I am looking to create a macro that allows me to delete a specific amount of text until it finds an open paragraph. For example:
gateways (...)
ABC
DEF
GHJ
IKL
Other
I want to delete by searching the word "gateways" and then delete all text until it finds the line before "Other". The problem is that the amount of text until "other" is variable from document to document, and I cannot find a macro that allows me to do that wothout errors.
To delete the text Gateway down to the empty line, leaving only Other
Public Sub DeleteFromTextToEmptyLine()
Selection.Find.ClearFormatting
With Selection.Find
.Text = "gateways"
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindContinue
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
'find text
If Selection.Find.Execute() Then
'Selection.Collapse wdCollapseEnd 'remove comment to leave the word gateway intact
'Turn on Extended select mode
Selection.ExtendMode = True
'find two paragraph marks together, i.e. empty line
With Selection.Find
.Text = "^p^p"
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindContinue
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
'if found, delete the selection
If Selection.Find.Execute() Then Selection.Delete
End If
End Sub
If you want to leave the word gateway, remove the comment from the indicated line
Related
I am new to VBA in MS Word, and only somewhat comfortable with VBA in Excel.
I am trying to process a Word file that has some special information attached to many words. The information is conveyed by changing the color, size, and position of the text, but there is no space between the word and the information.
In order to do some other processing, I need to add a space between the information text and the actual word:
Using VBA I am able to find the informational text by color, using a slight variation on the method described here. I can replace it with "Test". But what I want to do is replace it with a space plus the infromational text. Or perhaps just magically insert a space right before the informational text.
Ideas?
This code:
Sub ChangeColorWithReplace()
Selection.Find.ClearFormatting
Selection.Find.Font.Color = RGB(120, 48, 191)
Selection.Find.Replacement.ClearFormatting
'Selection.Find.Replacement.Font.Color = wdColorRed
With Selection.Find
.Text = ""
.Replacement.Text = "test"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchKashida = False
.MatchDiacritics = False
.MatchAlefHamza = False
.MatchControl = False
.MatchByte = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Selection.Find.Execute Replace:=wdReplaceAll
End Sub
Produces this result:
A neat thing with Word's Find/Replace functionality is that you can use special characters to perform actions like this. Click the "More" button, the "Special" to see a list.
In this case, what you want is to retain the "Find text" in your replacement, which is the character set: ^& which I've added to your sample code, below.
Sub ChangeColorWithReplace()
Selection.Find.ClearFormatting
Selection.Find.Font.Color = RGB(120, 48, 191)
Selection.Find.Replacement.ClearFormatting
With Selection.Find
.Text = ""
.Replacement.Text = "^&test"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchKashida = False
.MatchDiacritics = False
.MatchAlefHamza = False
.MatchControl = False
.MatchByte = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Selection.Find.Execute Replace:=wdReplaceAll
End Sub
I am trying to build a macro (as a noob) to find certain words and then change the formatting for that word (i.e. make it bold or italic)
This code 'sort of works.' It will find some words and change them but not change others. The weird thing is it works until I add another sub then it stops formatting on some words, while formatting others. The routine never breaks and runs until the end without error.
Can anyone teach why this is happening and what I am doing wrong? I am not a programmer. Thanks
Sub Macro2()
'
' Macro2 Macro
'
'
Selection.Find.ClearFormatting
Selection.Find.Replacement.ClearFormatting
With Selection.Find
.Text = "Printer"
.Replacement.Text = ""
.Replacement.Font.bold = True
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = True
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Selection.Find.Execute Replace:=wdReplaceAll
With Selection.Find
.Text = "Parameter Values"
.Replacement.Text = ""
.Replacement.Font.bold = True
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
With Selection.Find
.Text = "Use All Applicants Indicator"
.Replacement.Text = ""
.Replacement.Font.bold = True
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
With Selection.Find
.Text = "Next Section"
.Replacement.Text = ""
.Replacement.Font.bold = True
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
I would program the search macro as a separate sub, like this.
