Question

我正在使用php脚本从文件夹中获取所有图像并对它们进行分页并且一切正常，但是我想使用实际照片名称（filemname）作为alt标记，但我发现很难得到这个名字。可以使用以下代码完成此任务：

<?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,gif,png";

//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){
        //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='I NEED THIS TO BE THE FILE NAME'></a>
                </div>
            </div>

        ";
    }
}

//Now to display the page numbers!
for($p = 1; $p <= $totalPages; $p++){
    if($p == $page){
        $tmp_pages[] = "<strong class='pagination'>{$p}</strong>";
    }else{
        $tmp_pages[] = "<a class='pagination' href='?page={$p}'>{$p}</a>";
    }
}
?>

Answer 1

在你的循环中执行：

var_dump(basename($totalImages[$i]));

或者您也可以查看：

var_dump(pathinfo($totalImages[$i]));

还会有basename，dirname，extension等。

Answer 2

使用basename()。

$filename = explode('.', basename($images[$i]));
echo "
    <div class='galleryCellHolder'>
        <div class='galleryCell'>
            <a class='fancybox' rel='group' href='{$images[$i]}'><img class='galleryPhoto' src='{$images[$i]}' alt='" . $filename[0] . "'></a>
        </div>
    </div>    
";

使用PHP获取文件名

2 个答案: