I want a web page to display several random images on load. I've thought of several solutions but I would really like to be able to dump images into a folder without renaming them, and the web page will choose from those images in the folder and display them
I can imagine a PHP solution that will display random images from a folder, but it will need to look for certain names.
So my next step was to have an SQL database where every image got a key, and 10 keys would be chosen random by a query - the images associated with them would then be passed into an array that the document will load the elements of.
But now I guess I need to know how to automatically populate an SQL database by having it read a folder?
Insight appreciated, if I don't have to reinvent the wheel the better
Glob works like on the filesystem, e.g. supports wildcards
$files = glob('/path/to/files/*.jpg');
$yourRandomFile = array_rand($files);
This will return a random JPG file, based on it's extension.
This should do it :
function your_dir ($directory)
{
$results = array();
$handler = opendir($directory);
while ($file = readdir($handler)) {
if ($file != "." && $file != "..") {
$results[] = $file;
//or sql query for each file
$sql = mysql_query("SELECT * FROM your_table WHERE your_file = '".$file."'");
if(mysql_num_rows !== 0){
//your query...
}
//
}
}
closedir($handler);
return $results;
}
$directory = '/path/to/your/directory';
your_dir($directory);
Would probably be better to select existing files from db first put them into an array and exclude them rather than checking for each one.
You can do as below.
1. read complete dir from where you want the page to load the image
2. store the image names in array
3. now generate a number using random function from 1 to size of the array
4. consider the generated number as key for the array and use that image name to put the image.
Related
I have a directory full of images (40,000 +) that I need sorted. I have designed a script to sort them into knew proper directories, however, I am having issues with the file name.
The images urls with the id they belong to are stored in a database, and I am using the database in conjunction with the script to sort the images.
My Problem:
The image url's in the database are shortened. An example of such corresponding images are like this:
dsc_0107-367.jpg
dsc_0107-367-5478-2354-0014.jpg
The first part of the filenames are the same, but the actual file contains more info. I'd like a way to move the file from the database with the known part of the file name.
I have a basic code:
<?php
$sfiles = mysqli_query($dbconn, "SELECT * FROM files WHERE gal_id = '$_GET[id']");
while($file = mysqli_fetch_assoc($sfiles)){
$folder = $file['gal_id'];
$fileToMove = $file['filename'];
$origDir = "mypath/to/dir";
$newDir = "mypath/to/new/dir/$file['gal_id']";
mkdir "$newDir";
mv "$fileToMove" "$newDir";
}
Im just confused on how to select the file based on the small part from the database.
NOTE: It's not as simple as changing the number of chars in the db, because the db was given to me from an external site thats been deleted. So this is all the data I have.
PHP can open files using the function glob() . Glob searches your server, or specified directory, for any files containing a "match" to a pattern you specify.
Using glob() like this will pull your images from a partial name.
Run this query separate from the second:
$update = mysqli($dbconn, "UPDATE files
SET filename = REPLACE(filename, '.info', ''));
filename should be the column in your database that contains the list of images. The reason we are removing the .jpg from the db columns is if your names are partial, the .jpg may not match with the given name in your directory. With it removed, we can search solely for the pattern of the name.
Build the query to select and move the folders:
$sfiles = mysqli_query($dbconn, "SELECT * FROM files");
while($file = mysqli_fetch_assoc($sfiles)){
$fileToMove = $file['filename'];
// because glob outputs the result set into an array,
// we will use foreach to run each result from the array individually.
foreach(glob("$fileToMove*") as filename){
echo "$filename <br>";
// I'm echoing this out to see that the results are being run
// one line at a time and to confirm the photo's are
// matching the pattern.
$folder = $file['gal_id'];
// pulling the id from the db of the gallery the photo belongs to.
// This will specify which folder to move the pic to.
// Replace gal_id with the name of your column.
$newDir = $_SERVER['DOCUMENT_ROOT']."/admin/wysiwyg/kcfinder/upload/images/gallery/old/".$folder;
copy($filename,$newDir."/".$filename);
// I would recommend copy rather than move.
// This will leave the original photo in its place.
// This measure is to ensure the photo made it to the new directory so you don't lose it.
// You could go back and delete the photos after if you'd prefer.
