Php MySQL将多个图像加入相册，并使用foreach在页面上显示它们

我正在尝试将多个图像加入相册，并使用foreach在页面上显示它们

+----+----------+----------------------------------+-----------+----------+
| id | album_id | name                             | url       | theme_id |
+----+----------+----------------------------------+-----------+----------+
|  1 |        1 | 60b0603f40993bfe86d54756505416cb | project_1 | 1        |
|  2 |        1 | 5efd6e08fed71ac6ad18bdde997e97bd | project_1 | 1        |
|  3 |        2 | fe4ac6806722e7cd74a60476da9ec4ea | project_2 | 1        |
|  4 |        2 | 87028bcd233b5ca5133fd0be335ff4b7 | project_2 | 1        |
|  5 |        2 | 9326d956c5cd30a75ff8b90d59e18273 | project_2 | 1        |
|  6 |        3 | 325925af3c04d936aa0e292b007b19e9 | project_3 | 1        |
|  7 |        3 | 0e087fb5e2038a746398d8dbe362032d | project_3 | 1        |
|  8 |        3 | 8be793e3e619771bb81e55c805d23b7d | project_3 | 1        |
+----+----------+----------------------------------+-----------+----------+

当我运行下面的代码时，我在第一个相册\u id中获得了正确的图像名称和url，但例如，在相册\u id=2中，我仍将获得项目\u 1 url，在相册\u id=3中，我将获得项目\u 2 url。我完全迷路了

       $album = find_album($_GET['id']);
       $albums = find_albums_by_theme($album['theme_id']);
       $image = find_image($_GET['id']);
       $images = find_images_by_album($image['album_id']);
       $themes = find_themes();

<ul>
        <?php foreach($images as $image): ?>
            <li><img src="photos/<?php echo $image['url']; ?>/large/<?php echo $image['name']; ?>.jpg" alt="<?php echo safe_output($album['title']); ?>" /></li>
   <?php endforeach; ?>
               </ul>

---

在这两行中，您对album.id和image.id都使用了

id

。我想这会引起一个问题

   $album = find_album($_GET['id']);
   $image = find_image($_GET['id']);

   $album = find_album($_GET['id']);
   $image = find_image($_GET['id']);