[SOLVED] Display all items in a directory

adam291086 · December 11, 2007

I have a folder full of pictures. Is it possible to display all the folders content as thumbnails in one go. Without a database? If so what should i google. I have tried various search and really got nothing.

tapos · December 11, 2007

first read the directory for files then make thumbnail view of the image and show the thumbnail image.

Thanks

--

Tapos Pal

adam291086 · December 11, 2007

how do i read a directory?

I suppose once i read the directory i will need a while loop to say while there is informatio echo img scr=blah blah blah.

tapos · December 11, 2007

u can use readdir function.

see http://www.php.net/manual/en/function.readdir.php

--

Tapos Pal

~n[EO]n~ · December 11, 2007

This is the code which i am working, it reads the file and displays it, it is still not complete. Maybe you can modify it and use... (and later share with me )

<?php
$dir = "homio/";
$i=0;
if ($handle = opendir($dir)) 
{
while (false !== ($file = readdir($handle))) 
{
$all_files[$i] = $file;
$i++;
}
    closedir($handle);
}

foreach ($all_files as $value) 
{
$path = $dir.$value;
print_r($value);
echo '<img src="'.$path.'" width="250" height="150"><br />';

}
?>

delphi123 · December 11, 2007

Here's the code I've developed to show it in a listbox.

The only thing I've got to fix is stopping it showing the full directory location and just show the file titles in the list box - not sure if anyone can help me with this?

???

<select>
<?php
$ImageDirectory = "../Content/Reviews/images";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
echo "<option>".$image."</option>";
} 
?>
</select>

adam291086 · December 11, 2007

neon i have used you script. It works well. The only problem is that it echo out two other images with the name . and .. When i look in the directory there is no such file. Why is this happening?

kenrbnsn · December 11, 2007

The files with the names "." and ".." are special directory files in the UNIX world, they point to the current directory and the parent directory respectively.

Ken

~n[EO]n~ · December 11, 2007

I could not remove that DOTS... I switched to Delphi's code, i found it more efficient... does the same thing, just modify TR and TD's to get the look as you want...

<?php
$images = "images";
foreach (glob("$images/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
echo "<table><tr>";
echo '<td><img src="'.$image.'" width="250" height="150"></td>';
echo "</tr></table>";
} 
?>

adam291086 · December 11, 2007

ok i have modified the code

<?php
error_reporting(E_ALL);
$ImageDirectory = "../adam/cycling/admin/upload/pictures/thumbnails/";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
echo "<html>";
echo "<body>";
echo "<table><tr>";
echo '<td><img src="'.$image.'" width="250" height="150"></td>';
echo "</tr></table>";
echo "</html>";
echo "</body>";
} 
?>

Nothing is being echoed and i get no errors. Any ideas?

cooldude832 · December 11, 2007

verify you are getting data first


<?php
error_reporting(E_ALL);
$ImageDirectory = "../adam/cycling/admin/upload/pictures/thumbnails/";
$files = glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE);
print_r($files);
?>

If taht returns an empty array you know your $ImagDirectory isn't right, try $_SERVER['DOCUMENTS_ROOT']."LOcATION"

adam291086 · December 11, 2007

ok i have changed to

<?php
error_reporting(E_ALL);
$ImageDirectory = dirname(__FILE__)."/pictures/thumbnail/";
$files = glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE);
print_r($files);


?>

i know the direct is correct as that is where my upload script uploads the images to. All that's printed is array()

cooldude832 · December 11, 2007

lets just try

<?php
$ImageDirectory = dirname(__FILE__)."/pictures/thumbnail/";
$files = glob($ImageDirectory."*");
print_r($files);
?>

adam291086 · December 11, 2007

that still only displays ''Aray ()''

cooldude832 · December 11, 2007

try the last thing I posted with a folder path you know is good.

this shoudl work and then just get contex from it

<?php
$dir = $_SERVER['DOCUMENT_ROOT'];
$data = glob($dir."*");
print_r($data)

mr_mind · December 11, 2007

This function will do the following,

Select only the filetypes you specify.