}
}
Your MySQL query is ripe for SQL Injection, and your GET statement needs to be sanitized, if I went to your page with something similar to :
pagename.php?id=' DROP TABLE; #--
this is going to end extremely badly for you.
So;
OVerall it's much better to use Prepared Statements. THere's LOTS and LOTS of data about how to use them all over SO and the wider internet. What I show below is only a stopgap measure.
$id = (int)$_GET['id'] //This forces the id value to be numeric.
$sfiles = mysqli_query($dbconn, "SELECT * FROM files WHERE gal_id = ".$id);
Also keep note of closing your ' and " quotes as your original doesn't close the array key wrapper quotes.
I never used mysqli_fetch_assoc and always used mysqli_fetch_array so will use that as it fits the same syntax :
while($file = mysqli_fetch_array($sfiles)){
$folder = $id //same thing.
$fileToMove = $file['filename'];
$origDir = "mypath/to/dir/".$fileToMove;
//This directory shold always start with Server['DOCUMENT_ROOT'].
//Please read the manual for it.
$newDir = $_SERVER['DOCUMENT_ROOT']."/mypath/to/new/dir/".$folder;
if(!is_dir($newDir)){
mkdir $newDir;
}
// Now the magic happens, copies the file to the new directory.
// Then (optionally) delete the original.
copy($origDir,$newDir."/".$fileToMove);
unlink($origDir); //removes original.
// Add a flag to your Database to know that this file has been copied,
// ideally you should resave the filepath to the correct new one.
//MySQL update saving the new filepath.
}
Read up on PHP Copy and PHP unlink.
And; please use Prepared Statements for PHP and Database interactions.!
I have a directory of images which are all named *.jpg, where * is the ID of each record in a MySQL table. e.g. 1394.jpg
What I'd like to be able to do is pull 5 random images from that directory using PHP and for each image, then query the MySQL table for records matching that ID / image filename, so I can then display details of each record.
The part I am looking for some guidance on is how to a) randomly choose 5 images using php, then b) identifying the filenames. Once I have the filename as a variable, I am sure I should be able to write the MySQL query. It's the PHP element I'm struggling with.
Could anyone offer any advice?
If the file info in your db is update-to-date regarding the image files in your file system, you can get the 5 random filename ids along with other info directly from the db:
$sql = 'SELECT * FROM image_files ORDER BY RAND() LIMIT 5';
This will return you file ids along with other info you have there in the table.
EDIT: Apparently all files in db does not exist in file system. So below is updated answer:
// Get files info from db in random order
$sql = 'SELECT * FROM image_files ORDER BY RAND()';
$result = $conn->query($sql);
// Pick first 5 files that does exist in file system
$five_files=Array();
while($row = $result->fetch_assoc()) {
$filepath = '/dir/images/'.$row['id'].'.jpg';
if (is_file($filepath)) {
$five_files[] = $row;
if (count($five_files) == 5) break;
}
}
// This will have info on five randomly picked files from db that does exist
print_r($five_files);
I'll do that, first getting all the images of the folder
$files = glob("$dir/*.{jpg,jpeg}", GLOB_BRACE);
Then, I'll get one (you can make a loop) random pic using array_rand, like this
$file = array_rand($files); // You can check if is repeated or something too
Finally, you can preg_replace the name of the file to get only the name (without extension) and query to your DB with that name... (Maybe there are other ways, but this is simple and fast)
Okay, We will do it in these steps :
Get File Name Lists from the required Directory
Function : scandir() look for more details here
Store in Array
Get Random 5 Data
Function array_rand() look for more details here
For Each File Run Query For Details
Loop - Foreach() - Store MySQL-fetched-data in multi-dimensional Array
Display
I have an admin system which adds profile pages dynamically - and part of that adds images to a directory. When I add the images they all have the name format like 12_1.jpg, 12_2.jpg, 32_1.jpg, 32_9.jpg where the number before the underscore is the id $cid and the number after the underscore is the image number.
I'm trying to find a way to list the images on the edit-images.php page with an option to delete them (maybe a link next to the image name, or another way that is better).
Here is the code to find the images I need to have the option of deleting:
if ($handle = opendir('../images/company')) {
while (false !== ($entry = readdir($handle))) {
if (substr(basename($entry), 0, 2) == $cid) {
echo $entry . "<br>";
}
}
closedir($handle);
}
How would I go about deleting specified images from here? Any help would be appreciated.
