Re: Variable won't work in function, even when I global it?
| From: | Miguel Cruz | Date: | Mon, 20 May 2002 06:12:54 +0000 |
| Subject: | Re: Variable won't work in function, even when I global it? | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-98429@lists.php.net to get a copy of this message | ||
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?
> >>>
> >
> >
>
>