[SOLVED] Array problem 2

[Ashoar]

Wildteen just gave me a fix to an error i was having.

Any it worked fine untill i logged in as a user.

 

Here is the code:

$online = mysql_query("SELECT username FROM online") or die(mysql_error());

if(mysql_num_rows($online) > 0)
{
    while($row = mysql_fetch_assoc($online))
    {
        $online[] = '<A href="member_profile.php?username='.$row['username'].'">'.$row['username'].'</a>';
    }

    echo '<tr class="mainrow"><td>Currently active: ' . implode(',', $online) . '</td></tr>';
}
else
{
    echo '<tr class="mainrow"><td>There are currently no users logged in</td></tr>';
}

echo '</table>';

 

And here are the errors:

Warning: Cannot use a scalar value as an array

 

and

 

Warning: implode() [function.implode]: Invalid arguments passed

 

Another way to do this?

[ratcateme]

you need to rename the $online you are using inside the loop to something else say $onlinearray

then change the var in the implode to $onlinearray

 

Scott.

[wildteen88]

Yeah I have made mistake in that code

 

Change $online[] to $onlineUsers[] and then change implode(',', $online) to implode(',', $onlineUsers)

 

Should sort the error out.

[Ashoar]

Ah fixed.

Thanks guys