I am currently creating an e-commerce website where I retrieve the vehicle details from the database and search for its image in a folder. I used the code below for searching the folder for the image, but it only show the name of the JPG file and not the image of it. Could anyone help to to determine how I could display the image of the vehicle
$imageName=$_POST['car'];
if ($handle = opendir('Images/$imageName')) {
$dir_array = array();
while (false !== ($file = readdir($handle))) {
if($file!="." && $file!=".."){
$dir_array[] = $file;
}
}
echo $dir_array[rand(0,count($dir_array)-1)];
closedir($handle);
}
You have to put html tags around the result. Try:
echo "<img src='PATH_TO_IMAGE/".$dir_array[rand(0,count($dir_array)-1)]."'>";
where PATH_TO_IMAGE is the relative/absolute url address to your images folder
Ex
echo "<img src='/images/".$dir_array[rand(0,count($dir_array)-1)]."'>";
You need to print the path to the image inside an img HTML tag, otherwise it is just plain text.
echo '<img src="Images/' . $imageName . $dir_array[rand(0,count($dir_array)-1)] . '" alt="" />';
I hope you're doing some validation/sanitisation on $_POST['car'] somewhere?
Hmm... If you have the name of the JPG, use <img src="NAME_OF_THE_IMAGE.JPG" /> ...
Related
I am trying to create a gallery where images are automatically pulled from a images directory.
When I foreach() each image it returns however it pulls the entire root path as the image src
/home/dev/public_html/assets/images/
this causes the images not to show and shows dead linked images on screen. I need the relative path like this:
/assets/images/imagename.jpg
How would I set a base path/URL?
$dirname = "/home/dev/public_html/assets/images/";
$images = glob($dirname."*.jpg");
foreach($images as $image) {
echo '<img src="'.$image.'" /><br />';
}
I tried setting $dirname to this
$dirname = "/assets/images/";
But I get no result and nothing pulls through at all.
The file i am creating this in, is located outside the public_html folder, i believe this could be casing the issue but the file cannot be relocated.
I would edit the loop in order to remove unnecessary string from path:
foreach($images as $image) {
$src = str_replace('/home/dev/public_html' ,'', $image) ;
echo '<img src="'.$src.'" /><br />';
}
You could may make use of scandir() http://php.net/manual/en/function.scandir.php
Somethink like:
Edit: Changed the code a bit, so scandir only take files, otherwise the first two placed in the array are folders for the parent folders.
$dir = "assets/images/";
$images = scandir($dir);
foreach($images as $image) {
if($image != '.' && $image != '..') {
echo ' <img src="assets/images/'.$image.'"><br>';
}
}
Hope this helped. If not, let me know, so I can change my answer. But I think it should be fine, as far as I understood what you're trying.
I have a form that uploads data to the DB and this includes the path to the directory where images are uploaded. Everything works, except for the fact that the image won't display.
Viewing the source in my browser tells me that the image is found but I keep getting the broken image icon.
Here's my code:
$dir = "../uploaded_images/";
$filePath = $row['images_path'];
$fileArray = explode("*", $filePath);
if (count($fileArray) > 0) {
$image = $fileArray[0];
echo "<img src='$image' width='300px'>";
}
In the form you can upload multiple files. In the DB, the files get a random prefix then file name, like 3456456745654_imageName.jpg.
If multiple files are uploaded, they are split with an asterisk (*), which is why I'm exploding.
Then, to print only one image, I'm checking for the number of images relevant to a specific record then displaying only the first one.
PS. This code works for displaying all the images relevant to a selected image:
$dir = "uploaded_images/";
$filePath = $row['images_path'];
$fileArray = explode("*", $filePath);
foreach ($fileArray as $file) {
if (file_exists($dir . $file)) {
echo "<img class='images' src='$dir/$file' width='300px;'>";
}
}
But that's for a different page that displays a selected vehicle's information, including all images.
I need to show only one image per vehicle on the landing page that lists all vehicles.
Had to use the directory:
$dir = "../uploaded_images/";
$filePath = $row['images_path'];
$fileArray = explode("*", $filePath);
if (count($fileArray) > 0) {
$image = $fileArray[0];
if (file_exists($dir . $image)) {
echo "<img class='images' src='$dir/$image' width='300px;'>";
}
}
I have a php script which grabs all of the images in a set directory. The script works without fault. My issue is that I want it to only get none retina images. All of the retina images are named in the format 'image1#2x.jpg' whereas all of the regular images are named in the format 'image1.jpg'. With this in mind I want my script to discount any image that contains '#2x' in it's name.
My script is:
<?php
$directory = 'images/';
$dh = opendir($directory);
while (false !== ($filename = readdir($dh)))
{
$files[] = $filename;
};
$images = preg_grep ('/(\.gif|\.GIF|\.jpg|\.jpeg|\.JPG|\.JPEG|\.png|\.PNG)$/i', $files);
foreach ($images as $image)
{
echo '<img src="'.$directory.$image.'" alt="">';
}
?>
The problem is I don't have the vocabulary to know what to search for in Google so if anyone could point me in the right direction I would appreciate it.
