[SOLVED] Multiple Query Error

[steveangelis]

<table width="700" border="0" align="center" cellpadding="0" cellspacing="0">
  <tr>
    <td height="30" align="center" valign="middle"><table width="100%" border="1" cellspacing="0" cellpadding="0">
      <tr>
        <td align="center" valign="middle" background="styles/<?PHP echo $cfgcontent['style']; ?>/images/ranks_top.gif" height='30'><span class="style3">Members</span></td>
      </tr>
    </table></td>
  </tr>
  <tr>
    <td align="left" valign="top"> </td>
  </tr>
  <tr>
    <td align="left" valign="top"><?PHP
$queryrnk1 = "SELECT * FROM ranks order by rnk_order asc";
$resultrnk1 = mysql_query($queryrnk1) or die("$query does not make any sence;<br>" . mysql_error());
if 
(mysql_numrows($resultrnk1)>0)
{
for($n=0;$n<mysql_numrows($resultrnk1);$n++) 
{
$rnk1content = mysql_fetch_array($resultrnk1); 
$rank_id = $rnk1content['id'];
?><table width="100%" border="0">
  <tr>
    <td align="center" valign="top" class="style4"><?PHP echo $rnk1content['rnk_name'];?></td>
  </tr>
  <tr>
    <td><table width="100%" border="0" cellspacing="0" cellpadding="0">
      <tr>
        <td align="left" valign="top" class="style4">Rank</td>
        <td align="left" valign="top">Name</td>
        <td align="left" valign="top">Primary Game</td>
        <td align="left" valign="top">Active</td>
      </tr>
          <?PHP
$queryrnk2 = "SELECT * FROM members where mbr_rank='$rank_id' order by mbr_name asc";
$resultrnk2 = mysql_query($queryrnk2) or die("$query does not make any sence;<br>" . mysql_error());
if 
(mysql_numrows($resultrnk2)>0)
{
for($n=0;$n<mysql_numrows($resultrnk2);$n++) 
{
$rnk2content = mysql_fetch_array($resultrnk2); 
?><tr>
            <td width="150" align="left" valign="top" class="style4"><img src="<?PHP echo $rnk1content ['rnk_image']; ?>" alt="<?PHP echo $rnk1content ['rnk_name']; ?>"></td>
            <td align="left" valign="top"><a href="index.php?p=profiles&id=<?PHP echo $rnk2content ['id']; ?>"><?PHP echo $rnk2content ['mbr_name']; ?></a><br></td>
            <td align="left" valign="top"><?PHP echo $rnk2content ['mbr_gameprimary']; ?></td>
            <td align="left" valign="top"><?PHP echo $rnk2content ['mbr_status']; ?></td>
          </tr>
    <?PHP 
} 
} 
?>
        </table>
        </td>
  </tr>
</table>
        <hr width="100%">
      <?PHP 
} 
} 
?>
    </td>
  </tr>
</table>

 

This code is not too complicated.  What it is susposed to do is first of all get all the ranks in the database and order them by a field rnk_order until there are no more ranks.  Next with each rank it is to show all the members with that rank.  It shows all the members from the first rank but it not going onto the next rank and I know there is more than one rank in the database.  Can anyone see where I am going wrong?  It does now show any other rank except the first.

[steveangelis]

<?PHP

$queryrnk1 = "SELECT * FROM ranks order by rnk_order asc";

$resultrnk1 = mysql_query($queryrnk1) or die("$query does not make any sence;<br>" . mysql_error());

if

(mysql_numrows($resultrnk1)>0)

{

for($n=0;$n<mysql_numrows($resultrnk1);$n++)

{

$rnk1content = mysql_fetch_array($resultrnk1);

$rank_id = $rnk1content['id'];

?>

 

That is where the error is happening and I do not see the problem.

[sasa]

change

for($n=0;$n<mysql_numrows($resultrnk2);$n++) 

to

for($nn=0;$nn<mysql_numrows($resultrnk2);$nn++)//change variable name