Default to another table for search if results not found innitially

[zzz]

I'm passing a variable from a search form. The current code looks like this:

if ($type == 't1') {
	$SRCHwhere = "table1 WHERE orgType='association' AND NameX";
	$foundname = "NameX";
} elseif ($type == 't2') {
	$SRCHwhere = "table2 WHERE NameY";
	$foundname = "NameY";
} elseif ($type == 't3') {
	$SRCHwhere = "table3 WHERE NameZ";
	$foundname = "NameZ";


$foundResults = $db->get_results ("SELECT * FROM ".$SRCHwhere." LIKE '%".$srch."%'");

foreach ( $foundResults as $fr )
		echo "<tr>
		      	<td>".$fr->$foundname."</td>
			  </tr>";
		}

Everything works fine. Now, I need to expand a bit. For example, if $type == 't2' and the values not found in table2, I need it to automatically look into table4. I'm not sure how to do it taking into consideration my current code...

[devinemke]

i don't know exactly what the get_results() method does but your code suggests that it returns an array. simply check to see if that array is empty, if so then query to other table.

[zzz]

I think I understand what the steps need to be taken, I'm just having difficulties getting it in my code.

Oh... get_results() is just part of the class I'm using. In fact I could rewrite the code like this:

if ($type == 't1') {
	$SRCHwhere = "table1 WHERE orgType='association' AND NameX";
	$foundname = "NameX";
} elseif ($type == 't2') {
	$SRCHwhere = "table2 WHERE NameY";
	$foundname = "NameY";
} elseif ($type == 't3') {
	$SRCHwhere = "table3 WHERE NameZ";
	$foundname = "NameZ";


if($foundResults = $db->get_results ("SELECT * FROM ".$SRCHwhere." LIKE '%".$srch."%'"))
{
foreach ( $foundResults as $fr )
		echo "<tr>
		      	<td>".$fr->$foundname."</td>
			  </tr>";
		}
} else {
	// nothing found do something else, like run another query.   

}

But it's starting to look a bit convoluted, doesn't it?