatting
' disable events
Application.Calculation = xlManual
Application.ScreenUpdating = False
Application.EnableEvents = False
' iterate each cell
Set ws = target.Worksheet
For Each cell In target.Cells
' copy the cell to Word and disable the cut
cell.Copy
content.Delete
content.Paste
Application.CutCopyMode = False
' iterate each text to replace
For i = 2 To UBound(dictionary)
If Trim(dictionary(i, 1)) <> Empty Then
replaceCount = 0
strFind = dictionary(i, 1)
strReplace = dictionary(i, 2)
' replace in the document
diffCount = content.Characters.count
find.Execute FindText:=strFind, ReplaceWith:=strReplace, format:=True, Replace:=2
' count number of replacements
diffCount = diffCount - content.Characters.count
If diffCount Then
replaceCount = diffCount \ (Len(strFind) - Len(strReplace))
End If
Debug.Print replaceCount
End If
Next
' copy the text back to Excel
content.Copy
ws.Paste cell
Next
' terminate Word
appWord.Quit False
' restore events
Application.Calculation = xlCalculationAutomatic
Application.ScreenUpdating = True
Application.EnableEvents = True
End Sub
How about change it from: activesheet.paste
to:
activesheet.activate
activecell.pastespecial xlpasteAll
This post seems to explain the problem and provide two solutions:
http://www.excelforum.com/excel-programming-vba-macros/376722-runtime-error-1004-paste-method-of-worksheet-class-failed.html
Two items come to light in this post:
Try using Paste Special
Specify the range you wish to paste to.
Another solution would be to extract the targeted cells as XML, replace the text with a regular expression and then write the XML back to the sheet.
While it's much faster than working with Word, it might require some knowledge with regular expressions if the formats were to be handled. Moreover it only works with Excel 2007 and superior.
I've assemble an example that replaces all the occurences with the same style:
Sub FindAndReplace()
Dim area As Range, dictionary(), xml$, i&
Dim matchCount&, replaceCount&, strFind$, strReplace$
' create the regex object
Dim re As Object, match As Object
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.MultiLine = True
' copy the dictionary to an array with column1=search and column2=replacement
dictionary = [Sheet1!A1].CurrentRegion.Value
'iterate each area
For Each area In ActiveSheet.Cells.SpecialCells(xlCellTypeConstants)
' read the cells as XML
xml = area.Value(xlRangeValueXMLSpreadsheet)
' iterate each text to replace
For i = 2 To UBound(dictionary)
If Trim(dictionary(i, 1)) <> Empty Then
strFind = dictionary(i, 1)
strReplace = dictionary(i, 2)
' set the pattern
re.pattern = "(>[^<]*)" & strFind
' count the number of occurences
matchCount = re.Execute(xml).count
If matchCount Then
' replace each occurence
xml = re.Replace(xml, "$1" & strReplace)
replaceCount = replaceCount + matchCount
End If
End If
Next
' write the XML back to the sheet
area.Value(xlRangeValueXMLSpreadsheet) = xml
Next
' print the number of replacement
Debug.Print replaceCount
End Sub
DDuffy's answer is useful.
I found the code can run normally at slowly cpu PC .
add the bellow code before paste, the problem is sloved:
Application.Wait (Now + TimeValue("0:00:1"))'wait 1s or more
ActiveSheet.Paste
I'm trying to write some VBA code that will unhide an entire row if another specific row is hidden. This macro also hides a range of rows based on the value in a specific column. This aspect works fine - I have reliable code. I can't get the first function I described to work. Should be easy to do, just don't know the syntax. This subroutine should execute upon opening the workbook.
Private Sub Workbook_Open()
Application.ScreenUpdating = False
Dim targ As Range
Dim msg As Range
targ = "DETAILS!B6"
msg = "DETAILS!B42"
msg.EntireRow.Hidden = True
With Range("DETAILS!B6:B40")
.EntireRow.Hidden = False
For Each cell In Range("DETAILS!B6:B40")
Select Case cell.Value
Case Is = 0
cell.EntireRow.Hidden = True
End Select
Next cell
End With
If targ.EntireRow.Hidden = True Then
msg.EntireRow.Hidden = False
End If
Application.ScreenUpdating = True
End Sub
You need to set the variables like below
Private Sub Workbook_Open()
Application.ScreenUpdating = False
Dim targ As Range
Dim msg As Range
Set targ = "DETAILS!B6"
Set msg = "DETAILS!B42"
msg.EntireRow.Hidden = True
With Range("DETAILS!B6:B40")
.EntireRow.Hidden = False
For Each cell In Range("DETAILS!B6:B40")
Select Case cell.Value
Case Is = 0
cell.EntireRow.Hidden = True
End Select
Next cell
End With
If targ.EntireRow.Hidden = True Then
msg.EntireRow.Hidden = False
End If
Application.ScreenUpdating = True
End Sub
Oh! Just put Set before targ and msg since they're a Range. When declaring ranges, you have to have Set, i.e. Set myRng = Range("A1:A10").
