I want to be able to open the provided URL (which is done via a form) that is an URL that will allow the server to save the file into a directory, for example:
http://www.google.co.uk/intl/en_com/images/srpr/logo1w.png
I want to save that logo into this directory:
img/logos/
Then it will add it to the database by giving it a random file name before so, e.g.
827489734.png
It will now be inserted to the database with the following:
img/logos/827489734.png
I do not want to use cURL for this, I like to work with fopen, file_get_contents, etc...
Cheers.
EDIT
$logo = safeInput($_POST['logo']);
if(filter_var($avatar, FILTER_VALIDATE_URL))
{
$get_logo = file_get_contents($logo);
$logo_directory = 'img/logos/';
$save_logo = file_put_contents($logo_directory, $logo);
if($save_logo)
{
$logo_path = $logo_directory . $save_logo;
A part of this code I need helping...
You need to specify a full file name when doing a file_put_contents(). A pure directory name won't cut it.
Related
I'm trying to download a file stored in my Larval 8 Storage folder. The path is correct I have checked it multiple times. And If I open the path in the search bar. It takes me to the needed image. What am I doing wrong here?
Controller
public function show($_id)
{
$currentUrl = URL::current();
$qrCode = QrCode::format('svg')->size(100)->generate($currentUrl."/".$_id);
$path = storage_path('app/public/'.$_id.'.png');
$asset = Asset::findorFail($_id);
return view('assets::asset',compact('asset', 'qrCode', 'path'));
}
asset.blade.php
<a href="{{storage_path('app\public/'.$asset->_id.'.png')}}" download>{{$qrCode}}</a>
The path of the image is: file:///D:/Facility_Management_System/storage/app/public/6155a180001300004e005e22.png
Which is same path as href="{{storage_path('app\public/'.$asset->_id.'.png')}}"
But, When I click to download the file to download it does nothing.
The download attribute specifies that the target will be downloaded when a user clicks on the hyperlink.
So in this case, if the downloaded file extension is .htm, it means that the response from specified link {{storage_path('app\public/'.$asset->_id.'.png')}} is html, instead of image file(maybe 404).
You need to fix the public url of the image file.
I will completely clarify my question, sorry to everybody.
I have code writed in files from a website that now is not working, the html code is on pages with php extension, in a folder of a Virtual Host in my PC using Wampserever. C:\wamp\1VH\PRU1, when the site was online there was a folder where was a file called image.php. This file was called from other pages inside the site like this: (a little code of a file, C:\wamp\1VH\PRU1\example.php)
"<div><img src="https://www.example.com/img/image.php?f=images&folder=foods&type=salads&desc=green&dim=50&id=23" alt="green salad 23"></div>"
And the result was that the images was showed correctly.
Now, like i have this proyect in local, and i HAVE NOT the code of that image.php file i must to write it myself, this way the images will be showed the same way that when the site was online.
When the site was online and i open a image returned by that image.php file the URL was, following the example, https://example.com/images/foods/salads/green_50/23.png.
Now how the site is only local and i have not that image.php file writed because i'm bot sure how to write it, the images obviously are not showed.
In C:\wamp\1VH\PRU1\example.php the code of files was changed deleting "https://www.example.com/img/image.php?" for a local path "img/image.php?".
And in the same folder there is anothers: "img" folder (here must be allocated the image.php file), and "images" folder, inside it /foods/salads/green_50/23.png, 24.png.25.png..............
So i have exactly the same folder architecture that the online site and i changed the code that i could only, for example replacing with Jquery "https://www.example.com/img/image.php?" for "img/image.php?" but wich i can not do is replace all the code after the image.php file to obtain a image file.
So i think that the easiest way to can obtain the images normally is creating that IMAGE.PHP file that i have not here in my virtual host.
I'd like to know how to obtain the parameters and return the correct URL in the image,php file.
The image of the DIV EXAMPLE must be C:/wamp/1VH/PRU1/images/foods/salads/green_50/23.png
I have in my PC the correct folders and the images, i only need to write the image.php file.
Note that there are "&" and i must to unite the values of "desc=green&dim=50&" being the result: green_50 (a folder in my PC).
TVM.
