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嗨，我想弄清楚如何在用户上传的相册中显示第一张图片。我有一个网站可以让用户将图像上传到相册，现在我想将相册中的第一张图片显示为相册封面，现在当用户点击相册页面时，只显示相册信息，如名称、描述、等等，我使用相册名称作为查看相册中图像的链接，我想要的是相册中的第一张图像显示为该相册的链接。我有很多事情要做，但我会尽力提供一切，以便您更容易理解。这是我的相册页面代码：

$albums = get_albums();

if(empty($albums)){
echo 'No Albums';
}else{

foreach($albums as $album){
    echo '<p><a href="view_album.php?album_id=', $album['id'] ,'">', $album['name'],'</a>(', $album['count'], ' images) <br />
    ', $album['description'], '...<br />
    <a href="edit_album.php?album_id=', $album['id'] , '">Edit</a> / <a href="delete_album.php?album_id=', $album['id'] , '">Delete</a>
    </p>';
}
}

这是我包含在此文件中的函数 php 文件中的函数：

function get_albums(){
$albums = array();

$albums_query = mysql_query("
SELECT `albums`.`album_id`, `albums`.`timestamp`, `albums`.`name`, LEFT(`albums`.`description`, 50) as `description`, COUNT(`images`.`image_id`) as `image_count`
FROM `albums`
LEFT JOIN `images`
ON `albums`.`album_id` = `images`.`album_id`
WHERE `albums`.`user_id` = " . $_SESSION['user_id'] . "
GROUP BY `albums`.`album_id`
");

while($albums_row = mysql_fetch_assoc($albums_query)){
    $albums[] = array(
        'id'            => $albums_row['album_id'],
        'timestamp'     => $albums_row['timestamp'],
        'name'          => $albums_row['name'],
        'description'   => $albums_row['description'],
        'count'         => $albums_row['image_count']
    );
}

return $albums;

}

Now I am a little lost and confused on what route to take to change the link to open the albums from the album name to the first image in the album. If the albums do not have any pics in them I plan on using a simple if statement to show a generic empty folder image. My database using two table to upload images, the first is the albums table and the second is the images table. I also have a timestamp field that inserts the image upload timestamp, so can i return the images from that album and the image with the lowest timestamp so to return the first image uploaded to the album? Is there a SELECT statement to return this? Also I did create a folder that stores these images in the corresponding albums, can i return the first image in that album? What would be the easiest way to achieve this?

Here is how I get the images path:

function get_images($album_id){
$album_id = (int)$album_id;

$images = array();

$image_query = mysql_query("SELECT `image_id`, `album_id`, `timestamp`, `ext` FROM `images` WHERE `album_id` = $album_id AND `user_id` = ". $_SESSION['user_id']);
while($images_row = mysql_fetch_assoc($image_query)){
    $images[] = array(
        'id'            => $images_row['image_id'],
        'album'         => $images_row['album_id'],
        'timestamp'     => $images_row['timestamp'],
        'ext'           => $images_row['ext']
    );
}
return $images;
}

Here is my view_album.php file:

if(!isset($_GET['album_id']) || empty($_GET['album_id']) || album_check($_GET['album_id']) === false){
header('Location: albums.php');
exit();
}

$album_id = $_GET['album_id'];
$album_data = album_data($album_id, 'name');

echo '<h1 class="logo2">', $album_data['name'], '</h1>';
include 'includes/menu_album.php';

$images = get_images($album_id);

if(empty($images)){
echo 'No images.';
}else{
foreach($images as $image){
    echo '<a href="uploads/', $image['album'], '/', $image['id'], '.',     $image['ext'], '"><img src="uploads/thumbs/', $image['album'], '/', $image['id'], '.', $image['ext'], '" title="Uploaded ', date('D M Y', $image['timestamp']), '"></a> [<a       href="delete_image.php?image_id=', $image['id'], '">x</a>] ';
}
}
4

1 回答 1

2

You are actually pretty close to getting the first image with your existing query. All you need to do as add the image name or path to the SELECT and the image timestamp to an ORDER BY to make sure you get the earliest one:

   $albums_query = mysql_query("
        SELECT 
          `albums`.`album_id`, 
          `albums`.`timestamp`, 
          `albums`.`name`, 
          LEFT(`albums`.`description`, 50) as `description`, 
          COUNT(`images`.`image_id`) as `image_count`, 
          CONCAT(`images`.`image_id`, '.', `images`.`ext`) AS `first_image`
        FROM `albums`
        LEFT JOIN `images`
        ON `albums`.`album_id` = `images`.`album_id`
        WHERE `albums`.`user_id` = " . $_SESSION['user_id'] . "
        GROUP BY `albums`.`album_id`
        ORDER BY `albums`.`timestamp` ASC, `images`.`timestamp` ASC
    ");

    if (!$albums_query) {
      die('Invalid query: ' . mysql_error());
    }

    while($albums_row = mysql_fetch_assoc($albums_query)){
        $albums[] = array(
            'id'            => $albums_row['album_id'],
            'timestamp'     => $albums_row['timestamp'],
            'name'          => $albums_row['name'],
            'description'   => $albums_row['description'],
            'count'         => $albums_row['image_count'],
            'image'         => $albums_row['first_image']
        );
    }

You can then update your output to include the image:

foreach($albums as $album){
    if (empty($album['image'])) {
        $album['image'] = 'default.jpg';
    }
    echo '<p>
      <a href="view_album.php?album_id=', $album['id'] ,'">
        <img src="uploads/thumbs/', $album['id'] ,'/', $album['image'] ,'" />', $album['name'],'
      </a>(', $album['count'], ' images) <br />
      ', $album['description'], '...<br />
      <a href="edit_album.php?album_id=', $album['id'] , '">Edit</a> / <a href="delete_album.php?album_id=', $album['id'] , '">Delete</a>
    </p>';
}
于 2012-07-31T01:28:44.250 回答