Mysql fetch array, table results 2 - php

Question

I've recently posted a similar question but that was to create one row of cells across and after it reaches 3 columns, it creates a new row, so I can have 3 columns of infinite rows. That was solved.
Now what I need is this (GIVEN A and B should be records from the database using $row = mysql_fetch_array($results) so basically it would be something like $row['username']; for A and B ).
<tr>
<td><img src="images/ava/A.png" /></td>
<td>A</td>
<td width="2px" rowspan="3"></td>
<td><img src="images/ava/B.png" /></td>
<td>B</td>
</tr>
<tr>
<td><div class="gauge"><div class="innergauge"></div>A</div></td>
<td><div class="gauge"><div class="innergauge"></div>B</div></td>
</tr>
<tr>
<td>A</td>
<td>B</td>
</tr>
As what you can see, the record I got from my database has to fill in to this format, where A is one record and B is another, then if there is record C, this whole thing should repeat again to form a second row.

Answer 1

Build each section as you go, combine them at the end. Here's how you do sections 1 and 3, the middle section is left as an exercise:
$section1 = '<tr>';
$section2 = // start $section2
$section3 = '<tr>';
$i = 0;
while($row = mysql_fetch_array(...)){
if($i > 0){
// won't happen the first time through the loop
$section1.= '<td width="2px" rowspan="3"></td>';
}
$section1 .= '<td><img src="images/ava/'.$row['username'].'"/></td>';
$section1 .= '<td>'.$row['username'].'</td>';
// now do $section2
$section3 .= '<td>'.$row['username'].'</td>';
$i++;
}
$section1 .= '</tr>';
// finish $section 2
$section3 .= '</tr>';
// now output
echo $section1 . $section2 . $section3;