Regex to leave only numbers and the Last period

Targets Web
1

I am currenty using this regex search pattern.
reg.searchPattern = “[^.0-9]” But I want to add a pattern that will leave the numbers and Only the last period in the string. Does anyone have a search pattern for that?
Thanks

2

this is not very clear, can you give some examples of what should be extracted or not ?

3

In a texfield, the user inputs abc$123.45g.67
I want a regex to return 12345.67 everything except the numbers and the last dot is removed.
Thanks

4

(I have no idea if this works or not but this is what chat GPT says)

CleanShot 2024-02-09 at 10.41.16
CleanShot 2024-02-09 at 10.41.16762×841 88.5 KB

Dim pattern As String = "(\d+)(\.\d+)?"
Dim input As String = "abc$123.45g.67"
Dim rx As New RegEx
Dim match As RegExMatch
Dim result As String

rx.SearchPattern = pattern

match = rx.Search(input)

While match IsNot Nil
    result = result + match.SubExpressionString(1) // Capture digits
    If match.SubExpressionString(2) <> "" Then
        result = result + match.SubExpressionString(2) // Capture dot and digits if present
    End If
    match = match.NextMatch
Wend

// result now contains "12345.67"

CleanShot 2024-02-09 at 10.42.35
CleanShot 2024-02-09 at 10.42.35751×1099 47.4 KB

5

Is that the only possible input pattern? Or might they add more random characters, such as €7/we12,4;5.31

6

Gemini gives this
\d+(?:\.\d*)?
as a pattern for this regex.

or also
\d+(?:\.\d*)?(?:[^\d.]|\b)*\.?$
if you want more than two groups of numbers to be possible in the source text

7

Yes, it could be any combination of letters, numbers, and characters, locale EN-US. Thanks

8

I will give it a try. Thanks

9

I’d be inclined to just brute force it. Count the number of periods with CountFields. Then character by character, copy the numbers and the nth period to another string.

10

I agree with Tim: you could probably craft a RegEx to do this - although I think it would require more than one, right, @Kem_Tekinay ? But I think the easiest, most understandable way is to simply loop through the string and analyze each character:

dim input as string

input="234.34oij5.345lkj42.3432jkl.67kj"

dim inputCharacters() as string

inputCharacters=input.Split("")

dim characterCount as integer
dim foundFinalPeriod as boolean
dim currentCharacter as string

dim digits as string

digits="0123456789"

characterCount=UBound(inputCharacters)

for i as integer=characterCount downto 0
  currentCharacter=inputCharacters(i)
  
  if currentCharacter="." and not(foundFinalPeriod) then
    'We found the final period, set the flag
    foundFinalPeriod=true
    
  ElseIf digits.IndexOf(currentCharacter)>-1 then
    'Allow all digits
    
  else
    'Erase undesired character
    inputCharacters(i)=""
  end
next

dim output as string

output=string.FromArray(inputCharacters, "")
11 
  1.  [^0-9.\n] search for this and replace with nothing to clear out all but the digits and dot

  2.  (.*)\.(.*\..*)  replace with \1\2

Repeat step number 2 until all but the last dot are gone.

12

This works for that input:

Var input As String = "abc$123.45g.67"
Var rx As New RegEx
rx.SearchPattern="\.(?![^.]+$)|[^0-9.]"
rx.ReplacementPattern=""
rx.Options.ReplaceAllMatches=True
Var result As String = rx.Replace(input) //12345.67
1 Like
13

This is how I’d do it.

1 Like
14

Technically, that pattern does not meet his requirements because if the text ends with a period, that period is omitted.

15

I’m sure OP can check if the result has no period and do a + "." to the result (if needed). Or use another pattern.

Edit: Maybe change +$ with *$