I was wondering how to remove duplicate names/text's in a cell. For example
Jean Donea Jean Doneasee
R.L. Foye R.L. Foyesee
J.E. Zimmer J.E. Zimmersee
R.P. Reed R.P. Reedsee D.E. Munson D.E. Munsonsee
While googling, I stumbled upon a macro/code, it's like:
Function RemoveDupes1(pWorkRng As Range) As String
'Updateby20140924
Dim xValue As String
Dim xChar As String
Dim xOutValue As String
Set xDic = CreateObject("Scripting.Dictionary")
xValue = pWorkRng.Value
For i = 1 To VBA.Len(xValue)
xChar = VBA.Mid(xValue, i, 1)
If xDic.exists(xChar) Then
Else
xDic(xChar) = ""
xOutValue = xOutValue & xChar
End If
Next
RemoveDupes1 = xOutValue
End Function
The macro is working, but it is comparing every letter, and if it finds any repeated letters, it's removing that.
When I use the code over those names, the result is somewhat like this:
Jean Dos
R.L Foyes
J.E Zimers
R.P edsDEMuno
By looking at the result I can make out it is not what I want, yet I got no clue how to correct the code.
The desired output should look like:
Jean Donea
R.L. Foye
J.E. Zimmer
R.P. Reed
Any suggestions?
Thanks in Advance.
Input
With the input on the image:
Result
The Debug.Print output
Regex
A regex can be used dynamically iterating on the cell, to work as a Find tool. So it will extract only the shortest match. \w*( OUTPUT_OF_EXTRACTELEMENT )\w*, e.g.: \w*(Jean)\w*
The Regex's reference must be enabled.
Code
Function EXTRACTELEMENT(Txt As String, n, Separator As String) As String
On Error GoTo ErrHandler:
EXTRACTELEMENT = Split(Application.Trim(Mid(Txt, 1)), Separator)(n - 1)
Exit Function
ErrHandler:
' error handling code
EXTRACTELEMENT = 0
On Error GoTo 0
End Function
Sub test()
Dim str As String
Dim objMatches As Object
Set objRegExp = CreateObject("VBScript.RegExp") 'New regexp
lastrow = ActiveSheet.Cells(ActiveSheet.Rows.Count, "A").End(xlUp).Row
For Row = 1 To lastrow
str = Range("A" & Row)
F_str = ""
N_Elements = UBound(Split(str, " "))
If N_Elements > 0 Then
For k = 1 To N_Elements + 1
strPattern = "\w*(" & EXTRACTELEMENT(CStr(str), k, " ") & ")\w*"
With objRegExp
.Pattern = strPattern
.Global = True
End With
If objRegExp.test(strPattern) Then
Set objMatches = objRegExp.Execute(str)
If objMatches.Count > 1 Then
If objRegExp.test(F_str) = False Then
F_str = F_str & " " & objMatches(0).Submatches(0)
End If
ElseIf k <= 2 And objMatches.Count = 1 Then
F_str = F_str & " " & objMatches(0).Submatches(0)
End If
End If
Next k
Else
F_str = str
End If
Debug.Print Trim(F_str)
Next Row
End Sub
Note that you can Replace the Debug.Print to write on the target
cell, if it is column B to Cells(Row,2)=Trim(F_str)
Explanation
Function
You can use this UDF, that uses the Split Function to obtain the element separated by spaces (" "). So it can get every element to compare on the cell.
Loops
It will loop from 1 to the number of elements k in each cell and from row 1 to lastrow.
Regex
The Regex is used to find the matches on the cell and Join a new string with the shortest element of each match.
This solution operates on the assumption that 'see' (or some other three-letter string) will always be on the end of the cell value. If that isn't the case then this won't work.
Function RemoveDupeInCell(dString As String) As String
Dim x As Long, ct As Long
Dim str As String
'define str as half the length of the cell, minus the right three characters
str = Trim(Left(dString, WorksheetFunction.RoundUp((Len(dString) - 3) / 2, 0)))
'loop through the entire cell and count the number of instances of str
For x = 1 To Len(dString)
If Mid(dString, x, Len(str)) = str Then ct = ct + 1
Next x
'if it's more than one, set to str, otherwise error
If ct > 1 Then
RemoveDupeInCell = str
Else
RemoveDupeInCell = "#N/A"
End If
End Function
Related
I'm pretty new to VBA. So I have a string with a bunch of random words in it like this:
"foo Foo FOO Bar FoO Faz FAZ"
How do I find the number of unique words in the string? In the above example, it would be simply 3, as there's only foo, bar, and faz. I have some code here but it outputs the wrong value, I'm not sure if I should be adding it to an array or some sort of database instead and then iterate through the database to check for duplicates.
