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if ident & nick same = no connect
Posted: Mon May 23, 2005 3:30 pm
by rek1n
(my english verry little.......)
if ident & nick same = client no connect
sample:
* ASDEFR (
[email protected]) has joined #Channel
* EHSHEJEJSHEHS (
[email protected]) has joined #Channel
note: ident & nick no take number. all character is harf letter.
Posted: Mon May 23, 2005 3:47 pm
by TigerKatziTatzi
isn't possible to check this with spamfilter.you are only able to trigger user (u). means regexpression in nick, ident or realname.
you may use a script for this which is comparing nickname with ident. But its dangerous. Cause alot of users using nickname and ident same, so you need to script this carefully.
Posted: Mon May 23, 2005 4:27 pm
by Jason
Here is your regular expression. Use type u in /spamfilter.
([^!]+)!\1@.*
.
Posted: Mon May 23, 2005 4:47 pm
by rek1m
i have:
Code: Select all
spamfilter {
regex "^(.+)!~?\1@.+:\1";
target user;
action kill;
reason "Nicki ile identi aynı olanlar baglanamaz. Ident'i değiştiriniz.";
};
but spoilt
Posted: Mon May 23, 2005 10:20 pm
by Stealth
Take off the last \1
Your regex should be:
Posted: Mon Jun 06, 2005 10:29 am
by White_Magic
i have a question say i want to add the case sensitivity to this, how would it work...
would it be..
or would it be
Posted: Mon Jun 06, 2005 10:08 pm
by codemastr
White_Magic wrote:i have a question say i want to add the case sensitivity to this, how would it work...
would it be..
or would it be
The latter. The former is illegal.
Posted: Tue Jun 07, 2005 4:12 am
by aquanight
I'm curious to know how you have a user with a username/ident longer than the normal USERLEN of 10.
Here's a regex that will handle cases where the nickname is too long for the username (since NICKLEN is 30 which is > USERLEN of 10):
^((?i).{1,10}).*!\1@
AND
^((?i).{1,9}).*!~\1@
(Although I suppose you could get away with ^((?i).{1,10}).*!~?\1@ )
Posted: Tue Jun 07, 2005 1:59 pm
by White_Magic
thanks
Posted: Tue Jun 07, 2005 3:04 pm
by Jason
I cant be the only one who is Jason!Jason@*