[RESOLVED] displaying (table cell?) content as desired

[bonio]

To explain-

I have about 30 images and category titles and I want to display them in 3's from left to right across the page as:

print ' <img src =\"$maincat_image\">';
print $maincat_title;

so ending up with 10 or so rows.

can anyone provide a solution please.

[dcav]

Most people would use two FOR loops, but I'm not sure how you feed in your data.
Play with it !

<table>
<?
	$numRows = 30;
	for ($i = 1; $i < $numRows; $i++) {
		if (($i % 3) == 1)
			echo "<tr>\n"; 
		echo "<td>Picture $i</td>";
		if (($i % 3) == 0)
			echo "\n</tr>\n";
		$current = $i;
		}
	if (($current % 3) <> 0)
		echo "</tr>\n"; 
	?>
</table>

[bonio]

Put a variable above the table
Got there with much help in case anybody else stuck...

[COLOR=Red]<?php $td = 0; ?>[/COLOR]
<table width="706"  border="0">
  <tr>

<?php 
$getsubcatbox="SELECT s.maincategory_id,s.subcategory_name,
etc"
$getsubcatbox_res=mysql_query($getsubcatbox) or die(mysql_error());
//while there are rows produce a table and fill it with the selected array items
while ($row=mysql_fetch_row($getsubcatbox_res)) {

[COLOR=Red]$td = $td + 1;[/COLOR]

echo '[COLOR=Red]<td>[/COLOR]
<table width="170" height=" etc etc

[COLOR=Red]</td>';[/COLOR]
 [COLOR=Red]if ($td == 4) {$td = 0;
      echo "</tr><tr>";}[/COLOR]
}
?>