PHP - Unable To Access Images Outside Public Folder

I have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens   http://www.jayg.co.u...oad_gallery.php   <form> <?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>   any ideas please?   thanks

I run a dev site locally on Windows and the real site on a hosting provider on LINUX. PHP5, XAMP, etc.

Locally my fopen works.
On the web server it throws an error: "Warning: fopen() [function.fopen]: Unable to access" filepath/name

The file exists on both servers; upper/lower case is correct; so are access rights.

I noticed the error only today; this was working for the last 7 months; the function serves a range of content-types; just tested XLS and it works.

Now I am stuck 

This line throws the error:
$handle = fopen($strPathFileName, 'rb');

$strPathFileName uses / only

Any pointers appreciated... thanks.

Hi all,

Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else?
Appreciate your help with this.
I have included both of the pages codes i am using.

Thanks
Tony

image2.php

<?php
include("common.php");
error_reporting(E_ALL);
$link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error());
mysql_select_db(db) or die(mysql_error());
$sql = "SELECT id FROM photos";
$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
?>
<table border="1"><tr><td>id</td><td>image</td></tr>
<?php
while($row=mysql_fetch_assoc($result)){
print '<tr><td>'.$row['id'].'</td><td>';
print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">';
}
echo '</td></tr></table>'
?>

image1.php

<?php
ob_start();
include("common.php");    
   mysql_connect(host,username,password) or die(mysql_error());
   mysql_select_db(db) or die(mysql_error()); 

$query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}";
$row = mysql_fetch_array($query);
$content = $row['image']; 
header('Content-type: image/jpg');
     echo $content;
 }
 ob_end_flush();
?>

Hi, I want to be able to click on the photo and go to the next one in a folder. I have this code already, I just am not quite sure how to finish it.

-George
Code: [Select]
<?php
$count = $_GET['count'];
$dir = "images";
$names = array();
$handle = opendir($dir);
while ($name = readdir($handle)){
if(is_dir("$dir/$name")) {
if($name != '.' && $name != '..') {
echo "directory: $name\n";
}
} elseif ($name != '.DS_Store' ) {

$names[] = $name;
}

}
closedir($handle);
$numberofitems =  count($names);
$numberofitems--;


if ($count <= $numberofitems){


echo "<p>";
echo "<img src='images/".$names[$count]."'>";

} else {echo "end";}



?>

I made an upload image system, the images are stored in a folder, while the image name is stored in database. When I try to execute the image name, it works successfully, but when I try to disply the image from the folder by using the image name in the database, I fail each time.

The folder where the images are being stored is named: saveimage

Here is the query:

Code: [Select]
<?php
mysql_connect('localhost', 'root', '') or die ('Connection Failed');
mysql_select_db('imagedatabase');
$images = mysql_query("SELECT * FROM img WHERE email='$lemail'");
while($row = mysql_fetch_array($images))
{
echo "<img src='saveimage/'".$row['img_description'];
 }
?>

how do i get the image to show as the code i have is echoing what i have been told is binary, not confirmed.  and not the images.

all images are .jpg

Code: [Select]
$allowed_types = array('png','jpg','jpeg','gif');
$imgdir = '/home/mysite/ftpfolder';
$imgFilesArray = scandir($imgdir);
foreach ($imgFilesArray as $imgkey => $imgvalue)
{
$imgInfo =  pathinfo($imgdir . '/' . $imgvalue);
$imgExtension = $imgInfo['extension'];
if(in_array($imgExtension, $allowed_types))
{
readfile($imgdir . '/' . $imgvalue);
//echo '<img src = "'. $imgdir  .'/'. $imgInfo['basename'].'" alt="" />';
}
}

images are FTP'd to the ftpfolder and can not be placed anywhere else but this folder due to security reasons.

Can someone tell me how I can remove or delete an image from a folder on a server using PHP? I tried this:


Code: [Select]
unlink("http://midwestcreativeconsulting.com/jhrevell/wp-content/themes/twentyten_3/upload/" . $location);


before my delete MySQL statement, but I keep getting this error:


Warning: unlink() [function.unlink]: http does not allow unlinking in /home/midwestc/public_html/jhrevell/wp-content/themes/twentyten_3/removejewelry.php on line 22

Can anyone help and tell me how I can make it work?

I'm trying to write code that will let me pull 10 out of 15 images out of a folder and display them on my site. The images are all different, and I don't want dupes to show. So far, I have the following code figured out:

---------------

$s = array ("image.jpg", "image2.jpg"); // as many images as you want
$n = rand(1,len($s)); // randomly pick a number between 1 and the
length of the array
echo "<img src='". $s[$n] .'">"; // create an image tag for the randomly selected imagine (value of the randomly defined key)
array_pop($s, $n); // This piece isn't right, it needs to EXTRACT and delete the $n array element.