Function UniqueWordCount(TextString As String) As Integer
TextString = LCase(TextString)
Dim Result() As String
Dim Count As Integer
Result = Split(TextString, " ")
Count = UBound(Result()) + 1
Dim k As Integer
Dim repeat As Integer
repeat = 0
For i = LBound(Result) To UBound(Result)
For k = 0 To i
If Result(i) = Result(k) Then
repeat = repeat + 1
End If
Next k
Next i
If repeat > 1 Then
repeat = repeat - 1
End If
repeat = repeat - i
UniqueWordCount = Count - repeat
End Function
Count Unique Words in a Sentence
Be careful with this simple Regex pattern, it may not work as expected.
Uncomment the Debug.Print lines to get a better feel of what is happening.
Option Explicit
Sub UniqueWordsCountTEST()
Dim Sentence As String
Sentence = "I am using this Regex to get rid off the punctuation,?!:;-. " _
& "I am using a dictionary to get rid off duplicates, and the " _
& "dictionary's CompareMode for 'word' to be the same as 'WORD'."
' 12 Dupes: I,am,using,to,get,rid,off,the,dictionary,to,the,WORD
Debug.Print "Unique Words Count = " & UniqueWordsCount(Sentence)
End Sub
Function UniqueWordsCount( _
ByVal Sentence As String) _
As Long
Dim Matches As Variant
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\w+"
Set Matches = .Execute(Sentence)
'Debug.Print "All Words Count = " & Matches.Count
End With
With CreateObject("Scripting.Dictionary")
.CompareMode = vbTextCompare
Dim Match As Variant
For Each Match In Matches
.Item(Match.Value) = Empty
'Debug.Print "All Words: " & Match.Value
Next Match
UniqueWordsCount = .Count
'Debug.Print "Unique Words:" & vbLf & Join(.Keys, vbLf)
End With
End Function
Use a Scripting.Dictionary (add a reference to Microsoft Scripting Runtime):
Sub foo()
Dim d As Scripting.Dictionary
Set d = New Scripting.Dictionary
Dim s As String
s = "foo Foo FOO Bar FoO Faz FAZ"
Dim x As Variant
x = Split(s)
Dim i As Long
For i = LBound(x) To UBound(x)
If Not d.Exists(UCase$(x(i))) Then
d.Add UCase$(x(i)), "whatever"
End If
Next
Debug.Print d.Count ' returns 3
End Sub
Or even simpler:
For i = LBound(x) To UBound(x)
d(UCase$(x(i))) = "whatever"
Next
I have a column containing formulas as "strings", i.e. "=+I11+I192+I245+I280"
I need to replace the cells (I11, I192,I245andI280`) ID with the content (strings) contained in the cells themselves.
Example:
Cell X --> "=+I11+I192+I245+I280"
Cell I11 = 'A'
Cell I192 = 'B'
Cell I245 = 'C'
Cell I280 = 'D'
The formula should generate "=+A+B+C+D".
This?
="=+" & I11 &"+" & I192 &"+" & I245 & "+" & I280
Well, how about :
=I11 & I192 & I245 & I280
Or you can include spaces
=I11 & " " & I192
But straight quotes - my phone is being funny...
The formula should generate --> "=+A+B+C+D"
Try,
="=+"&textjoin("+", true, I11, I192, I245, I280)
Don't know what you will be doing with empty cells so here is draft
Public Sub test()
[I11] = "A": [I192] = "B": [I245] = "C": [I280] = "D"
Debug.Print ConvertedString("=+I11+I192+I245+I280")
End Sub
Public Function ConvertedString(ByVal inputString As String) As Variant
Dim arr() As String, i As Long
On Error GoTo errHand
If Not InStr(inputString, Chr$(43)) > 0 Then
ConvertedString = CVErr(xlErrNA)
Exit Function
End If
arr = Split(inputString, Chr$(43))
For i = 1 To UBound(arr)
arr(i) = Range(arr(i))
Next i
ConvertedString = Join(arr, Chr$(43))
Exit Function
errHand:
ConvertedString = CVErr(xlErrNA)
End Function
I think you mean something like
=INDIRECT(I11,TRUE)+INDIRECT(I192,TRUE)+INDIRECT(I245,TRUE)+INDIRECT(I280,TRUE)
but please note that Indirect is a volatile function, and can slow your calculations down if used extensively.
