我正在为一个需要照片库的客户建立网站,原本打算使用文件名作为alt标签,但他要求我使用他在EXIF数据中输入的关键字。由于我不是摄影师,对此技术方面并不了解。但我已经编写了一个脚本来获取文件名,希望只需更改几行代码就能获取EXIF数据。以下是我的代码:
<?php
//The directory to your images folder, with trailing slash
$dir = "cms/gallery/photo/";
//Set the extensions you want to load, seperate by a comma.
$extensions = "jpeg,jpg";
//Set the number of images you want to display per page
$imagesPerPage = 3;
//Set the $page variable
if(!isset($_GET['page'])){
$page = 1;
}else{
$page = $_GET['page'];
}
//Load all images into an array
$images = glob($dir."*.{".$extensions."}", GLOB_BRACE);
//Count the number of images
$totalImages = count($images);
//Get the total pages
$totalPages = ceil($totalImages / $imagesPerPage);
//Make sure the page you are on is not greater then the total pages available.
if($page > $totalPages){
//Set the currnet page to the total pages.
$page = $totalPages;
}
//Now find where to start the loading from
$from = ($page * $imagesPerPage) - $imagesPerPage;
//Now start looping
for($i = $from; $i < ($from + $imagesPerPage); $i++){
//We need to make sure that its within the range of totalImages.
if($i < $totalImages){
$filename = explode('.', basename($images[$i])); // GET EXIF DESCRIPTION AS $FILENAME
//Now we can display the image!
echo "
<div class='galleryCellHolder'>
<div class='galleryCell'>
<a class='fancybox' rel='group' href='{$images[$i]}'><img class='galleryPhoto' src='{$images[$i]}' alt='" . $filename[0] . "'></a>
</div>
</div>
";
}
}
//Now to display the page numbers!
for($p = 1; $p <= $totalPages; $p++){
if($p == $page){
$tmp_pages[] = "<a class='noPagination'>{$p}</a>";
}else{
$tmp_pages[] = "<a class='pagination' href='?page={$p}'>{$p}</a>";
}
}
?>
<div class="clearLeft"></div>
<div id="pagination">
<?php
//Now display pages, seperated by a hyphon.
echo "<br />" . implode("", $tmp_pages);
?>
</div>
$exif = exif_read_data($images[$i], 'COMMENT'); echo implode($exif['COMMENT'], ', ');
- Reeno