RegEx numbers at end of string

[knightstalker]

I am looking for help with a StringRegExpReplace function that only grabs the numbers plus a "." and "%"

currently I have "[^0-9.%]" which mostly works, but I and looking to only to match if it is at the end of the string.

Example string:  "DW-4 2.50%"

my expression also leaves the preceding "4"

I have tried using $ \b \z and a few others with no success.

Any suggestions?

[FrancescoDiMuro]

@knightstalker
Could you please post more examples with the expected results?

[faustf]

i am not specialist for  regexp  , (you know a maxim ? ...... A programmer has a problem. He decides to solve it using regular expressions. Now he has two problems. )

somthing like this    -(.*?)%

and when call array  add %

 

[knightstalker]

additional examples

"Ford Prius 1%"

 "Cotton Candy Non Circus7.5%"

 "Francos Absolute .5%"

results would be

1%

7.5%

.5%

 

[FrancescoDiMuro]

@knightstalker

Maybe something like this:

'([\d.%]+$)'

Edited April 29, 2020 by FrancescoDiMuro

[knightstalker]

10 minutes ago, FrancescoDiMuro said:

@knightstalker

Maybe something like this:

'([\d.%]+$)'

Well done!  works like a charm.  Thanks!

[mikell]

@FrancescoDiMuro

Your expression allows 5.% - which is inconsistent
Please try this

#Include <Array.au3>

$s = "Ford Prius 1%" & @crlf & _ 
    "DW-4 6.66% test" & @crlf & _ 
    "DW-4 6.%" & @crlf & _ 
    "DW-4 2.50%" & @crlf & _ 
    "Cotton Candy Non Circus7.5%" & @crlf& _ 
    "Francos Absolute .5%" 

$res = StringRegExp($s, '(?m)((?:\d*\.?\d+)%)$', 3)
_ArrayDisplay($res)

 

Edited April 29, 2020 by mikell

[Deye]

$a = StringRegExp($var, '(\d.*)', 3)
_ArrayDisplay($a)

 

[FrancescoDiMuro]

@mikell

It was kinda late when I posted that reply, and the OP request was not clear enough since it was like he wanted to get other numbers too.

Thanks by the way