Using VBA (with only single delimiter):
Function ReplaceAddr(sInput As String, Optional sDelimiter As String = "+") As String
Dim sArr
Dim i As Long
sArr = Split(sInput, sDelimiter)
For i = 1 To UBound(sArr)
sArr = Range(sArr(i))
Next i
ReplaceAddr = Join(sArr, sDelimiter)
End Function
From OP's comment:
The problem is that formulas changes, so I can't only change manually. The one I gave you is only an example, but I have so many different ones with all math operators.
You can try finding cell addresses with regular expression and replace with cell's value:
Function ReplaceAddr2(sInput As String) As String
Dim oRegEx As Object
Dim oMatches As Object
Dim i As Long, lStart As Long, lLength As Long
Set oRegEx = CreateObject("vbscript.regexp")
oRegEx.Pattern = "[A-Za-z]{1,3}\d{1,7}"
oRegEx.Global = True
oRegEx.MultiLine = True
Set oMatches = oRegEx.Execute(sInput)
lStart = 0
For i = 0 To oMatches.Count - 1
lLength = oMatches(i).FirstIndex - lStart
ReplaceAddr2 = ReplaceAddr2 & Mid$(sInput, lStart + 1, lLength) & Range(oMatches(i).Value)
lStart = lStart + lLength + oMatches(i).length
Next
ReplaceAddr2 = ReplaceAddr2 & Mid(sInput, lStart + 1, Len(sInput) - lStart)
End Function
Pattern is 1-3 letters followed by 1-7 digits.
Both functions are not volatile - will be recalculated only when input string changes, but not when cells addressed there change. Adding this line:
Application.Volatile True
will make it recalculate on every change, but it may affect performance.
I want to extract individual numbers from a string. So for:
x = " 99 1.2 99.25 "
I want to get three individual numbers: 99, 1.2, and 99.25.
Here is my current code. It extracts the first occurring number, but I do not know how to use loops to get the three individual numbers.
Sub ExtractNumber()
Dim rng As Range
Dim TestChar As String
Dim IsNumber As Boolean
Dim i, StartChar, LastChar, NumChars As Integer
For Each rng In Selection
IsNumber = False
i = 1
Do While IsNumber = False And i <= Len(rng)
TestChar = Mid(rng, i, 1)
If IsNumeric(TestChar) = True Then
StartChar = i
IsNumber = True
End If
i = i + 1
Loop
IsNumber = False
Do While IsNumber = False And i <= Len(rng)
TestChar = Mid(rng, i, 1)
If IsNumeric(TestChar) = False Or i = Len(rng) Then
If i = Len(rng) Then
LastChar = i
Else
LastChar = i - 1
End If
IsNumber = True
End If
i = i + 1
Loop
NumChars = LastChar - StartChar + 1
rng.Offset(0, 1).Value = Mid(rng, StartChar, NumChars)
Next rng
End Sub
My previous attempt (input is stored in cell A6):
Dim x, y, z As String
x = Range("A6")
y = Len(x)
For i = 1 To Len(x)
If IsNumeric(Mid(x, i, 1)) Then
z = z & Mid(x, i, 1)
End If
Next i
MsgBox z
If speed is not an issue (if the task is not intensive, etc) then you can use this
Public Sub splitme()
Dim a As Variant
Dim x As String
Dim i, j As Integer
Dim b() As Double
x = "1.2 9.0 0.8"
a = Split(x, " ")
j = 0
ReDim b(100)
For i = 0 To UBound(a)
If (a(i) <> "") Then
b(j) = CDbl(a(i))
j = j + 1
End If
Next i
ReDim Preserve b(j - 1)
End Sub
Error checking needs to be included for b(100), to suit your particular needs - and with CDbl.
If this is to be used as part of a loop, or for large x - or both, consider other options like RegEx (previous answer) - as repeated calls to ReDim Preserve are generally best avoided.
Rather than writing your own code to extract the numbers, why not try using Regular Expressions? This website has a lot of great info and tutorials on regular expressions. It can be a bit baffling at first but once you get the hang of it it's a very powerful tool for solving problems of this type.
Below is an example of extracting the information you're after using a regular expression object.
