How to replace certain character in a string - excel

Question

I am trying to replace not each space in a single string with line break. String is taken from specific cell, and looks like:
Now, Im trying to replace each space after abbreviation to line break. The abbreviation can be any, so the best way for precaching which space I intend to replace is like: each space after number and before a letter?
The output I want to get is like:
Below is my code, but it will change every space to line break in cell.
Private Sub Workbook_SheetChange(ByVal Sh As Object, ByVal Target As Range)
On Error GoTo Exitsub
If Not Intersect(Target, .Columns(6)) Is Nothing Then
Application.EnableEvents = False
Target.Value = Replace(Target, " ", Chr(10))
End If
Application.EnableEvents = True
Exitsub:
Application.EnableEvents = True
End Sub

Answer 1

You can try
Target.Value = Replace(Target, "kg ", "kg" & Chr(10))
If you can have other abbreviations like "g" or "t", do something similar for them (maybe in a Sub), just be cautious with the order (replace first "kg", then "g")
Update: If you don't know in advance the possible abbreviations, one attempt is to use regular expressions. I'm not really good with them, but the following routine seems to do:
Function replaceAbbr(s As String) As String
Dim regex As New RegExp
regex.Global = True
regex.Pattern = "([a-z]+) "
replaceAbbr = regex.Replace(s, "$1" & Chr(10))
End Function

Answer 2

The below will replace every 2nd space with a carriage return. For reason unknown to me The worksheet function Replace will work as intended, but the VBA Replace doesnt
This will loop through every character in the defined area, you can change this to whatever you want.
The if statement is broken down as such
(SpaceCount Mod 2) = 0 this part is what enable it to get every 2nd character.
As a side note (SpaceCount Mod 3) = 0 will get the 3rd character and (SpaceCount Mod 2) = 1 will do the first character then every other character
Cells(1, 1).Characters(CountChr, 1).Text = " " is to make sure we are replacing a space, if the users enters something funny that looks like a space but isn't, that's on them
I believe something like this will work as intended for you
Private Sub Workbook_SheetChange(ByVal Sh As Object, ByVal Target As Range)
On Error GoTo Exitsub
Application.EnableEvents = False
For CountChr = 1 To Len(Target.Value)
If Target.Characters(CountChr, 1).Text = " " Then
Dim SpaceCount As Integer
SpaceCount = SpaceCount + 1
If (SpaceCount Mod 2) = 0 Then
Target.Value = WorksheetFunction.Replace(Target.Value, CountChr, 1, Chr(10))
End If
End If
Next CountChr
Application.EnableEvents = True
Exitsub:
Application.EnableEvents = True
End Sub

Answer 3

Identify arbitrary abbreviation first
"abbreviations aren't determined ..."
Knowing the varying abbreviation which, however is the same within each string (here e.g. kg ) actually helps following the initial idea to look at the blanks first: but instead of replacing them all by vbLf or Chr(10), this approach
a) splits the string at this " " delimiter into a zero-based tmp array and immediately identifies the arbitrary abbreviation abbr as second token, i.e. tmp(1)
b) executes a negative filtering to get the numeric data and eventually
c) joins them together using the abbreviation which is known now for the given string.
So you could change your assignment to
'...
Target.Value = repl(Target) ' << calling help function repl()
Possible help function
Function repl(ByVal s As String) As String
'a) split into tokens and identify arbitrary abbreviation
Dim tmp, abbr As String
tmp = Split(s, " "): abbr = tmp(1)
'b) filter out abbreviation
tmp = Filter(tmp, abbr, Include:=False)
'c) return result string
repl = Join(tmp, " " & abbr & vbLf) & abbr
End Function
Edit // responding to FunThomas ' comment
ad a): If there might be missing spaces between number and abbreviation, the above approach could be modified as follows:
Function repl(ByVal s As String) As String
'a) split into tokens and identify arbitrary abbreviation
Dim tmp, abbr As String
tmp = Split(s, " "): abbr = tmp(1)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'b) renew splitting via found abbreviation (plus blank)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
tmp = Split(s & " ", abbr & " ")
'c) return result string
repl = Join(tmp, abbr & vbLf): repl = Left(repl, Len(repl) - 1)
End Function
ad b): following OP citing e.g. "10 kg 20 kg 30,5kg 15kg 130,5 kg" (and as already remarked above) assumption is made that the abbreviation is the same for all values within one string, but can vary from item to item.