You probably want something like this.
image.php
$id = intval($_GET['id']);
echo '<div><img src="images/foods/salads/green_50/'.$id.'.png" alt="green salad '.$id.'"></div>';
Then you would call this page
www.example.com/image.php?id=23
So you can see here in the url we have id=23 in the query part of the url. And we access this in PHP using $_GET['id']. Pretty simple. In this case it equals 23 if it was id=52 it would be that number instead.
Now the intval part is very important for security reasons you should never put user input directly into file paths. I won't get into the details of Directory Transversal attacks. But if you just allow anything in there that's what you would be vulnerable to. It's often overlooked, so you wouldn't be the first.
https://en.wikipedia.org/wiki/Directory_traversal_attack
Now granted the Server should have user permissions setup properly, but I say why gamble when we can be safe with 1 line of code.
This should get you started. For the rest of them I would setup a white list like this:
For
folder=foods
You would make an array with the permissible values,
$allowedFolders = [
'food',
'clothes'
'kids'
];
etc...
Then you would check it like this
///set a default
$folder = '';
if(!empty($_GET['folder'])){
if(in_array($_GET['folder'], $allowedFolders)){
$folder = $_GET['folder'].'/';
}else{
throw new Exception('Invalid value for "folder"');
}
}
etc...
Then at the end you would stitch all the "cleaned" values together. As I said before a lot of people simply neglect this and just put the stuff right in the path. But, it's not the right way to do it.
Anyway hope that helps.
You essentially just need to parse the $_GET parameters, then do a few checks that the file is found, a real image and then just serve the file by setting the appropriate content type header and then outputting the files contents.
This should do the trick:
<?php
// define expected GET parameters
$params = ['f', 'folder', 'type', 'desc', 'dim', 'id'];
// loop over parameters in order to build path: /imagenes/foods/salads/green_50/23.png
$path = null;
foreach ($params as $key => $param) {
if (isset($_GET[$param])) {
$path .= ($param == 'dim' ? '_' : '/').basename($_GET[$param]);
unset($params[$key]);
}
}
$path .= '.png';
// check all params were passed
if (!empty($params)) {
die('Invalid request');
}
// check file exists
if (!file_exists($path)) {
die('File does not exist');
}
// check file is image
if (!getimagesize($path)) {
die('Invalid image');
}
// all good serve file
header("Content-Type: image/png");
header('Content-Length: '.filesize($path));
readfile($path);
https://3v4l.org/tTALQ
use $_GET[];
<?php
$yourParam = $_GET['param_name'];
?>
I can obtain the values of parameters in the image.php file tis way:
<?php
$f = $_GET['f'];
$folder = $_GET['folder'];
$type = $_GET['type'];
$desc = $_GET['desc'];
$dim = $_GET['dim'];
$id = $_GET['id'];
?>
But what must i do for the image:
C:/wamp/1VH/PRU1/images/foods/salads/green_50/23.png
can be showed correctly in the DIV with IMG SRC atribute?
I have visited this article previously and found it useful, but i would like to add more functionality to it by having it save an image file name according to the URL name.
This is what I've done so far and it works.
$contents=file_get_contents('http://www.domain.com/logo.png');
$save_path="C:/xampp/htdocs/proj1/download/[logo.png]";
file_put_contents($save_path,$contents);
Basically, where I have put square brackets around I want to have that dynamic based on the URL file name. For example, if i have an image url such as this: https://cf.dropboxstatic.com/static/images/brand/glyph-vflK-Wlfk.png, I would like it to save the image into the directory with that exact image name which in this case is glyph-vflK-Wlfk.png.
Is this possible to do?
I would do this way
$url = "http://www.domain.com/logo.png";
$file = file_get_contents($url);
$path = "C:/xampp/htdocs/proj1/download/". basename($url);
return !file_exists($path) ? file_put_contents($path, $file) : false;
From what I understand, what you're trying to do is the following :
$url = 'http://www.domain.com/logo.png';
$contents=file_get_contents($url);
$posSlash = strrpos($url,'/')+1);
$fileName = substr($url,$posSlash);
$save_path="C:/xampp/htdocs/proj1/download/".$fileName;
file_put_contents($save_path,$contents);
There is a function for that, basename():
$url="http://www.domain.com/logo.png";
$contents=file_get_contents($url);
$save_path="C:/xampp/htdocs/proj1/download/".basename($url);
file_put_contents($save_path,$contents);
You might want to check if it already exists with file_exists().
