如何使用php显示文件夹中的图像 - PHP
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How to display images from a folder using php - PHP
提问by user2837048
It would be great if someone could help me figure out why the browser cannot load the images (error 404). The code works, and the image source is correct, but I cannot figure out what is wrong. (using localhost)
如果有人能帮我弄清楚为什么浏览器无法加载图像（错误 404），那就太好了。代码有效，图像来源正确，但我无法弄清楚出了什么问题。（使用本地主机）
$dir = '/home/user/Pictures';
$file_display = array(
'jpg',
'jpeg',
'png',
'gif'
);
if (file_exists($dir) == false) {
echo 'Directory \'', $dir, '\' not found!';
} else {
$dir_contents = scandir($dir);
foreach ($dir_contents as $file) {
$file_type = strtolower(end(explode('.', $file)));
if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true) {
echo '<img src="', $dir, '/', $file, '" alt="', $file, '" />';
}
}
}
回答by Bill
You had a mistake on the statement below. Use . not ,
你在下面的陈述中有一个错误。用 。不是 ，
echo '<img src="', $dir, '/', $file, '" alt="', $file, $
to
到
echo '<img src="'. $dir. '/'. $file. '" alt="'. $file. $
and
和
echo 'Directory \'', $dir, '\' not found!';
to
到
echo 'Directory \''. $dir. '\' not found!';
回答by jerjer
Here is a possible solution the solution #3 on my comments to blubill's answer:
这是我对 blubil 回答的评论中的解决方案 #3 的可能解决方案：
yourscript.php
========================
<?php
$dir = '/home/user/Pictures';
$file_display = array('jpg', 'jpeg', 'png', 'gif');
if (file_exists($dir) == false)
{
echo 'Directory "', $dir, '" not found!';
}
else
{
$dir_contents = scandir($dir);
foreach ($dir_contents as $file)
{
$file_type = strtolower(end(explode('.', $file)));
if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true)
{
$name = basename($file);
echo "<img src='img.php?name={$name}' />";
}
}
}
?>
img.php
========================
<?php
$name = $_GET['name'];
$mimes = array
(
'jpg' => 'image/jpg',
'jpeg' => 'image/jpg',
'gif' => 'image/gif',
'png' => 'image/png'
);
$ext = strtolower(end(explode('.', $name)));
$file = '/home/users/Pictures/'.$name;
header('content-type: '. $mimes[$ext]);
header('content-disposition: inline; filename="'.$name.'";');
readfile($file);
?>
回答by Rully Hendrawan
You have two ways to do that:
你有两种方法可以做到这一点：
METHOD 1. The secure way.
方法 1. 安全的方式。
Put the images on /www/htdocs/
将图像放在 /www/htdocs/
<?php
$www_root = 'http://localhost/images';
$dir = '/var/www/images';
$file_display = array('jpg', 'jpeg', 'png', 'gif');
if ( file_exists( $dir ) == false ) {
echo 'Directory \'', $dir, '\' not found!';
} else {
$dir_contents = scandir( $dir );
foreach ( $dir_contents as $file ) {
$file_type = strtolower( end( explode('.', $file ) ) );
if ( ($file !== '.') && ($file !== '..') && (in_array( $file_type, $file_display)) ) {
echo '<img src="', $www_root, '/', $file, '" alt="', $file, '"/>';
break;
}
}
}
?>
METHOD 2. Unsecure but more flexible.
方法 2. 不安全但更灵活。
Put the images on any directory (apache must have permission to read the file).
将图像放在任何目录中（apache 必须具有读取文件的权限）。
<?php
$dir = '/home/user/Pictures';
$file_display = array('jpg', 'jpeg', 'png', 'gif');
if ( file_exists( $dir ) == false ) {
echo 'Directory \'', $dir, '\' not found!';
} else {
$dir_contents = scandir( $dir );
foreach ( $dir_contents as $file ) {
$file_type = strtolower( end( explode('.', $file ) ) );
if ( ($file !== '.') && ($file !== '..') && (in_array( $file_type, $file_display)) ) {
echo '<img src="file_viewer.php?file=', base64_encode($dir . '/' . $file), '" alt="', $file, '"/>';
break;
}
}
}
?>
And create another script to read the image file.
并创建另一个脚本来读取图像文件。
<?php
$filename = base64_decode($_GET['file']);
// Check the folder location to avoid exploit
if (dirname($filename) == '/home/user/Pictures')
echo file_get_contents($filename);
?>