Public Sub ExtractNumbers()
'Regular Expression Objects
Dim objRegEx As Object
Dim objMatches As Object
Dim Match As Object
'String variable for source string
Dim strSource As String
'Iteration variable
Dim i As Integer
'Create Regular Expression Object
Set objRegEx = CreateObject("VBScript.RegExp")
'Set objRegEx properties
objRegEx.Global = True '<~~ We want to find all matches
objRegEx.MultiLine = True '<~~ Allow line breaks in source string
objRegEx.IgnoreCase = False '<~~ Not strictly necessary for this example
'Below pattern matches an integer or decimal number 'word' within a string
' \b matches the start of the word
' [+-]? optionally matches a + or - symbol
' [0-9]+ matches one or more digits in sequence
' (\.[0-9]+)? optionally matches a period/decimal point followed by one or more digits
' \b matches the end of the word
objRegEx.Pattern = "\b[+-]?[0-9]+(\.[0-9]+)?\b"
'Example String
strSource = "x= 99 10.1 20.6 Aardvark"
'Ensure that at least one match exists
If objRegEx.Test(strSource) Then
'Capture all matches in objMatches
Set objMatches = objRegEx.Execute(strSource)
'TODO: Do what you want to do with them
'In this example I'm just printing them to the Immediate Window
'Print using Match object and For..Each
For Each Match In objMatches
Debug.Print Match.Value
Next Match
'Print using numeric iteration (objMatches.Items is a 0-based collection)
For i = 0 To (objMatches.Count - 1)
Debug.Print objMatches.Item(i)
Next i
End If
End Sub
Both of the print variations shown in this example would print the following output to the Immediate window
99
10.1
20.6
I am getting the impression that this is not possible in word but I figure if you are looking for any 3-4 words that come in the same sequence anywhere in a very long paper I could find duplicates of the same phrases.
I copy and pasted a lot of documentation from past papers and was hoping to find a simple way to find any repeated information in this 40+ page document there is a lot of different formatting but I would be willing to temporarily get rid of formatting in order to find repeated information.
To highlight all duplicate sentences, you can also use ActiveDocument.Sentences(i). Here is an example
LOGIC
1) Get all the sentences from the word document in an array
2) Sort the array
3) Extract Duplicates
4) Highlight duplicates
CODE
Option Explicit
Sub Sample()
Dim MyArray() As String
Dim n As Long, i As Long
Dim Col As New Collection
Dim itm
n = 0
'~~> Get all the sentences from the word document in an array
For i = 1 To ActiveDocument.Sentences.Count
n = n + 1
ReDim Preserve MyArray(n)
MyArray(n) = Trim(ActiveDocument.Sentences(i).Text)
Next
'~~> Sort the array
SortArray MyArray, 0, UBound(MyArray)
'~~> Extract Duplicates
For i = 1 To UBound(MyArray)
If i = UBound(MyArray) Then Exit For
If InStr(1, MyArray(i + 1), MyArray(i), vbTextCompare) Then
On Error Resume Next
Col.Add MyArray(i), """" & MyArray(i) & """"
On Error GoTo 0
End If
Next i
'~~> Highlight duplicates
For Each itm In Col
Selection.Find.ClearFormatting
Selection.HomeKey wdStory, wdMove
Selection.Find.Execute itm
Do Until Selection.Find.Found = False
Selection.Range.HighlightColorIndex = wdPink
Selection.Find.Execute
Loop
Next
End Sub
'~~> Sort the array
Public Sub SortArray(vArray As Variant, i As Long, j As Long)
Dim tmp As Variant, tmpSwap As Variant
Dim ii As Long, jj As Long
ii = i: jj = j: tmp = vArray((i + j) \ 2)
While (ii <= jj)
While (vArray(ii) < tmp And ii < j)
ii = ii + 1
Wend
While (tmp < vArray(jj) And jj > i)
jj = jj - 1
Wend
If (ii <= jj) Then
tmpSwap = vArray(ii)
vArray(ii) = vArray(jj): vArray(jj) = tmpSwap
ii = ii + 1: jj = jj - 1
End If
Wend
If (i < jj) Then SortArray vArray, i, jj
If (ii < j) Then SortArray vArray, ii, j
End Sub
SNAPSHOTS
BEFORE
AFTER
I did not use my own DAWG suggestion, and I am still interested in seeing if someone else has a way to do this, but I was able to come up with this:
Option Explicit
Sub test()
Dim ABC As Scripting.Dictionary
Dim v As Range
Dim n As Integer
n = 5
Set ABC = FindRepeatingWordChains(n, ActiveDocument)
' This is a dictionary of word ranges (not the same as an Excel range) that contains the listing of each word chain/phrase of length n (5 from the above example).
