This is the code I'm using to get and display images from a folder on my server:
$root = realpath($_SERVER["DOCUMENT_ROOT"]);
$dirname = "$root/Folder/".$row['city']."/";
$images = glob($dirname."".$row['id']."#*.jpg");
foreach($images as $image) {
echo "<img src=\"".$image."\">";
}
The "normal" path of a picture is /Folder/Berlin/1#1.jpg. In the rendered HTML source code I can see PHP makes this link: /var/www/user_name/html/Folder/Berlin/1#1.jpg
But unfortunately the image doesn't get loaded.
What am I doing wrong?
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The code I have should output a jpg from a list of files in a directory however it is not. I have trawled this site and tried different methods but not helped. I am a relative beginner at php so looking for any help at all.
I have tried using img src in the php code but I am trying to get the image to display within a Wordpress post so I cannot echo the img src within the script. I have tried file_get_contents and read file as well but it may be my lack of knowledge holding me back.
<?php
$imagepath = htmlspecialchars($_GET["image"]);
$imagenum = htmlspecialchars($_GET["num"]);
define('LOCALHOST', 'localhost' === $_SERVER['SERVER_NAME'] );
If(LOCALHOST){
define('PATH_IMAGES', 'this_path');
}else{
define('PATH_IMAGES', '../../../Images/');
}
$arrnum = $GLOBALS[imagenum] - 1;
$dirname = PATH_IMAGES . $GLOBALS[imagepath]."/";
$images = scandir($dirname);
rsort($images);
$ignore = Array(".", "..");
foreach($images as $curimg){
if(!in_array($curimg, $ignore)) {
header('Content-type: image/jpeg');
file_get_contents('$dirname$images[$arrnum]');
}
}
?>
Have you tried readfile(...); should read and output the file. In your example you are not outputting the image data
http://php.net/manual/en/function.readfile.php
I need to pull all images from a URL directory (they are not displayed...just sitting in a folder on a server that I do not have access to) and display them within a Bootstrap Image gallery.
http://www.electrictoolbox.com/extract-images-web-page-php/
<?php
require_once('./simple_html_dom.php');
require_once('./url_to_absolute.php');
$url = 'http://www.bbc.co.uk';
$html = file_get_html($url);
foreach($html->find('img') as $element) {
echo url_to_absolute($url, $element->src), "\n";
}
?>
The URL for the folder where all the images are stored is:
http://masterplan.imgix.net/Slimming_Book/
Is it possible for php to scan this URL directory and pull the images to another website that is being hosted on another server?
Bit late but I figured I'd answer this. The below PHP code loads all ".png" images from the directory and then echos the image tag. You would replace the plain html tag for the equivalent bootstrap one.
dirname = "media/images/cats/";
$images = glob($dirname."*.png");
foreach($images as $image) {
echo '<img src="'.$image.'" /><br />';
}
I am trying to echo out all of the images in a folder directory with a couple of exceptions/ignores.
This is working ok apart from it also echoes out a blank photo for every photo it echoes out?
why is this happening can someone please show me where I'm going wrong thanks.
<?php
$dirname = "./data/photos/".$profile_id."/";
$images = scandir($dirname);
$ignore = Array("_cover.jpg", "_default.jpg");
foreach($images as $curimg){
if(!in_array($curimg, $ignore)) {
echo "<img src='./data/photos/".$profile_id."/$curimg'/ class=\"profile_photos\"><br>\n";
};
}
?>
You appear to have an extra slash after your image source:
echo "<img src='./data/photos/".$profile_id."/$curimg'/ class=\"profile_photos\"><br>\n";
//--------------------------------------------------------^ here
This may be interefering with how the browser parses the DOM and causing an extra image to appear.
Also, small suggestion, try using this line instead:
echo "<img src='".$dirname.$curimg."' class=\"profile_photos\"><br>\n";
You should ensure that $curimg is actually a jpg file:
if(!in_array($curimg, $ignore) && preg_match("/\.jpg$/i", $curimg)) {
scandir returns not just files but subdirectories, including . and ...
I have a script that scans a directory of thumbnails and echoes them to the page. It works nicely, but the thumbnails are not clickable, and i would really like this to be the case. echo "<img src='$thumbnail' class='resizesmall'>"; is the line where the thumbnails are echoed. I'm not sure how to write the path to the larger image inside the php without breaking it. Maybe this should be done inside the foreach statement? thanks for your help?
$dir = "../mysite/thumbnails/";
$dh = opendir($dir);
// echo "$dh";
$gallery = array();
while($filename = readdir($dh))
{
$filepath = $dir.$filename;
//pregmatch used to be ereg
if (is_file($filepath) and preg_match("/\.png/",$filename))
{
$gallery[] = $filepath;
}
}
sort($gallery);
foreach($gallery as $thumbnail)
{
echo "<img src='$thumbnail' class='resizesmall'>";
}
?>
</div>
<??>
The easiest way would be to setup a situation where your thumbs and your full size images were named the same. So you may have thumbs/image1.png and full/image1.png. Then instead of using $thumbnail use a variable $image, or something similar just so the code reads better. You'll also want to leave the $filepath out of the mix so that $image ends up as just the file name.
foreach($gallery as $image)
{
echo "<a href='full/$image'><img src='thumb/$image' class='resizesmall'></a>";
}
You may want to throw in some checks to make sure there is a matching image just to prevent errors or bad UX. However, the code above should work.
I am trying to generate some HTML code to list some images for a slide show.
I arrived at the following idea:
function galGetTopPhotos()
{
//path to directory to scan
$directory = SITE_ROOT_PATH."/gallery/best/";
//get all files
$images = glob($directory . "*.*");
//print each file name
$ret = "";
$ret .= '<div id="myslides">';
foreach($images as $image)
{
$ret .= '<img src="'.$image.'" />';
}
$ret .= '</div>';
return $ret;
}
The problem is that it only works when I use root path for $directory...if I use URL it will not work. And it causes the images to not load. Here is what this code generates:
<div id="myslides">
<img src="D:/xampp/htdocs/mrasti/gallery/best/1.jpg" />
<img src="D:/xampp/htdocs/mrasti/gallery/best/10.jpg" />
</div>
So the question is, how to get the list of files so it generates img sources in http://127.0.0.1/.... format?
What I mean if I use the code like this it returns no file!
$directory ="http://127.0.0.1/mrasti/gallery/best/";
This looks like a job for PHP function basename. This takes a file path and returns only the final element of the path - in this case the actual name of the jpeg image.
You could amend your code so that it looks something like this:
$urlPath = "http://127.0.0.1/mrasti/gallery/best/";
...
...
foreach($images as $image)
{
$relative_path = $urlPath.basename($image);
$ret .= '<img src="'.$relative_path.'" />';
}
The above takes the path and appends the filename "example.jpg" to your image directory url
glob does only work for local files and not on remote files. Have a look here:
http://php.net/manual/en/function.glob.php
For remote files have a look here:
http://www.php.net/manual/en/features.remote-files.php
But i do not think that you need remote files. It seems like you want to go through a local directory and display this images.
Try something like this
...
$ret .= '<img src="http://127.0.0.1/my/path/'.basename($image).'" />';
...
You need to have some functionality to translate the file path on disk to the correct URI so that your browser can understand it.
In your specific case as outlined and with the exact data given in your question, the following could work:
foreach($images as $image)
{
$src = '/mrasti/gallery/best/'.substr($image, strlen($directory));
$ret .= '<img src="'.$src.'" />';
}