Thank you all that works just fine, my last step is to create an if
statement to only enter to do this if the condition matches.
if($rU=~"XYZ") //where xyz is anything not 0-9 * and #
{
$rU = $(rU{re.subst,/[^0-9*#]//g});
}
Regards
On Tue, Feb 12, 2013 at 10:42 AM, Daniel-Constantin Mierla <
mico...@gmail.com> wrote:
> Hello,
>
>
> On 2/12/13 9:28 AM, Ali Jawad wrote:
>
> Hi
> Thanks, what I actually need is keep numbers and * and # and remove the
> rest, something like
>
> $rU = $(rU{re.subst,/![0-9*#]//g});
>
> But I am not sure that ! works for negation.
>
> use [^0-9*#] for negation, see:
>
> - http://www.regular-expressions.info/charclass.html
>
> Cheers,
> Daniel
>
>
> Regards
>
> On Mon, Feb 11, 2013 at 5:43 PM, Daniel Tryba <dan...@pocos.nl> wrote:
>
>> On Monday 11 February 2013 11:24:36 Ali Jawad wrote:
>> > I am using the below to strip "-" from destination numbers, it does work
>> > just fine "I replace - with nothing", what i need to do now is to strip
>> all
>> > characters except 0-9 * and #. Is there anyway to do it using the
>> following
>> > configuration part
>>
>> You are using regular expression substitution, so yes this can be done:
>> /[^0-9*#]//g
>>
>> Take a look at http://en.wikipedia.org/wiki/Regular_expression
>> for more info.
>>
>
>
>
> *
> *
>
>
> _______________________________________________
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> --
> Daniel-Constantin Mierla - http://www.asipto.comhttp://twitter.com/#!/miconda
> - http://www.linkedin.com/in/miconda
> Kamailio World Conference, April 16-17, 2013, Berlin
> - http://conference.kamailio.com -
>
>
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