我有工作代码，它从动态创建的文件夹中提取所有图像。但是我只想在我的显示页面中显示来自特定文件夹的 1 张图像。有什么建议吗？我的代码：
<?php
$search_dir = "$directory/{$row['name']}{$row['hotel_address']}";
$images = glob("$search_dir/*.jpg");
sort($images);
//display images
foreach ($images as $img) {
echo "<img src='$img' height='150' width='150' /> ";
}
?>
您可以使用以下命令显示单个图像：
<?php
$search_dir = "$directory/{$row['name']}{$row['hotel_address']}";
$images = glob("$search_dir/*.jpg");
sort($images);
// Image selection and display:
//display first image
if (count($images) > 0) { // make sure at least one image exists
$img = $images[0]; // first image
echo "<img src='$img' height='150' width='150' /> ";
} else {
// possibly display a placeholder image?
}
?>
如果您想要随机图像，请执行以下操作：
// Image selection and display:
//display random image
if (count($images) > 0) { // make sure at least one image exists
// Get a random index in the array with rand(min, max) which is inclusive
$randomImageIndex = rand(0, count($images)-1);
$img = $images[$randomImageIndex]; // random image
echo "<img src='$img' height='150' width='150' /> ";
} else {
// possibly display a placeholder image
}
您可以使用
current从数组中获取第一个图像。
<?php
$search_dir = "$directory/{$row['name']}{$row['hotel_address']}";
$images = glob("$search_dir/*.jpg");
sort($images);
//display one image:
echo "<img src='current($images)' height='150' width='150' /> ";
?>