Private Sub FindAndReplace(ByVal Txt As String, _
Optional ByVal NewTxt As String, _
Optional ByVal Fmt As Boolean = False, _
Optional ByVal BldFmt As Boolean = False)
With ActiveDocument.Content
With .Find
.ClearFormatting
.Text = Txt
.Format = Fmt Or BldFmt
With .Replacement
.ClearFormatting
.Text = NewTxt
.Font.Bold = BldFmt
End With
.Forward = True
.Wrap = wdFindContinue
.MatchCase = False
.MatchWholeWord = True
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
.Find.Execute Replace:=wdReplaceAll
End With
End Sub
All the optional parameters you may use but don't have to.
Then I would call the sub repeatedly with different parameters, perhaps like this:-
Sub MakeReplacements1()
FindAndReplace "Printer", BldFmt:=True
FindAndReplace "Parameter values", BldFmt:=True
FindAndReplace "Use All Applicants Indicator", BldFmt:=True
FindAndReplace "Next Section", BldFmt:=True
End Sub
or even like this:-
Sub MakeReplacements2()
Dim Fnd() As String
Dim i As Long
Fnd = Split("Printer|Parameter values|Use All Applicants Indicator|Next Section", "|")
For i = 0 To UBound(Fnd)
FindAndReplace Fnd(i), BldFmt:=True
Next i
End Sub
Selection is an object comprising the part of the document currently selected.
Find is a property of the Selection object defining the Find object (same name but one is a property, the other an object). The Find object has properties such as Text, Forward, Wrap, etc. and it has methods like ClearFormatting or Execute. All of this you can read up in the MSDN library.
Now, when you define the Find object you are describing something you want to find. With the Execute command you start looking for it. Your code is missing this command in some places.
The search is limited to the Selection. If you have selected nothing Word will presume you want to search the whole document. But Selection.Find will change the Selection to highlight the found item. Therefore, if you want to continue searching the whole document you would need to reset the Selection after each search with, for example, Activedocument.Content.Select.
In a nutshell, if you clear the Find object after each use, set a new description before each repeated use, define the Selection object for each search and don't forget to Execute each separate search your code should work just as you intend it to work.
What I need to do is add a Rich Text Content Control to all "xx"'s in a large word document.
I saw a similar thread on highlighting multiple instances. I based the solution that I've tried so far on this:
Options.DefaultHighlightColorIndex = wdYellow
Selection.Find.ClearFormatting
Selection.Find.Replacement.ClearFormatting
Selection.Find.Replacement.Highlight = True
With Selection.Find
.Text = "target1"
.Replacement.Text = "target1"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Selection.Find.Execute Replace:=wdReplaceAll
Now just testing in word to see what command it uses to add the content controls individually I found this command:
Selection.Range.ContentControls.Add (wdContentControlRichText)
What I came up with was:
Sub StandardLanguageVariableSearch()
Selection.Find.ClearFormatting
Selection.Find.Replacement.ClearFormatting
Selection.Range.ContentControls.Add (wdContentControlRichText)
With Selection.Find
.Text = "xx"
.Replacement.Text = "xx"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Selection.Find.Execute Replace:=wdReplaceAll
End Sub
Swapping out Selection.Range with Selection.Find simply resulted in a "method or data field not found" or something like that. I think Range refers to the currently highlighted range of characters, whereas find refers to whatever the with block... well, finds.
Whatever find is doesn't appear to have the ability to throw some content controls onto it.
You were quite close:
Sub StandardLanguageVariableSearch()
Selection.Find.ClearFormatting
Selection.Find.Replacement.ClearFormatting
With Selection.Find
.Text = "xx"
'.Replacement.Text = "xx"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
End With
Do While Selection.Find.Execute
'Selection.Text = "" 'uncomment if you want to remove xx
Selection.Range.ContentControls.Add (wdContentControlRichText)
Loop
End Sub
Our Word documents have several occurrences of the text "Division XX" where XX ranges from 00 thru 99. I need to boldface these using VBA. Below is some code I adopted which gets me part way
With mDoc.Tables(1).Cell(1, 1).Range.Find
.Forward = True
.Wrap = wdFindStop
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
.Replacement.Font.Bold = True
.Execute FindText:="Division", Format:=True, ReplaceWith:="Division", Replace:=wdReplaceAll
End With
However, it doesn't select or highlight the " XX". I tried and failed to use wildcards because it wasn't clear to me what to use for the ReplaceWith value?