Sort the files according to filetype

Show the image as a thumbnail

And it is easy to use

<?php
function show_images($directory) {
	$images_types = array('gif','jpg','png');
	foreach($images_types as images_type) {
		$images_files = glob($directory . "*." . $images_type);
		if($images_files != '') {
			foreach($images_files as $images_file) {
				print '<img src=' . $directory . $images_file . ' width=100px height=100px /><br />';
			}
		}
	}
}
$images = show_images($_SERVER['DOCUMENT_ROOT']);
if($images != '') {
	print $images;
}
?>

adam291086 · December 11, 2007

ok i am using mr_mind code. The only thing echoed is

ï»¿

. I am confuse with the code. Any help appreciated.

<?php
function show_images($directory) {
	$images_types = array('gif','jpg','png');
	foreach($images_types as $images_type) {
		$images_files = glob($directory . "*." . $images_type);
		if($images_files != '') {
			foreach($images_files as $images_file) {
				print '<img src=' . $directory . $images_file . ' width=100px height=100px /><br />';
			}
		}
	}
}
$images = show_images($_SERVER['DOCUMENT_ROOT']."/pictures/thumbnails/");
if($images != '') {
	print $images;
}
?>

adam291086 · December 12, 2007

ok my script has stopped working.

<?php
$dir = dirname(__FILE__)."/pictures/thumbnail";
foreach (glob("$dir/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{

echo '<img src="'.$image.'" width="250" height="150">';
echo '<br />';

} 
?>

When i run the script it show 4 images boxes. Which is correct. The only problem is that the image itself is not being echoed. It just shows a red cross.

http://adamplowman.co.uk/cycling/admin/upload/view_pics.php view whats happening here

~n[EO]n~ · December 12, 2007

Your path is wrong i think ... full path is like this

http://adamplowman.co.uk/cycling/admin/upload/pictures/thumbnail/BOTTOM-LINE.jpg (seen in the browser address bar)

and

http://adamplowman.co.uk/homepages/12/d214897219/htdocs/adam/cycling/admin/upload/pictures/thumbnail/BOTTOM-LINE.jpg

this comes while checking image properties

adam291086 · December 12, 2007

the problem is when i try and manually set the location it doesn't work at all. When is use dirname(_file_) it works.

adam291086 · December 12, 2007

finally got it working. The code below shows how. How would i go about echoing the file name as well?

<?php
$ImageDirectory = "../upload/pictures/thumbnail/";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
echo "<table><tr>";
echo '<td><img src="'.$image.'"></td>';
echo '<td>$ImageDirectory</td>';
echo "</tr></table>";
} 
?>

cooldude832 · December 12, 2007

Try this (the glob returns a complete path so you need to get just the image name)

<?php
$ImageDirectory = "../upload/pictures/thumbnail/";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
echo "<table><tr>";
echo '<td><img src="'.$image.'"></td>';
echo '<td>$ImageDirectory</td>';
$temp = explode("/",$image);
echo "<td>".$temp[count($count)-1]."</td>";
echo "</tr></table>";
}

that should get you just the last item in the array exploded by the / removing all the folder contexes

adam291086 · December 12, 2007

i did it the less complicated way, which i understand

<?php
$ImageDirectory = "../upload/pictures/thumbnail";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
//remove .. for url
$dir = str_replace("..", "", "$ImageDirectory");

//get name for url
$name = str_replace("$ImageDirectory", "", "$image");

//get name on it's own
$fullname = str_replace("/", "", $name);


echo '<br />';
echo $fullname;
echo '<br />';
echo '<img src="'.$image.'">';
echo '<br />';
echo "www.adamplowman.co.uk".$dir.$name;
echo '<br />';




}


?>

cooldude832 · December 12, 2007

yeah if you want that full path a better idea your way, that or what makes more sense (a universal one is)

<?php
$ImageDirectory = "../upload/pictures/thumbnail";
foreach (glob("$ImageDirectory/{*.gif,*.jpg,*.png,*.GIF,*.JPG,*.PNG}", GLOB_BRACE) as $image)
{
$path = str_replace($_SERVER['DOCUMENT_ROOT'],"",$image);
echo '<br />';
echo $fullname;
echo '<br />';
echo '<img src="'.$image.'">';
echo '<br />';
echo $path;
echo '<br />';




}


?>

That will work on any server instead of only yours, and its simpler

adam291086 · December 12, 2007

ok, i can see what your doing. One last question. How would i go about deleting one pic from the directory?

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[SOLVED] Display all items in a directory

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