' Loop through this collection to make your selections/highlights/whatever you want to do.
If Not ABC Is Nothing Then
For Each v In ABC
v.Font.Color = wdColorRed
Next v
End If
End Sub
' This is where the real code begins.
Function FindRepeatingWordChains(ChainLenth As Integer, DocToCheck As Document) As Scripting.Dictionary
Dim DictWords As New Scripting.Dictionary, DictMatches As New Scripting.Dictionary
Dim sChain As String
Dim CurWord As Range
Dim MatchCount As Integer
Dim i As Integer
MatchCount = 0
For Each CurWord In DocToCheck.Words
' Make sure there are enough remaining words in our document to handle a chain of the length specified.
If Not CurWord.Next(wdWord, ChainLenth - 1) Is Nothing Then
' Check for non-printing characters in the first/last word of the chain.
' This code will read a vbCr, etc. as a word, which is probably not desired.
' However, this check does not exclude these 'words' inside the chain, but it can be modified.
If CurWord <> vbCr And CurWord <> vbNewLine And CurWord <> vbCrLf And CurWord <> vbLf And CurWord <> vbTab And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbCr And CurWord.Next(wdWord, ChainLenth - 1) <> vbNewLine And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbCrLf And CurWord.Next(wdWord, ChainLenth - 1) <> vbLf And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbTab Then
sChain = CurWord
For i = 1 To ChainLenth - 1
' Add each word from the current word through the next ChainLength # of words to a temporary string.
sChain = sChain & " " & CurWord.Next(wdWord, i)
Next i
' If we already have our temporary string stored in the dictionary, then we have a match, assign the word range to the returned dictionary.
' If not, then add it to the dictionary and increment our index.
If DictWords.Exists(sChain) Then
MatchCount = MatchCount + 1
DictMatches.Add DocToCheck.Range(CurWord.Start, CurWord.Next(wdWord, ChainLenth - 1).End), MatchCount
Else
DictWords.Add sChain, sChain
End If
End If
End If
Next CurWord
' If we found any matching results, then return that list, otherwise return nothing (to be caught by the calling function).
If DictMatches.Count > 0 Then
Set FindRepeatingWordChains = DictMatches
Else
Set FindRepeatingWordChains = Nothing
End If
End Function
I have tested this on a 258 page document (TheStory.txt) from this source, and it ran in just a few minutes.
See the test() sub for usage.
You will need to reference the Microsoft Scripting Runtime to use the Scripting.Dictionary objects. If that is undesirable, small modifications can be made to use Collections instead, but I prefer the Dictionary as it has the useful .Exists() method.
I chose a rather lame theory, but it seems to work (at least if I got the question right cuz sometimes I'm a slow understander).
I load the entire text into a string, load the individual words into an array, loop through the array and concatenate the string, containing each time three consecutive words.
Because the results are already included in 3 word groups, 4 word groups or more will automatically be recognized.
Option Explicit
Sub Find_Duplicates()
On Error GoTo errHandler
Dim pSingleLine As Paragraph
Dim sLine As String
Dim sFull_Text As String
Dim vArray_Full_Text As Variant
Dim sSearch_3 As String
Dim lSize_Array As Long
Dim lCnt As Long
Dim lCnt_Occurence As Long
'Create a string from the entire text
For Each pSingleLine In ActiveDocument.Paragraphs
sLine = pSingleLine.Range.Text
sFull_Text = sFull_Text & sLine
Next pSingleLine
'Load the text into an array
vArray_Full_Text = sFull_Text
vArray_Full_Text = Split(sFull_Text, " ")
lSize_Array = UBound(vArray_Full_Text)
For lCnt = 1 To lSize_Array - 1
lCnt_Occurence = 0
sSearch_3 = Trim(fRemove_Punctuation(vArray_Full_Text(lCnt - 1) & _
" " & vArray_Full_Text(lCnt) & _
" " & vArray_Full_Text(lCnt + 1)))
With Selection.Find
.Text = sSearch_3
.Forward = True
.Replacement.Text = ""
.Wrap = wdFindContinue
.Format = False
.MatchCase = False
Do While .Execute
lCnt_Occurence = lCnt_Occurence + 1
If lCnt_Occurence > 1 Then
Selection.Range.Font.Color = vbRed
End If
Selection.MoveRight
Loop
End With
Application.StatusBar = lCnt & "/" & lSize_Array
Next lCnt
errHandler:
Stop
End Sub
Public Function fRemove_Punctuation(sString As String) As String
Dim vArray(0 To 8) As String
Dim lCnt As Long
vArray(0) = "."