This worked for me:
Sub Tester()
With ThisDocument.Range.Find
.Forward = True
.Wrap = wdFindStop
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchWildcards = True '<<**
.MatchSoundsLike = False
.MatchAllWordForms = False
.Replacement.Font.Bold = True
'Find instances of "Division" followed by space then at least one digit
.Execute FindText:="Division [0-9]{1,}", Format:=True, Replace:=wdReplaceAll
End With
End Sub
See: http://word.mvps.org/faqs/general/usingwildcards.htm
Give this a shot
Option Compare Text
Dim Search as Variant
For Each Search In ActiveSheet.UsedRange
If Search.Value Like "Division*" Then
Search.Font.Bold = True
End If
Next Search
Hope this helps
in my word file there are hundrens of paragraph which like the format below. There is a single letter Y here. It can be other letter except "A","T","C","G". I want to remove the white space in it then create a new line.
AAATGGGCCC CACAGAAGTG AGAATGGGTG AAGTCAGAAT TCCTGGTAAT GAAGTGCTTG
AACTTGGATT CCTCCCGACA TGTGCAGTAC AATGAGATGA TTTTCTCCTT AATGAGATTA
GGAAATTCTA TTAGCGCTCC CAGCTGCTGA CCCGATTCCA TGAGGCTGAG GCTCCAGGGC
TGAACCTGCC TGGTT
Y
AGTGTTCCTG GAAACTAGAC ACCCCACCCT TCAGATGGGC CAGGGCCTCC CCAGCTCTAC
CTAAAGCTGT GGTCTGCCCC CAGGGGTGCC CAGTTTCCTC CCTTCACCCT GTGCTCCAGA
GGAGTGTGGG GCCCTGGGCA TTCTGCAGTG TACCCCAGGA TCCTCACTCC TTCCTGCTTA
The new line's format is
AAATGGGCCCCACAGAAGTGAGAATGGGTGAAGTCAGAATTCCTGGTAATGAAGTGCTTGAACTTGGATTCCTCCCGACATGTGCAGTACAATGAGATGATTTTCTCCTTAATGAGATTAGGAAATTCTATTAGCGCTCCCAGCTGCTGACCCGATTCCATGAGGCTGAGGCTCCAGGGCTGAACCTGCCTGGTT[Y]AGTGTTCCTGGAAACTAGACACCCCACCCTTCAGATGGGCCAGGGCCTCCCCAGCTCTACCTAAAGCTGTGGTCTGCCCCCAGGGGTGCCCAGTTTCCTCCCTTCACCCTGTGCTCCAGAGGAGTGTGGGGCCCTGGGCATTCTGCAGTGTACCCCAGGATCCTCACTCCTTCCTGCTTA
Notice Y becomes [Y].
The final result will be saved as a text file. Thanks for help.
You don't need to write a program. The “Replace” tool is sufficient for this:
Replace Y with [Y] (EDIT: see the comments below, because it's a little more complex than that indeed)
Replace ^w with nothing (^w means whitespace)
Replace ^p with nothing (^p means paragraph markers)
EDIT: if you need a macro, just do the above once while recording a macro.
EDIT: by applying the method discussed in the comments, I get the following VBA macro:
Sub ProcessATCG()
Selection.Find.ClearFormatting
Selection.Find.Replacement.ClearFormatting
With Selection.Find
.Text = "([!ACGT^13^32])"
.Replacement.Text = "[\1]"
.Forward = True
.Wrap = wdFindContinue
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchAllWordForms = False
.MatchSoundsLike = False
.MatchWildcards = True
End With
Selection.Find.Execute Replace:=wdReplaceAll
With Selection.Find
.Text = "[^13^32]"
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindContinue
.Format = False
.MatchCase = False
.MatchWholeWord = False
.MatchAllWordForms = False
.MatchSoundsLike = False
.MatchWildcards = True
End With
Selection.Find.Execute Replace:=wdReplaceAll
End Sub