[SR-Users] subst query

[mahesh b]

Thanks John for the Reply.

So if i understand properly :
group 1 is here  <sip:anonymous@
group 2 is empty
group 3 is  ;tag=41008079_nab_FFFF_isp_FFFF_cco_FFFF_igo_FFFF_mgt_78DD

the previous expression is working for IPv4.


Now for IPv6

Input         :           From: <sip:anonymous@
[2008:abcd:1234:2260:208:5dff:fe93:5b65]>;tag=42006C96_nab_FFFF_isp_FFFF_cco_FFFF_igo_FFFF_mgt_8092
subst('/^From:(.*)[2008:abcd:1234:2260:208:5dff:fe93:5b65](.*)>(.*)/From:\1[2008:abcd:1234:2260:20c:29ff:fe9a:b9cb]>\3/ig');
Output      :           From:
<sip:anonymous@[2008:abcd:1234:2260:208:5dff:fe93:5b6[2008:abcd:1234:2260:20c:29ff:fe9a:b9cb]>;tag=42006C96_nab_FFFF_isp_FFFF_cco_FFFF_igo_FFFF_mgt_8092
  //what is going wrong here?

Here
group 1 is   <sip:anonymous@
group 2 is empty
group 3 is  ;tag=42006C96_nab_FFFF_isp_FFFF_cco_FFFF_igo_FFFF_mgt_8092

what mistake am i making for IPv6 ? if you see the output.

Regards,
Mahesh.b


On Mon, Jul 2, 2018 at 6:54 PM John Petrini <jpetrini at coredial.com> wrote:

> They are references to groups in the regex match. The groups are in
> parenthesis. Whatever is matched by group 1 and 3 will be included in the
> substitution.
>
> https://regexr.com/ and https://regex101.com/ are both good sites for
> building and explaining regexes.
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