PHP Code:
new query[254],i[254],name[MAX_PLAYER_NAME],nr[128],string2[512];
format(query,sizeof(query),"SELECT * FROM `drgbaze` WHERE `VardasD` = '%s'",GetPlayerNameEx(playerid));
mysql_query(query);
mysql_store_result();
if(mysql_num_rows() != 0)
{
format(string2,sizeof(string2),"{82171f}[+] {ffffff}new friend/number add");
while(mysql_fetch_row_format(query,"|"))
{
FriendsData[playerid][TotalFriends] ++;
mysql_fetch_field_row(i,"Fvardas"); format(name,sizeof(name),i);
mysql_fetch_field_row(i,"numeris"); format(nr,sizeof(nr),i);
if(IsPlayerOnline(name))
{
format(name,sizeof(name),"{ffc38c}%s\t{ffffff}%s\t{5c8141}on\n",name,nr);
strcat(string2,name);
}
else
{
format(name,sizeof(name),"%s %s- off\n", name);
strcat(string2,name);
}
}
ShowPlayerDialog(playerid, 9999, DIALOG_STYLE_TABLIST_HEADERS, "Test", string2, "Select", "");
}
You have to count your friends, and if the player click on "[+] new friend/number add" you have to check if the variable with the count is 10 or over 10. Sorry, but my english isn't very well. I hope you know what I mean.
, i need if i add second player or more it show but not create again and again new friend/number add