render image only when a database entry exists

[original89]

im trying to display a link as an image or url on a page only when the URL record exists in my database. If there is no URl in the dsatabase i dont want the url/image to display on the page. I have dynamic pages for a catalogue and only want urls to appear on the pages for products that i have a url for them.

At the moment i can only get it to render an image for all occasions but this is no good. Im doign something like this but as i said i only want it to display something if i have a URL and not display anything if i dont.

	function getURL($ID)
	{
		$sql = "SELECT URL etc
		if ($recordSet === false)
		{
			log_error($sql);
		}		
		while (!$recordSet->EOF)
		{
			$URL = make_db_unsafe ($recordSet->fields[URL]);
			$recordSet->MoveNext();
		} // end while
		echo " <a href=\"$URL\"><img src=\"/gazania.jpg\" border=\"0\"></a><br>";
	} // function display url

anyone help me..??

thanks..

[cmancone]

ummm..... at the risk of sounding stupid, wouldn't $URL be empty if there is no database record? So can't you just say:

if ( $URL )
echo " <a href="$URL\"><img src=\"/gazania.jpg\" border=\"0\"></a><br>";

[LoganK]

Yeah... if ($URL == null) { // don't do image } else { // do image }

You have a parse error in your echo statement, you didn't escape those first double quotes.