สอบถามผู้รู้ค่ะ ถ้าinsertข้อมูลแล้ว ต้องการให้แสดงผล url แบบhref="show-resume.php?rid=.$rid. ค่ะ ต้องทำยังไง

	$sql = "SELECT * FROM resume ORDER BY resume_id DESC";
	$result = page_query($link, $sql, 10);
	$first = true;
	while($resume = mysqli_fetch_array($result)) {
		if(!$first) {
			echo "<hr>";
		}
		$rid = $resume['resume_id'];
		$sql = "SELECT COUNT(*) FROM image WHERE resume_id = $rid";
		$r = mysqli_query($link, $sql);
		$row= mysqli_fetch_array($r);
		$src = "images/no-photo.jpg";
		if($row[0]!=0) {
			$src = "read-img.php?id=$rid";
		}
		echo '<img src="'.$src.'" class="photo">';
		echo '<div class="info">';
		echo '<a href="show-resume.php?rid='.$rid.'" target="_blank">'.$resume['name'].'</a>';
		echo '<a href="#" class="delete">ลบ</a>';
		
		$sql = "SELECT * FROM experience WHERE resume_id = $rid ORDER BY exp_id ASC";
		$r = mysqli_query($link, $sql);
		while($exp = mysqli_fetch_array($r)) {
			if(!empty($exp['position'])) {
				echo "<br><span>ประสบการณ์ล่าสุด:</span> ";
				echo $exp['position'] . "  - " . $exp['workplace']; 
				break;
			}
		}

		$sql = "SELECT * FROM education WHERE resume_id = $rid ORDER BY edu_id ASC";
		$r = mysqli_query($link, $sql);
		while($edu = mysqli_fetch_array($r)) {
			if(!empty($edu['level'])) {
				echo "<br><span>การศึกษาสูงสุด:</span> ";
				echo $edu['level'] . " " . $edu['major'] . " - " . $edu['academy'];
				break;
			}
		}
		echo '</div><br class="clear">';
		$first = false;		
	}