[SOLVED] Link-o-rama

[phatgreenbuds]

Below is my code. There are two lines in the middle for the html img src=. The first is a database entry that holds the path to the thumbnail. I want to make that a link to the second which is another database entry holding the path to the actual image and have it open in a new window. Any suggestions how I would make that happen?

 

 

<?php
$query = "SELECT * FROM images";
$target = mysql_query($query); // uses the query variable defined above.
confirm_query($target); // calls to the function file.

	while ($final_target = mysql_fetch_array($target)) {
	$submitter = "{$final_target["picsubmit"]}";
	$description = "{$final_target["picdesc"]}";
	$filename = "{$final_target["picfile"]}";
?>

	<img src="<?php echo $final_target["tnpath"]; ?>" alt=""  />
	<br>
	<img src="<?php echo $final_target["picpath"]; ?>" alt="" />

<?php
	echo "<br>" . "<h4>" . $description . "</h4>" . "<br>";
	echo "Submitted By: " . $submitter;
	echo "<h4>" . "<hr>" . "</h4>";
	echo "<br>";
	} 	
?>

[pocobueno1388]

<?php

echo "<a href='{$final_target['picpath']}' target='blank'><img src='{$final_target['tnpath']}'></a>";

?>

 

[phatgreenbuds]

Good Grief...I am an idiot. Thank you...i can't believe it was that simple.