vArray(1) = ","
vArray(2) = ","
vArray(3) = "?"
vArray(4) = "!"
vArray(5) = ";"
vArray(6) = ":"
vArray(7) = "("
vArray(8) = ")"
For lCnt = 0 To UBound(vArray)
If Left(sString, 1) = vArray(lCnt) Then
sString = Right(sString, Len(sString) - 1)
ElseIf Right(sString, 1) = vArray(lCnt) Then
sString = Left(sString, Len(sString) - 1)
End If
Next lCnt
fRemove_Punctuation = sString
End Function
The code assumes a continuous text without bullet points.
How do I remove special characters and alphabets in a string ?
qwert1234*90)! ' this might be my cell value
I have to convert it to
123490 ' I mean I have to remove everything but keep only the numbers in string
but it should allow spaces !
qwe123 4567*. 90 ' String with spaces
123 4567 90 ' output should be
I found the vba Replace - but writing a replace for each character makes my code big. Well let me tell you clearly without hiding anything from you:
input: qwe123 4567*. 90 ' String with spaces cells(1,"A").value
My idea to do these next: 123 4567 90 ' remove characters first keeping white spaces
final output in A1:A3
123
4567
90
(for every space it should insert row and fill that)
Could you tell me how do remove all characters except numbers and spaces in string?
Thanks In advance
You need to use a regular expression.
See this example:
Option Explicit
Sub Test()
Const strTest As String = "qwerty123 456 uiops"
MsgBox RE6(strTest)
End Sub
Function RE6(strData As String) As String
Dim RE As Object, REMatches As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.MultiLine = False
.Global = True
.IgnoreCase = True
.Pattern = "([0-9]| )+"
End With
Set REMatches = RE.Execute(strData)
RE6 = REMatches(0)
End Function
Explanation:
Pattern = "([0-9]| )+" will match any 0 or more group (+) containing a number ([0-9]) or (|) a space ().
Some more info on the regexp:
a thread on ozgrid
a very good reference about regexp
Non-re alternative;
Public Function fmt(sValue As String) As String
Dim i As Long
For i = 1 To Len(sValue) '//loop each char
Select Case Mid$(sValue, i, 1) '//examine current char
Case "0" To "9", " " '//permitted chars
'//ok
Case Else
Mid$(sValue, i, 1) = "!" '//overwrite char in-place with "!"
End Select
Next
fmt = Replace$(sValue, "!", "") '//strip invalids & return
End Function
For:
?fmt("qwe123 4567*. 90")
123 4567 90
Those two funny codes will do both of your whishes..
Sub MySplitter(strInput As String)
Row = 10 ' Start row
Col = "A" ' Column Letter
Range(Col & Row) = "" ' Clean the start cell
For i = 1 To Len(strInput) ' Do with each Character in input string...
c = Mid(strInput, i, 1) ' Get actual char
If IsNumeric(c) Then Range(Col & Row) = Range(Col & Row) & c ' If numeric then append to actual cell
If (c = " ") And (Range(Col & Row) <> "") Then 'If space and actual row is not empty then...
Row = Row + 1 ' Jump to next row
Range(Col & Row) = "" ' Clean the new cell
End If
Next
End Sub
Function KeepNumbersAndSpaces(ByVal strInput As String)
For i = 1 To Len(strInput) ' Do with each Character in input string...
c = Mid(strInput, i, 1) ' Get actual char
If IsNumeric(c) Or c = " " Then ' If numeric or a space then append to output
KeepNumbersAndSpaces = KeepNumbersAndSpaces & c
End If
Next
End Function
Sub Test()
strInput = "qwert1234*90)! qwe123 4567*. 90"
MySplitter (strInput)
Range("A5") = KeepNumbersAndSpaces(strInput)
End Sub
Something like this to
split the string using a regexp
place the matches into an array
dump the array to an automatically sized spreadsheet range
main sub
Sub CleanStr()
Dim strOut As String
Dim Arr
strOut = Trim(KillChar("qwe123 4567*. 90 "))
Arr = Split(strOut, Chr(32))
[a1].Resize(UBound(Arr) + 1, 1) = Application.Transpose(Arr)
End Sub
function
Function KillChar(strIn As String) As String
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^\d\s]+"
KillChar = .Replace(strIn, vbNullString)
End With
End Function