How to display a specific image?

Question

I am making a system of films that register the poster of the film and I want that when generate the report with all the registered movies, it shows me the image of the poster saved in the bank and not the name of the image. I found a code that shows all the images in a table of the bank, but shows all of them below the other and in the report I want to show them separately, next to the name, for example:

Title.................. Director............... Poster

13: The player..... Gela Babluani.... Image

The code that I found on the Internet and am basing myself on it is as follows:

<?php

include ('config.php');

$sql = mysql_query("SELECT * FROM filmes ORDER BY nome_filme");

while ($filmes = mysql_fetch_object($sql)) {

echo "<img src='fotos/".$filmes->foto."' alt='Foto de exibição' /><br />";

?>

Table name in the database: movies Poster picture variable name: $photo Movie name variable name: $movie name

Answer 1

I’d do it that way:

<?php
include ('config.php');
$sql = mysql_query("SELECT * FROM filmes ORDER BY nome_filme");

    echo "
    <table>
        <tr>
            <td>COD</td>
            <td>Título</td>
            <td>Diretor</td>
            <td>Imagem</td>
        </tr>
    ";
    while ($filmes = mysql_fetch_array($sql)) {
        echo "
        <tr>
            <td>".$filmes['id']."</td>
            <td>".$filmes['titulo']."</td>
            <td>".$filmes['diretor']."</td>
            <td><img src='fotos/".$filmes['foto']."'></td>
        </tr>
        ";
    }
    echo "</table>";
?>