Ashoar Posted April 9, 2009 Share Posted April 9, 2009 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? Link to comment https://forums.phpfreaks.com/topic/153301-solved-array-problem-2/ Share on other sites More sharing options...
ratcateme Posted April 9, 2009 Share Posted April 9, 2009 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. Link to comment https://forums.phpfreaks.com/topic/153301-solved-array-problem-2/#findComment-805373 Share on other sites More sharing options...
wildteen88 Posted April 9, 2009 Share Posted April 9, 2009 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. Link to comment https://forums.phpfreaks.com/topic/153301-solved-array-problem-2/#findComment-805374 Share on other sites More sharing options...
Ashoar Posted April 9, 2009 Author Share Posted April 9, 2009 Ah fixed. Thanks guys Link to comment https://forums.phpfreaks.com/topic/153301-solved-array-problem-2/#findComment-805378 Share on other sites More sharing options...
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