Thanks
I would use stripos put this conditional in your foreach loop
if(stripos($image,'#2x') === false){
echo '<img src="'.$directory.$image.'" alt="">';
}
You can make your script a little shorter by using:
$directory = 'images/';
$allImages = glob($directory.'*.{jpg,JPG,jpeg,JPEG,gif,GIF,png,PNG}', GLOB_BRACE);
$nonRetinaImages = preg_grep('/#2x/i', $allImages, PREG_GREP_INVERT);
foreach ($nonRetinaImages as $image)
{
echo '<img src="'.$image.'" alt="">';
}
I need to build a dynamic image gallery plugin for Joomla! that will go into a specific folder and pull out all images from the folder and show first of them as a large preview image and the rest in a list. Afterwards I will need to make the preview image open in a lightbox if clicked and in the lightbox I need to have the small thumbnails of listed images as well.
But know I need just php to go to the folder and pull out all the images from the folder specified. I have googled a bit and found some solution but this does not work for some reason I don't understand. Could someone tell me please, what is wrong with the code?
Thanks!
<div id="images">
<?php
$images_dir = 'images/';
$scan = scandir($images_dir);
echo '<img src="' . $images_dir . $scan[2] . '"alt="image" />';
?>
<ul id="smallimages">
<?php
for($i=0; $i<count($scan); $i++){
$ext = substr($scan[$i], strpos($scan[$i], '.'), strlen($scan[$i]-1));
$filetypes = array('.jpg', '.JPEG', '.jpeg');
if(in_array($ext, $filetypes)){
echo '<li><img src="' . $images_dir . $scan[$i] . '" alt="' . $scan[$i] . '"></li>';} }?></ul>
</div>
I think it's because you haven't defined the path correctly. Try using the following:
$images_dir = JUri::root() . 'plugins/content/plg_new/images';
JUri::root() is the root of your Joomla site so change the path from there on accordingly to where ever your images are located
Hope this helps
Does this work for you?
<?php
$image_directory="hrevert/images/";
$pictures = glob("$image_directory*.{gif,jpg,png}", GLOB_BRACE);
//displays the first image
$first_img=reset($pictures);
echo "<img src=\"".$first_img."\" width='20px' />";
//loops through other images and prints them
echo "<ul>";
foreach($pictures as $picture) {
echo "<a href='$picture'><img src=\"".$picture."\" width='20px' /></a>";
}
echo "</ul>"
?>
How can I display images from directory and get a corresponding description with each image, give the description exists.
in Directory //
01.png
01.txt
02.png
03.png
03.txt
etc.
to display as //
<img src="01.png"><br>This is the description from the text file named 01.txt
<img src="02.png"><br>
<img src="03.png"><br>This is the description from the text file named 03.txt
I've been searching and searching, but can't find anything, so if someone could point me in the right direction it would be greatly appreciated. Also this is a very useful thing to be able to do for people wanting to create very simple galleries or lists of images and names.
Thanks in advance!
This is what you're looking for, as the description must be dynamically captured from a corresponding .txt file:
$dir = './';
$files = glob( $dir . '*.png');
foreach( $files as $file) {
$filename = pathinfo( $file, PATHINFO_FILENAME) . '.txt';
$description = file_exists( $filename) ? file_get_contents( $filename) : '';
echo '<img src="' . $file . '"><br>' . $description;
}
What it does is grabs an array of *.png files using glob() from a given directory ($dir). Then, for each image, it gets the filename of the image (so 01.png would be 01), and appends .txt to get the name of the description file. Then, it loads the description file into the $description variable using file_get_contents() if the description file exists. It then outputs the desired HTML.
I assume you have the .php file in the same directory as the pictures and text files are.
You could use function glob() to read in all image-files as array from the directory, cut off the file extension (so '01.png' becomes '01') and append the file extension with string concatentation.
A working code example may look like this:
<?php
$path_to_directory = './';
$pics = glob($path_to_directory . '*.png');
foreach($pics as $pic)
{
$pic = basename($pic, '.png'); // remove file extension
echo '<img src=\"{$pic}.png\"><br>';
if(file_exists($pic . '.txt'))
{
echo file_get_contents("{$pic}.txt");
}
}
?>
So definitely have a look on these functions:
http://www.php.net/glob
http://www.php.net/basename
http://www.php.net/file_get_contents
Happy coding.
Your question is a bit confusing.
make an array with all information.
$pics = array('img' => '01.png', 'text' => 'This is the description');
foreach($pics as $pic) {
echo '<img src="'.$pic['name'].'" alt="">' . $pic['text'];
}
So you have to put your information in an array or a database otherwise you cannot map the desciption to your image.
When you want to read dynamicly the folder its a bit difficult.
You can look at readdir or glob then you can read all images get the name and load the textfile with file_get_contents but i think its not a really performant way.
Code Modified from here : http://php.net/manual/en/function.readdir.php
//path to directory to scan
$directory = "../images/team/harry/";
//get all image files with a .jpg extension.
$images = glob($directory . "*.jpg");
//print each file name
foreach($images as $image)
{
print "<img src=\"$image\"><br>This is the description from the text file named $image";
}
ok, so this won't print the contents of the text file, but I'm sure you can further modify the code above to figure that out
YEP
Updated version of ZnArKs code, as he missed that you wanted the content of the files
//path to directory to scan
$directory = "../images/team/harry/";
//get all image files with a .jpg extension.
$images = glob($directory . "*.png");
//print each file name
foreach($images as $image)
{
$textfile = substr($image, 0, -3) . "txt";
echo "<img src='{$image}'><br/>";
if(file_exists($textfile))
{
echo file_get_contents($textfile) . "<br/>";
}
}