I am using TinyMCE as a WYSIWYG editor.
It is working perfectly, except for the image upload directory. I want each user to have their own directory in the images directory, but I cannot get it to work.
I am passing the user id in the URL and have tried adding the code to get it from the URL in the config.php file where the directories are defined, but the $user_id value remains empty.
Any assistance would be great.
The URL:
http://mydomain.co.za/index.php?user_id=1
The Code:
<?php
$user_id= htmlspecialchars($_GET["user_id"]);
// The URL that points to the upload folder on your site.
// Can be a relative or full URL (include the protocol and domain)
$imageURL = 'http://mydomain.co.za/images/'.$user_id;
// Full upload system path. Make sure you have write permissions to this folder
$uploadPath = '/home/username/public_html/editor/images/'.$user_id;
//We create the directory if it does not exist - you can remove this if you consider it a security risk
if(!is_dir($uploadPath)) {
mkdir($uploadPath,0755,true);
}
//Create thumb directory if doesn't exist
if(!is_dir($uploadPath . 'thumbnail')) {
mkdir($uploadPath . 'thumbnail',0755,true);
}
//Allowed extenstions
$allowedExtensions = array('jpg','gif','jpeg','bmp','tif','png');
//Maximum upload limit
$sizeLimit = 2 * 1024 * 1024;
function isAuth() {
//Perform your own authorization to make sure user is allowed to upload
return true;
}
Is it possible the reason is because it is not in the main php file?
Or Can I get the variable from the URL?
They suggested on their Instructions that I add $userId = Auth::getId(); but id returns an empty value. Plus I have no idea what that command is executing.
PLEASE NOTE:
the file management is being done by TinyMCE Image Uploader & Manager
UPDATE:
By adding the $actual_link = "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]"; echo $actual_link; I noticed by the time the $_GET command is rung the URL has changed to http://mydomain.co.za/tinymce/plugins/lioniteimages/connector/php/gallery.php, but in the browser URL bar, the URL is still the same with the variable.
Is there anyway to access that URL instead of the one i am getting?
I found the solution.
Simple enough, just created a session and the problem was solved.
I was able to get the variable from the session.
How do I find the filename of an image on a MediaWiki site?
I don't want to put the filename in manually. I need PHP code which will fetch me the filename.
I can use $f = wfFindFile( '$filename' ); but HOW DO I GET $filename?
I've been looking at the FILE class but I can't figure out how to use File::getFilename(); I keep getting an error call to undefined method.
What am I doing wrong?
Explaining in more detail:
I would like to add the pin it button to my site so when you click on the button it post it on the pin it board with the image and description of the image. I need to use php to send the image information so it works on every page on my site. I can't code the image name manually each time.
So far I have the code:
<img border="0" src="//assets.pinterest.com/images/PinExt.png" title="Pin It" />
Which works great except I need to put in a value for $f (image name). My question is how do I get the value of $f without having to put in in eg $f = wfFindFile( 'Sunset.jpg' );
I would have thought this would be a really common request for anyone trying to add pinterest to their site.
Thanks
The $filename you are looking for is basically how it is named in MediaWiki when it got uploaded, for example Landscape-plain.jpg. You will just use the wfFindFile() helper function to get a File object. Then call the methods:
$ php maintenance/eval.php
> $file = wfFindFile( 'Landscape-plain.jpg' );
> print $file->getName();
Landscape-plain.jpg
> print $file->getPath();
mwstore://local-backend/local-public/b/b0/Landscape-plain.jpg
> print $file->getFullPath();
/path/to/images/b/b0/Landscape-plain.jpg
> print $file->getTitle();
File:Landscape-plain.jpg
> exit
API documentation:
http://svn.wikimedia.org/doc/classFile.html
http://svn.wikimedia.org/doc/classLocalFile.html
EDIT BELOW
The file informations are available through a File object, so you definitely need to use wfFindFile() to get such an object.
To actually find the filename for the page the user is browsing on, you want to use the query context and get its title:
$context = RequestContext::getMain();
$t = $context->getTitle();
if( $title->getNamespace == 'NS_FILE' ) {
$filename = $title->getPrefixedText;
// do your stuff.
}