How to show all the contents of a database in HTML?

Good I need to know how I do that the url lodged in my Database can extract them with php and place them in the "src" of the images so that they always load. You will see I have the following problem, I am doing a social network as a degree project, obviously not complete, but if some functions, I have a week left, and I need to create a gallery with PHP mysqli AJAX, and I can already host the url of the photos in the database, but I can not get those url for a query "fetch_assoc" and print a label "" for each url that has the database, really that's what I want.

I leave the codes to help me better.

PS: I'm a newbie in PHP.

<?php 
    session_start();
    $usuario = $_SESSION['usuario'];

    if (isset($_POST['upload'])) {
	    $conexion = mysqli_connect("localhost", "root", "","imagen"); 
	    if (!$conexion) { 
			die("Fallo la conexión a la Base de Datos: " . mysql_error()); 
	    }else{
		$imgAlbum = "tmpAlbums/"; 
		$archivo = $_FILES['fl_album']['tmp_name'];
		$nombrearchivo = $_FILES['fl_album']['name']; 

		move_uploaded_file($archivo, $imgAlbum."/".$nombrearchivo); 

		$imgAlbum = $imgAlbum."/".$nombrearchivo;
		$saveIMG = mysqli_query($conexion,"INSERT INTO imagephp VALUES('".$imgAlbum."')");
	    }      			
    }
?>

<div class="albumes">
			<img src="<?php echo ($imgAlbum); ?>" alt="">
			<?php     							
				$selectURL = mysqli_query($conexion, "SELECT URL_img FROM imagephp");
				while ($url = mysqli_fetch_assoc($selectURL)) {
					echo "<img src='$url'/>";
				}
				var_dump($url);
			 ?>
		</div>

I am ignoring a sea of code because it is obvious that you know and understand what it would take, like body etc, but what is there is because it is the only thing I want to be helped in that.

I thank you in advance and I ask you please understand that they speak with a newbie ...:)