I'm using this script to display all the images in a folder, but I can't figure out how to get each image's file name to display underneath it. Any suggestions?
<?php
$dirname = "images";
$images = scandir($dirname);
$ignore = Array(".", "..", "otherfiletoignore");
foreach($images as $curimg){
if (!in_array($curimg, $ignore)) {
echo "<img src='images/$curimg' /><br />\n";
}
}
?>
echo "<img src='images/$curimg' /><br />$curimg<br />\n";
I think nickf's suggestion is the simplest thing you can do achieve what you want without any css or complex structure..
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'm using PHP to put all the images in a given folder into a slideshow.
This works fine in the following code:
<?php
//get all image files with a .jpg extension.
$images = glob("" . $directory . "*.jpg");
$imgs = array();
// create array
foreach($images as $image){ $imgs[] = "$image"; }
/****** Display Images ******/
foreach ($imgs as $img) {
//I would like to get the image title here, to put in the echo below
echo "<div><img src='$img' border='0' width=\"100%\" height=\"100%\"/ /></div>";
}
?>
This all goes quite easy, but since I would now also like to add the Titles of the pictures as captions, I need to extract/get this information from the image.
I'm thinking I can get it with something along the lines of the function exif_read_data , but I'm not quite sure how to get the title and not all the meta data . . .
With a little help from the smart people using stackoverflow, this is the final and functional result, as seen in my answer below as well and again, made with bits and pieces from several answers.
<?php
/*
This script will print the image files in a given folder, along with their image description (title in windows file explorer).
*/
// Set the directory where the files reside
$directory = $GLOBALS['directory'];
//get all image files with a .jpg extension.
$imagenpath = glob("" . $directory . "*.jpg");
$imgs = array();
// create array
foreach($imagenpath as $image){ $imgs[] = "$image"; }
// Print each image file with the ImageDescription (the title/caption) in the ALT field
foreach ($imgs as $img) {
$exif_data = exif_read_data($img, 'IFD0');
$exif_description = "";
if (!empty($exif_data['ImageDescription']))
$exif_description = $exif_data['ImageDescription'];
echo "<div><img src='$img' alt='$exif_description' border='0' width=\"100%\" height=\"100%\"/ /></div>";
}
?>
thats everything that can be done, because most pictures are not supported with this function, though, you are doing it wrong because the error that you are getting is because the file is not found, the error you get when the file is found and is not supported is this:
Warning: exif_read_data(file.png): File not supported in
/path/to/file/file.php on line x
and its because you single quoted the variable in
$exif_data = exif_read_data($img, 'IFD0');
the code could be less characters, this is my solution:
<?php
/*
This script will print the image files in a given folder, along with their image description (title in windows file explorer).
*/
// Set the directory where the files reside
$directory = $GLOBALS['directory'];
//get all image files with a .jpg extension.
$imgs = glob($directory . "*.jpg");
// Print each image file with the ImageDescription (the title/caption) in the ALT field
foreach ($imgs as $img) {
$exif_data = exif_read_data($img, 'IFD0');
if (!empty($exif_data['ImageDescription']))
$exif_description = $exif_data['ImageDescription'];
echo "<div><img src=\"$img\" alt=\"$exif_description\" border=\"0\" width=\"100%\" height=\"100%\"/ /></div>";
}
?>
Try this :-
foreach (glob("*.jpg") as $filename) {
$name = basename($filename, '.jpg');
echo "<div><img src='$filename' title='$name' alt='$name' border='0' width=\"100%\" height=\"100%\"/ /></div>";
}
To solve the problem, I basically used the Windows file browser, selected the image file and entered the text in the document comments section.
Then I printed the $exif_data as suggested above by #RamRaider and found out this text ends up in the ImageDescription of the file.
Once I nailed down the errors in my script, along with help from #Abdallah Samman, there wasn't much to it anymore!
Thank you everyone...
The following works beautifully:
<?php
/*
This script will print the image files in a given folder, along with their image description.
*/
// Set the directory where the files reside
$directory = $GLOBALS['directory'];
//get all image files with a .jpg extension.
$imagenpath = glob("" . $directory . "*.jpg");
$imgs = array();
// create array
foreach($imagenpath as $image){ $imgs[] = "$image"; }
// Print each image file with the ImageDescription (the caption) in the ALT field
foreach ($imgs as $img) {
$exif_data = exif_read_data($img, 'IFD0');
$exif_description = "";
if (!empty($exif_data['ImageDescription']))
$exif_description = $exif_data['ImageDescription'];
echo "<div><img src='$img' alt='$exif_description' border='0' width=\"100%\" height=\"100%\"/ /></div>";
}
?>
I am using the following code from Timthumb to built an thumb gallery on my page from a directory.
However, I would like to have my image sort so that the latest image is on top.
I've try to implement some codes I found on this website, but with no luck.
Can anyone help? You can ignore all the stuff with regex.
<?php
$images = scandir($dirname);
$ignore = Array(".", "..", ".DS_Store", "index.php");
foreach($images as $curimg){
if(!in_array($curimg, $ignore)) {
echo "<li><img src=\"img.php?src=$dirname$curimg&w=500&zc=1&q=50\" alt='' style=\"width:100%;\" /></li>\n ";
}
}
?>
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/>";
}
}
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" /> ...