How to remove duplicate items in listbox - vba

I created this code to add found items enclosed with "[]" or "()" or "{}". If in my word document I have "ouch! [crying] That hurts! [crying] [laughing]" so the items enclosed with "[]" will be added to the listbox and there are 3 of it but the 2 are the same. I want to merge them.
How would I do that?
Sub cutsound()
Dim arrs, arrs2, c2 As Variant, pcnt, x2, x3, intItems as Integer
pcnt = ActiveDocument.Paragraphs.Count
arrs = Array("[", "(", "{")
arrs2 = Array("]", ")", "}")
UserForm1.Show False
Application.ScreenUpdating = False
With Selection
.WholeStory
For c2 = 0 To UBound(arrs)
.Find.Execute (arrs(c2))
Do While .Find.Found
.MoveEndUntil Cset:=arrs2(c2), Count:=wdForward
.MoveRight Unit:=wdCharacter, Count:=1, Extend:=wdExtend
UserForm1.ListBox1.AddItem Selection.Text
.MoveRight Unit:=wdCharacter, Count:=1
.EndKey Unit:=wdStory, Extend:=wdExtend
.Find.Execute
Loop
Next c2
End With
Application.ScreenUpdating = True
End Sub

Try to merge in a set rather then list, it maintains the duplication.

You could use the keys of a Dictionary to enforce uniqueness. Add a reference (Tools -> References...) to Microsoft Scripting Runtime. Then, do the following:
'I suggest searching using wildcards. The body of your loop will be much simpler
Dim patterns(3) As String, pattern As Variant
'Since these characters have special meaning in wildcards, they need a \ before them
patterns(0) = "\[*\]"
patterns(1) = "\(*\)"
patterns(2) = "\{*\}"
Dim rng As Range 'It's preferable to use a Range for blocks of text instead of Selection,
'unless you specifically want to change the selection
Dim found As New Scripting.Dictionary
For Each pattern In patterns
Set rng = ActiveDocument.Range
With rng
.WholeStory
.Find.Execute pattern, , , True
Do While .Find.found
found(rng.Text) = 1 'an arbitrary value
'If you want the number of times each text appears, the previous line could be modified
.Find.Execute
Loop
End With
Next
Dim key As Variant
For Each key In found.Keys
Debug.Print key
Next
Note: this code won't find entries in the order they appear in the document, but first entries with [], then with (), then with {}.
References:
Dictionary object
Find object
Range object

Related

How to print row of found string?

I'd like to find several strings within Word document and for each string found, I like to print (debug.print for example) the whole row content where the string is found, not the paragraph.
How can I do this? Thanks
Sub FindStrings
Dim StringsArr (1 to 3)
StringsArr = Array("string1","string2","string3")
For i=1 to 3
With
Selection.Find
.ClearFormatting
.Text = Strings(i)
Debug.Print CurrentRow 'here I need help
End With
Next
End Sub
The term Row in Word is used only in the context of a table. I assume the term you mean is Line, as in a line of text.
The Word object model has no concept of "line" (or "page") due to the dynamic layout algorithm: anything the user does, even changing the printer, could change where a line or a page breaks over. Since these things are dynamic, there's no object.
The only context where "line" can be used is in connection with a Selection. For example, it's possible to extend a Selection to the start and/or end of a line. Incorporating this into the code in the question it would look something like:
Sub FindStrings()
Dim StringsArr As Variant
Dim bFound As Boolean
Dim rng As Word.Range
Set rng = ActiveDocument.content
StringsArr = Array("string1", "string2", "string3")
For i = LBound(StringsArr) To UBound(StringsArr)
With rng.Find
.ClearFormatting
.Text = StringsArr(i)
.Wrap = wdFindStop
bFound = .Execute
'extend the selection to the start and end of the current line
Do While bFound
rng.Select
Selection.MoveStart wdLine, -1
Selection.MoveEnd wdLine, 1
Debug.Print Selection.Text
rng.Collapse wdCollapseEnd
bFound = .Execute
Loop
End With
Set rng = ActiveDocument.content
Next
End Sub
Notes
Since it's easier to control when having to loop numerous times, a Range object is used as the basic search object, rather than Selection. The found Range is only selected for the purpose of getting the entire line as these "Move" methods for lines only work on a Selection.
Before the loop can continue, the Range (or, if we were working with a selection, the selection) needs to be "collapsed" so that the code does not search and find the same instance of the search term, again. (This is also the reason for Wrap = wdFindStop).

Finding a "Heading" Style in a Word Document

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

Change the case of a Range text and type it

Possible scenario, let say we created a Range Object containing the following line:
Speculative BUY, FV: EGP19.59
Now I want to split the Range Object into two parts by ", " as delimiter so that the given Range will change into two Ranges containing "Speculative BUY" and ", FV: EGP19.59" (Two separate range).
Now I need to change the case of only the first range containing "Speculative BUY" into "Speculative Buy" using:
.Case = wdTitleWord
Previously I am using .Find to change the Ranges in the following code (this is not complete code as it is only changing the Range R, not splitting it into two):
Sub Range_into_Ranges()
selection.EndKey Unit:=wdLine
selection.MoveUp Unit:=wdParagraph, COUNT:=1, Extend:=wdExtend
Dim R, F As Word.Range
Set R = selection.Range
Set F = R.Duplicate
With F.Find
.Text = ", "
.Forward = True
.Wrap = wdFindStop
.Execute
End With
If F.Find.Found Then
R.SetRange Start:=R.Start, _
End:=F.Start
R.Case = wdTitleWord
Else
End If
End Sub
Note: There may be other ways of producing the same results. you are free to advice me another simple code.
You can assign a case to a Range using the WdCharacterCase enumeration. For title case:
R.Case = wdTitleWord
Put into the context of your sample code, something like as follows. I did some tweaking:
My version assumes you want to work with the paragraph where the selection currently is, which is why I commented out your first two lines
In VBA you need to declare the data type of every variable, otherwise it's a Variant. So: Dim R As Word.Range
VBA provides the Split function to divide up a string according to a delimiter. I use this to get the term to search, so that you can get the Range directly
I found when setting Title Case on text that has ALL CAPS that it doesn't reduce upper case to lower case. But first applying lower case, then title case, does work.
Sample code
Sub Range_into_Ranges()
' Selection.EndKey Unit:=wdLine
' Selection.MoveUp Unit:=wdParagraph, Count:=1, Extend:=wdExtend
Dim R As word.Range, F As word.Range
Dim sTerm As String, bFound As Boolean
Set R = Selection.Paragraphs(1).Range
R.MoveEnd wdCharacter, -1 'Trim off the paragraph mark
sTerm = R.Text
sTerm = Split(sTerm, ",")(0)
Set F = R.Duplicate
With F.Find
.Text = sTerm
.Forward = True
.wrap = wdFindStop
bFound = .Execute
End With
If bFound Then
F.Case = wdLowerCase
F.Case = wdTitleWord
Else
End If
End Sub

Smart quotations aren't recognized by InStr()

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.

In microsoft word for replacing words with blanks

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.