IF exchange placements - what have I done wrong?

[jasper1106]

Hi guys,

 

New to PHP but learning every day. This is probably a really newbie error but I've tried multiple different ways and can't get it to work.

 

$tableData = '';
if(mysql_num_rows($result)==0)
{
    $tableData = "<tr><td colspan=\"4\">No results returned.</td></tr>\n";
}
else
{
    while($row = mysql_fetch_assoc($result))
    {
    			
        $tableData .= "  <tr bgcolor='#FFFFFF' align='center'>\n";
			    $tableData .= "    <td>{$rank}</td>\n";
        $tableData .= "  </tr>\n";
  			
$rank     = ($row['rank']);
if ($rank == 0) { print "Cadet"; }
if ($rank == 1) { print "Sergeant"; }
if ($rank == 2) { print "2nd Lieutenant"; }
if ($rank == 3) { print "1st Lieutenant"; }
if ($rank == 4) { print "Captain"; }
if ($rank == 5) { print "Major"; }
if ($rank == 6) { print "Lt. Colonel"; }
if ($rank == 7) { print "Colonel"; }  
}
}

?>

  <table class="tables" cellspacing="0" cellpadding="0">
    <tr>
      <th>Rank</th>
    </tr>
    <?php echo $tableData; ?>
  </table>

 

When this script is run the $rank variable shows the numbers 1,2,3,4,5,6,7 instead of the textual translations for example if 0 is displayed then "cadet" should be shown instead. 

 

The current script shows the numbers in the table but prints the accurate translations at the top outside the table tag..... 

Can anyone show me what I've done wrong here? Thank you in advance.

[jcbones]

  Quote

$tableData .= "  <tr bgcolor='#FFFFFF' align='center'>\n";
    $tableData .= "    <td>{$rank}</td>\n"; //<--Printing out rank, before changing it.

 

Try it this way:

//top of script.
$ranks = array('Cadet','Sergeant','2nd Lieutenant','1st Lieutenant','Captain','Major','Lt. Colonel','Colonel');

//inside the while loop, replace these lines.
$tableData .= "  <tr bgcolor='#FFFFFF' align='center'>\n";
    $tableData .= '<td>' . $ranks[$rank] . '</td>' . "\n";

[jasper1106]

Ah I never thought about converting it so the array counts it, thank you jc!