// Next random image
$n = rand(1,len($s));
echo "<img src='". $s[$n] .'">";

---------------
Any ideas what array_pop should be to work properly?

Thank you for the help!

The problem:

I'm trying to create a page which outputs images from a folder which I have been
able to do, but the problem I'm having is not being able to get the page to display
the most recent image according to file modification date at the top.

The first set of code below outputs the image timestamps in descending order, from newest to oldest which is
what I want, but as soon as I change/add a couple lines of code (Shown in the second lot of code) to get the image
file name along with the timestamp, the echoed list (timestamp and file names) gets muddled up in a random order.

In short;
As soon as the file names are retrieved with the timestamp, the list
goes from being organised descendingly, to not.

Show timestamp only code (1st lot of code):

<?php

//Open images directory
$ignore = array("..",".");

$dir = opendir("images1");

$images = array();

$sortedimages = array();

//List files in images directory
while (($file = readdir($dir)) !== false) 

if (!in_array($file, $ignore))

$images[] = $file;

foreach ($images as $image)

{

$filetime = filemtime("images1/$image");

$sortedimages[] = $filetime;

}

rsort($sortedimages);

foreach ($sortedimages as $sorted)

{

//foreach ($sorted as $key => $value)

//{

echo "$sorted<br/>";

}

//}

closedir($dir);

?>

The show timestamp and file name code (2nd lot of code):

<?php

//Open images directory
$ignore = array("..",".");

$dir = opendir("images1");

$images = array();

$sortedimages = array();

//List files in images directory
while (($file = readdir($dir)) !== false) 

if (!in_array($file, $ignore))

$images[] = $file;

foreach ($images as $image)

{

$filetime = filemtime("images1/$image");

$sortedimages[] = array($filetime => $image);

}

rsort($sortedimages);

foreach ($sortedimages as $sorted)

{

foreach ($sorted as $key => $value)

{

echo "$key and $value<br/>";

}

}

closedir($dir);

?>

The changes made in the 2nd script from the 1st:

//Changed:
$sortedimages[] = $filetime; ---> $sortedimages[] = array($filetime => $image);

//Included the previously commented out:
foreach ($sorted as $key => $value)
{

}

//Changed:
echo "$sorted<br/>"; ---> echo "$key and $value<br/>";

Thanks for any help!

Im using this code to call all the images in a folder:

Code: [Select]
$handle = opendir(dirname(realpath(__FILE__)).'/images/');
while($file = readdir($handle)){
if($file !== '.' && $file !== '..'){
echo '<img src="admin/img/uploads/'.$file.'" border="0" />';
}
}
My html says the images are present but they aren't visable on screen:

Code: [Select]
<div id="contentbody">
<img src="admin/img/uploads/send-button-sprite copy.png" border="0" />
<img src="admin/img/uploads/test" border="0" />
<img src="admin/img/uploads/counter.jpg" border="0" />
<img src="admin/img/uploads/send-button-sprite.png" border="0" />
</div>
Any help is much appreciated!

Hi    I echo out images from a folder. The problem is that I have a html file in the same folder but I don't want to echo that out. How can I write my code to get rid of the html in the output?   Here my code:

<?php 
$filetype = '*.html';
$dirname = substr('$filetype');
$i=0;
if (isset($_POST['submit2']))
{
  //get the dir from POST
  $selected_dir = $_POST['myDirs'];
  //now get the files from within the selected dir and echo them to the screen
  foreach(glob($selected_dir . DIRECTORY_SEPARATOR . '*') as $dirname)
  {
    echo substr($dirname, 0, -4);
    echo '<img src="'.$dirname.'" />';
    echo "<label><div class=\"radiobt\"><input type='radio' name='radio1' value='$i'/></div></label>";
  }
}
?>

P.S. when I echo out : substr($dirname, 0, -4); I want to get ride of the html filename there too. How can I do so something like this?

Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this?

thanks

Hello once again, i got this code that takes all images from a folder and displays them close to each other.
Heres the code:

<?php
$dir = 'uploads/thumbs/watermarkedthumbs/';
$file_display = array ('jpg', 'jpeg', 'png', 'gif');

if (file_exists($dir) == false) {
echo 'tt';
} 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, '" />';
}

}
}
?>

How do i add margins in between them? i wanna make it smth like 'margin-left:10px, margin-top-10px'.
Also, how do i add an 'overflow'? i mean, my images currently are displayed in a div, and i wanna make it so that if the folder has lets say 9 images, it would only display 6 of them, and the rest would be displayed in the same div after i click a button or smth (like '1' for the first div and '2' would appear if theres an overflow, and when i click '2' the rest of the images would appear).

To make these adjustments do i need to make a different php file or do i change something in this code?
thanks in advance.