Extracting a substring from a cell at a certain index (different row length, multiple rows) - excel

Question

Working as an accountant, and I'm wondering if there's a way to extract a certain string of characters, at a certain index, from a cell of varying lengths. Sometimes, I receive a statement of account as a PDF, which doesn't convert nicely to an Excel spreadsheet.
This is what I want to achieve
The strings are at the same index within each cell. I can't quite figure out how to isolate them, though, because each string is a different length, and the substring I want doesn't start with the same character. I've tried LEN, MID, etc. to extract them, but I'm not sure how to do it.
I was able to extract items that start with specific characters (like INV and "45") with:
=TRIM(MID(SUBSTITUTE(A1," ",REPT(" ",99)),MAX(1,FIND("XXX",SUBSTITUTE(A1," ",REPT(" ",99)))-50),99))
But I can't figure out how to get the strings I'm looking for (in this case, the amounts for the invoices).

Answer 1

If you don't mind some VBA, you can try this:
The basic problem with the data is the dates included in the fields are variable length and include the only real delimiter (spaces) within those elements. Using the spaces from the right, can be used to return your fields.
Credit to Sumit Bansal at (https://trumpexcel.com) for the original LastPosition() function. However, this function only returns the last occurrence of a character.
But a couple of slight modifications, LastPositionN() now returns the nth occurrence of the search text, to be specified as a 3rd parameter.
Note: Follow here for the original function, and info on cutting/pasting into the VBA editor: https://trumpexcel.com/find-characters-last-position/
Function LastPosition(rCell As Range, rChar As String)
'This function gives the last nth position of the specified character
'This code has been developed by Sumit Bansal (https://trumpexcel.com)
Dim rLen As Integer
rLen = Len(rCell)
For i = rLen To 1 Step -1
If Mid(rCell, i - 1, 1) = rChar Then
'LastPosition = i ' original code returnd Last position +1
LastPosition = i - 1 ' correction returns codes original intent
Exit Function
End If
Next i
End Function
Function LastPositionN(rCell As Range, rChar As String, Optional nNthOccurance As Integer = 1)
'This function gives the last position of the specified character
'This code has been developed by Sumit Bansal
' (https://trumpexcel.com)
' and was modified my Jeff Bowman 9/11/2018 for nth position from right
'Returns zero (0) if not nth occurence is not found
Dim rLen As Integer, nNthCount As Integer, LastPosition As Integer
nNthCount = 0
rLen = Len(rCell)
For i = rLen To 1 Step -1
If Mid(rCell, i - 1, 1) = rChar Then
'LastPosition = i ' original code returnd Last position +1
LastPositionN = i - 1 ' correction returns codes original
'intent
nNthCount = nNthCount + 1
If nNthCount = nNthOccurance Then
Exit Function
End If
End If
Next i
End Function
`
Now, with left(), right(), and find() we can isolate and parse the end of the data string.
=LEFT(RIGHT(A8,LastPositionN(A8," ",1)),FIND(" ",RIGHT(A8,LastPositionN(A8," ",1))))