Error in PHP Code

[kingkong142]

Hi,

I am having error in following code,

<!-- START --> 
<p class="date-head"> 
<?php 
$sql_online_mem = "SELECT membername FROM ` trade_messengerd` WHERE status = 'Online'"; 
$db_sql_mem = mysql_query($sql_online_mem); 
while($online_mem = mysql_fetch_array($db_sql_mem)){ 
$members_online[] = $online_mem['membername']; 
} 
for($o=0; $o<count($members_online); $o++){ 
if($members_online[$o] == $uid){ 
$mem_online = "Online"; 
} 
} 
?> 
<a href="#" onClick="window.open('<?=$domain_url?>/messenger/members/_login.php','login','toolbar=0,location=0,directories=0,status=0,menubar=0,scrollbars=0,resizable=1,width=368,height=600,left=380,top=60');return false;" title="Chat Now"><?php if($mem_online == 'Online'){ ?>
<img src="<?=$domain_url?>/images/chat_now_online.jpg" alt="Online Chat Now" title="Online Chat Now" height="31" width="159" border="0" />
<?php }else{ ?>
<img src="<?=$domain_url?>/images/chat_now_offline.jpg" alt="Chat Offline" title="Chat Offline" height="31" width="159" border="0" />
<?php } ?></a>
</p>
 <!-- END -->






Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /public_html/offer-new11.php on line 1289
Line # 1289 is     while($online_mem = mysql_fetch_array($db_sql_mem)){ 

Could you please tell me why i am having error and how i can solve it?

Regards,
FHS

[Martyn_86]

Sounds like an error with your SQL syntax;

$sql_online_mem = "SELECT membername FROM ` trade_messengerd` WHERE status = 'Online'"; 

You appear to have an unintended space where trade_messengerd starts in your query where you have enclosed it.

$db_sql_mem = mysql_query($sql_online_mem) or die(mysql_error());

Modify your code to include that. It may help track down any errors in your syntax.

[kingkong142]

Hi,

Thank you for your email. Now it is not showing any error, But it is not showing member online status. Any idea why?

Regards,
FHS

[kingkong142]

Hi,

Now everything is fine. Thanks :)

Regards,
FHS