In php you use the unlink function to delete files. Just echo a link to a page that calls a function do to that and pass the image name. Only be sure to check the user input before actually perform deletion.
Why are you scanning the directory for the file? You can check if the file exists and delete it instead. For example:
$files = glob("../images/company/$cid_*.jpg");
foreach($files as $file) {
if(file_exists($file)) unlink($file);
}
This will unlink all files in ../images/company with the name starting with 123_ for example and ending in .jpg (you can provide more extensions by using GLOB_BRACE parameter)
Scenario :I am trying to build a Mobile Entertainment Portal. It will enable users to download Music & Movies to their Cell Phones...
Problem Exp : Suppose I upload 100 folders of Songs, each folder is for one Album. I want a way to generate a page with all the folders name (Album Name) in it. If user click on the page, they should be taken to a page where they get list of all songs in the album. Clicking on any song name will let them download it. Can it be done anyway or will I have to manually design each of the 3 pages for each album. If I do that, its time consuming and also will be difficult to change anything like footer, header...
First of all, this is a weighted question. But I will try to answer some of your questions to get you started.
You can scan for directories using scandir() in PHP.
$path = '/path/to/music';
$dir = scandir($path);
if (is_array($dir)) {
foreach ($dir as $directory) {
if (is_dir("{$path}/{$directory}")) {
// validate that it's among the directories you want
}
}
}
So, now that you know how to do that, perhaps instead of trying to create a separate page for each album folder you could use one script and based on the GET vars, display the appropriate content. e.g.
domain.com/index.php?album=Album+Name
Now let's see how that might work with the above example:
// assume your album folder names use underscores
$album = (isset($_GET['album']))
? str_replace('+', '_', $_GET['album'])
: null;
$path = '/path/to/music';
$dir = scandir($path);
if (is_array($dir)) {
foreach ($dir as $directory) {
if (is_dir("{$path}/{$directory}") && $album == $directory) {
// now, scan for files
}
}
}
Then to get the files, when you're looping through the directories, instead of checking if it is a directory, check that it's not a directory and that will give you your files. (e.g. if (!is_dir(...)
I am completely new to PHP so forgive me if this question seems very rudimentry. And thank you in advance.
I need to include a jpg that is generated from a webcam on another page. However I need to include only the latest jpg file. Unfortunately the webcam creates a unique filename for each jpg. How can I use include or another function to only include the latest image file?
(Typically the filename is something like this 2011011011231101.jpg where it stands for year_month_date_timestamp).
Easy way is to get the latest image with the help of the below code
$path = "/path/to/my/dir";
$latest_ctime = 0;
$latest_filename = '';
$d = dir($path);
while (false !== ($entry = $d->read())) {
$filepath = "{$path}/{$entry}";
// could do also other checks than just checking whether the entry is a file
if (is_file($filepath) && filectime($filepath) > $latest_ctime) {
$latest_ctime = filectime($filepath);
$latest_filename = $entry;
}
}
}
// now $latest_filename contains the filename of the newest file
give the source of latest image to <img> tag
Since the images are named via pattern which relates to the date, you should be able to just use:
$imgs = glob('C:\images\*.jpg');
rsort($imgs);
$newestImage = $imgs[0];
This is fairly straightforward, since your file names are in order.
The first thing you need is a list of files in the directory. The readdir (doc) function is what you are looking for. Example script that uses it: http://www.liamdelahunty.com/tips/php_list_a_directory.php
Once you have that, use substr() (doc) to chop off the file name extensions.
You're left with an array of numbers, essentially. From here, do a sort (doc) and specify the SORT_NUMERIC flag. Grab the number on the end, stick a .jpg back on it, and you have the last file.
Alternate Solution: Read the timestamps of files to get the last one. This would generally be a better answer, but perhaps not in your situation if you plan to edit any of the files.
I guess you will have to know a way to determine what the latest image file is called. Maybe you can make a textfile or something where every time a new image is created the webcam writes the latest filename in the text file (so the only text in the text file is the file name of the latest image file if it makes any sense). Of course you will have to have access to the script that generates the php file.
addition to #ken 's post, it's probably sorting alphabetically instead of numerically. perhaps you could try:
$imgs = glob('C:\images\*.jpg');
rsort($imgs, SORT_NUMERIC);
$newestImage = $imgs[0];