display the content of a folder - php

Question

I have the following PHP code to display the content of a directory in my website:
<?php
$conn = ftp_connect("host") or die("Could not connect");
ftp_login($conn,"username","password");
if ($_GET['dir'] != null) {
ftp_chdir($conn, "logs/{$_GET['dir']}");
}
else
{
ftp_chdir($conn, "logs");
}
$files = ftp_nlist($conn,"");
foreach($files as $value) {
echo "{$value}";
}
ftp_close($conn);
?>
This is my webpage
When I click on the subdirectory1 or subdirectory2, I get the content of it (some images). Then when I click on one of the images, I get the content of the root directory of my website.
How can I display only the image when a visitor clicks on it? Note that I don't want to download it or anything - I just want to display it in the browser when a visitor clicks on it.

Answer 1

You need to establish which of the returned items are files and which are directories. For that you are better of using ftp_rawlist as it allows to extract more data. Then create links for each case and so you can process them appropriately. Here's a possible implementation:
$ftpHost = 'hostname';
$ftpUser = 'username';
$ftpPass = 'password';
$startDir = 'logs';
if (isset($_GET['file'])) {
// Get file contents (can also be fetched with cURL)
$contents = file_get_contents("ftp://$ftpUser:$ftpPass#$ftpHost/$startDir/" . urldecode($_GET['file']));
// Get mime type (requires PHP 5.3+)
$finfo = new finfo(FILEINFO_MIME);
$mimeType = $finfo->buffer($contents);
// Set content type header and output file
header("Content-type: $mimeType");
echo $contents;
}
else {
$dir = (isset($_GET['dir'])) ? $_GET['dir'] : '';
$conn = ftp_connect($ftpHost) or die("Could not connect");
ftp_login($conn, $ftpUser, $ftpPass);
// change dir to avoid ftp_rawlist bug for directory names with spaces in them
ftp_chdir($conn, "$startDir/$dir");
// fetch the raw list
$list = ftp_rawlist($conn, '');
foreach ($list as $item) {
if(!empty($item)) {
// Split raw result into pieces
$pieces = preg_split("/[\s]+/", $item, 9);
// Get item name
$name = $pieces[8];
// Skip parent and current dots
if ($name == '.' || $name == '..')
continue;
// Is directory
if ($pieces[0]{0} == 'd') {
echo "<a href='?dir={$dir}/{$name}'><strong>{$name}</strong></a><br />";
}
// Is file
else {
echo "<a href='?file={$dir}/{$name}'>{$name}</a><br />";
}
}
}
ftp_close($conn);
}

Answer 2

You'll need to add a function which checks which type of file we're handling.
If it's a directory show the regular link (the one you're using now) and if it's an image
show a link with the path of the image.
Because $value contains the name of the file you can use end(explode('.',$value)); to find the ext. of the file (php , jpg , gif , png).
Use another condition with the piece of information and you can identify if it's a picture or not.
In order to build the path of the image you'll need to use the $_GET['dir'] variable's value.
For instance:
<a href='<?=$_GET['dir']?>/Flower.gif'>Flower.gif</a>
I hope you got the idea.