Re: [sr-dev] [kamailio/kamailio] uri.param and param.value error when last character is delimiter (#1991)

On 25 Jun 2019, at 18:05, Daniel-Constantin Mierla &lt;notifications(a)github.com&gt; wrote: From where did you get such URI? Because, afaik, by specs/RFC grammar, the uri must not be terminated by ; -- the last param value must be left without ;. I haven't implemented the parser for uri+params, but that's what I understood from the developer long time ago. I know that for some some modparams that use the uri-params parser, I had to check if the last char is ; and skip it when passing the value to the parser to avoid loosing the checks of the parser to break specs requirements. The solution right now is to check the last char in such value and skip it using other string transformations such as s.substr. — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub <https://github.com/kamailio/kamailio/issues/1991?email_source=notifications&email_token=ADBOF2RHFJIY2TLEBXBBONDP4IX4NA5CNFSM4H26BGUKYY3PNVWWK3TUL52HS4DFVREXG43VMVBW63LNMVXHJKTDN5WW2ZLOORPWSZGODYQR3PI#issuecomment-505486781>, or mute the thread <https://github.com/notifications/unsubscribe-auth/ADBOF2W77WPQLRS437HXUE3P4IX4NANCNFSM4H26BGUA>.