You might need to do Set targ = Range("Details!B6") if just Set Targ = "DetailsB6" doesn't work.
On second thought, I don't think Set Targ = "Details!B6" will work if you are Dim Targ as Range. You're dim'ing as a Range, but are declaring it as like a string. You need this to be a Range, to use it like targ.EntireRow.Hidden, etc.
Though you can call range objects like this : Range("DETAILS!B6:B40")
In vba it is better accepted to call it like this: Sheets("DETAILS").Range("B6:B40")
I fixed a few more syntax errors:
Private Sub Workbook_Open()
Application.ScreenUpdating = False
Dim targ As Range
Dim msg As Range
Set targ = Sheets("DETAILS").Range("B6")
Set msg = Sheets("DETAILS").Range("B42")
msg.EntireRow.Hidden = True
With Sheets("DETAILS").Range("B6:B40")
.EntireRow.Hidden = False
End With
For Each cell In Sheets("DETAILS").Range("B6:B40")
Select Case cell.Value
Case 0
cell.EntireRow.Hidden = True
End Select
Next cell
If targ.EntireRow.Hidden = True Then
msg.EntireRow.Hidden = False
End If
Application.ScreenUpdating = True
End Sub
This would be a very simple question.
But I am not sure why this is not working in my excel vba code.
Sheets("I- ABC").Select
If IsEmpty(Range("A3").Value) = True And _
IsEmpty(Range("A4").Value) = True And _
IsEmpty(Range("A5").Value) = True And _
IsEmpty(Range("A6").Value) = True Then
Sheets("I- ABC").Delete
End If
What type of error do you get? I tried this code and Excel displays only warning message:
You can avoid this message by adding:
Application.DisplayAlerts = False
and
Application.DisplayAlerts = True
at the beginning and at the end of your code respectively.
--Edited code
Sub Example()
Application.DisplayAlerts = False
With Sheets("I- ABC")
If Application.WorksheetFunction.CountA(.Range("A3:A6")) = 0 Then
.Delete
End If
End With
Application.DisplayAlerts = True
End Sub
Try Similiar to This
Sub Test()
Application.DisplayAlerts = False
With Sheets("Sheet1")
Columns("A:A").SpecialCells(xlCellTypeBlanks).EntireRow.Delete
End With
Application.DisplayAlerts = True
End Sub
PS: It works for me and deletes rows containg empty cells in `A:A``
Approach Suggested by #Tim Williams also works for me as per following code in my situation
Sub Test6()
Dim r As Range, rows As Long, i As Long
Set r = ActiveSheet.Range("A3:A6")
rows = r.rows.Count
For i = rows To 1 Step (-1)
If WorksheetFunction.CountA(r.rows(i)) = 0 Then r.rows(i).Delete
Next
End Sub
It works even if we use Application instead of WorksheetFunction
If If Application.CountA(Range("A3:A6")) = 0 Then is not working as Tim suggested then that means the cells have blank spaces or unprintable characters.
Try this
Sub Sample()
Dim pos As Long
With Sheets("I- ABC")
pos = Len(Trim(.Range("A3").Value)) + _
Len(Trim(.Range("A4").Value)) + _
Len(Trim(.Range("A5").Value)) + _
Len(Trim(.Range("A6").Value))
If pos = 0 Then
Application.DisplayAlerts = False
.Delete
Application.DisplayAlerts = True
Else
MsgBox "The cells are not empty"
End If
End With
End Sub
With skkakkar's idea expanded.
Sub Hello()
Dim rng As Range
Application.DisplayAlerts = 0
On Error GoTo er
Set rng = Range("A3:A6").SpecialCells(xlCellTypeConstants, 23)
Exit Sub
er: MsgBox "ActiveSheet.Delete" 'delete sheet
End Sub
If the spaces are the issue, then you can try this code:
Public Sub RemoveIfEmpty()
Application.DisplayAlerts = False
With Sheets("I- ABC")
If Trim(.Range("A3") & .Range("A4") & .Range("A5") & .Range("A6")) = "" Then
.Delete
End If
End With
Application.DisplayAlerts = True
End Sub