Re: Variable won't work in function, even when I global it?

So in function checkmember(), do you now have: global $REMOTE_ADDR, $userinfo, $loggedin, $time_start, $getlayout; ?? miguel On Mon, 20 May 2002, Leif K-Brooks wrote: > No good as in "that doesn't solve it". Thanks for trying to help. > > Miguel Cruz wrote: > > >No good as in "that doesn't solve it" or as in "I sure wish it > >didn't have > >to be that way"? > > > >miguel > > > >On Mon, 20 May 2002, Leif K-Brooks wrote: > > > >>No good. :-( > >> > >>Miguel Cruz wrote: > >> > >>>Okay, I'm pretty sure your problem is this: > >>> > >>>All those variables referenced in $footertext ($REMOTE_ADDR, $userinfo, > >>>etc.) need to be brought into the local context in the function calling > >>>eval(), using the global keyword. Otherwise they won't be available. > >>> > >>>miguel > >>> > >>>On Mon, 20 May 2002, Leif K-Brooks wrote: > >>> > >>>>Ok, I guess I'll send it. Thanks for trying to help. Here it is: > >>>> > >>>> $ip = $REMOTE_ADDR; > >>>> $query = mysql_query("select COUNT(*) as rowexists from ips where ip > >>>> = '$ip'"); > >>>> $result = mysql_fetch_array($query); > >>>> if(($result['rowexists'] == 0) && ($loggedin)){ > >>>> mysql_query("INSERT INTO ips > >>>> (ip,username) VALUES > >>>> ('$ip','$userinfo[username]')") or print > >>>> "Error: ".mysql_error(); > >>>> } > >>>> $time_end = getmicrotime(); > >>>> $time_taken = $time_end - $time_start; > >>>> $query = mysql_query("select COUNT(*) as cnt from ips"); > >>>> $result = mysql_fetch_array($query); > >>>> $visits = $result['cnt'] + 2800; > >>>> eval($getlayout['footer']); > >>>> > >>>> > >>>>Miguel Cruz wrote: > >>>> > >>>>>Rather than continuing to shroud this in mystery, could you at least give > >>>>>us a taste of what is in $footertext? > >>>>> > >>> > >> > > > > > > > >