Sorry, a newbie coder here, below code is really hacked.
I just came up with some code that ALMOST works for what I have it
planned for! This is used in a mobs spec_proc and is used to send a
message to the room. If the person is in a group, sends it to everyone
in the group, if not, just to the person. Everyone else doesn't see what
is being said, just that their is talking going on. The problem I am having
is
if there are 2 groups in the room at the same time, I can't figure out how
to display to the by-standing group what is going on. Understand????
void quest_say (struct char_data * ch, char *questname, char *message)
{
struct char_data *k;
struct char_data *i;
struct follow_type *f;
if (!AFF_FLAGGED(ch, AFF_GROUP)){
send_to_char("TEMP MESSAGE ** You are not grouped! ** TEMP MESSAGE\r\n",
ch);
sprintf(buf, "%s tells you, '%s'", questname, message);
act(buf, FALSE, ch, 0, 0, TO_CHAR);
sprintf(buf, "%s talks to $n.", questname);
act(buf, FALSE, ch, 0, 0, TO_ROOM);
}
if (AFF_FLAGGED(ch, AFF_GROUP)){
if (ch->master)
k = ch->master;
else
k = ch;
}
for (i = world[ch->in_room].people; i; i = i->next_in_room)
if (!AFF_FLAGGED(i, AFF_GROUP)) {
sprintf(buf, "%s talks to a group of people.\r\n", questname);
send_to_char(buf, i);
}
sprintf(buf, "%s tells the group, '%s'", questname, message);
if (AFF_FLAGGED(k, AFF_GROUP) && (k != ch->master))
act(buf, FALSE, ch, 0, k, TO_VICT | TO_SLEEP);
for (f = k->followers; f; f = f->next)
if (AFF_FLAGGED(f->follower, AFF_GROUP) && (f->follower != ch))
act(buf, FALSE, ch, 0, f->follower, TO_VICT | TO_SLEEP);
}
Corey Elliott
strmchsr@arn.net
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