So i have this one small thing im doing, and the thing is: It generates QRcodes using the google qrcode api. Id like to download the generated qr code with the press of a button into a /img folder i have inside the folder my php file is located at, and id need it to save with a numerical number which is the button's id. (it has like 120 buttons and each one has a 0-120 id, id like to save the qrcode that belongs to the button 10 with a name like 10.png) any ideas on how to do so?
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Little edit: as Man Manam pointed out my question is: I have the image link, i want to save it in my server. Sorry for the confusion.
As you're using Google for creating the QR code, I assume you have a code like this somewhere:
$ga = new \Google\Authenticator\GoogleAuthenticator();
$qr_link = $ga->getUrl($username, $_SERVER['HTTP_HOST'], $secret);
// then you'd display that code with <img src="$qr_link">
What you need to do is to simply download that QR code instead of showing in browser. You can use curl for this or, even simpler, just fopen/file_get_contents if allow_url_fopen is set to ON in php.ini
$image_bin = file_get_contents($qr_link);
file_put_contents($target_file_on_filesyste, $image_bin);
Related
I am looking to generate a QR code on a PHP webpage. The user starts on a page where they submit a form, the only input being the name of the new client. This page is a manager for VPN clients.
After the redirect, PHP generates the configuration file on the page to copy and paste. The user must save this file as wg.conf in their etc directory. I have been trying to use qrencode, a Linux based command line tool to generate the QR code on that same page with the configuration file. According to the documentation on qrencode, you can save the file as a .png or .svg.
I have been playing around with it, both trying to pass data through the URL as GET parameters in the redirect (the only issue is the QR code is sensitive) and by using the backtick operators to run bash commands to save it as a file in the user's local files. I can't seem to figure out where to save it as a file or whether there's another solution to display it on the webpage.
how i would do it:
check this library https://github.com/neocotic/qrious , all you need to do it load the js on your page
generate text you want to convert to qr code in a text area, div or other HTML element and assign it an id
use that id to extract the data via pure JS, now you have a text variable and you have the possibility to make qr code dynamic now
once you have the data dynamically extracted via JS you can use that variable to generate qr codes like this :
var x = 'sample text'
var qr = window.qr = new QRious({
element: document.getElementById('qrious'),
size: 200,
value: x
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/qrious/4.0.2/qrious.min.js"></script>
<canvas id="qrious">
I have an add form, where users can add an image, once they click submit I want the image to be saved into a certain folder located called Event_images (which I use to display the image on another page where I display the image along with other details they inputted. But once I submit the form, the image goes to the MYSQL database, but isn't in the folder where I direct it to. My code and a screenshot are provided. I want the image to be displayed in the Event_images folder I have created. Thank you in advance.
<?php
$event_img = $_FILES['event_img']['name'];
$tempimage = $_FILES['event_img']['tempname'];
move_uploaded_file($tempimage,"Event_images/$event_img");
?>
The screenshot I have provided is what happens when I try to display the information (because nothing is in the folder which I direct to)
['tempname'] isn't an index of $_FILES. See here.
Use $tempimage = $_FILES['event_img']['tmp_name']; instead. I tested locally and it worked fine. If you're using PHP 7 you may need to use { } around $event_img when you move the file so it'll process that variable.
Does this work?
$event_img = rand(1000,100000)."-".$_FILES['event_img']['name'];
$tempimage = $_FILES['event_img']['temp_name'];
$folder="Event_images/";
move_uploaded_file($file_loc,$folder.$event_img);
This is an odd question but I'm stuck on how I would achieve this and I am unable to find any methods of doing so.
I have a simple php script that takes variables (containing file names) from the URL, cleans then and then uses them to generate a single image from the inputted values. This works fine and outputs a new png to the webpage using:
imagepng($img);
I also have a facebook sharing script in PHP that takes a filepath as an input and then shares the image on the users feed where this statement is used to define the image variable:
$photo = './mypic.png'; // Path to the photo on the local filesystem
I don't know how I can link these two together though. I would like to use my generation script as the image to share.
Can anyone point me in the right direction of how to do this? I am not the master of PHP so go easy please.
-Tim
UPDATE
If it helps, here are the links to the two pages on my website containing the outputs. They are very ruff mind you:
The php script generating the image:
http://the8bitman.herobo.com/Download/download.php?face=a.png&color=b.png&hat=c.png
The html page with the img tag:
http://the8bitman.herobo.com/Share.html
Treat it as a simple image:
<img src="http://yourserve/yourscript.php?onlyImage=1" />
yourscript.php
if($_GET['onlyimage']) {
header('Content-type:image/png'); //or your image content type
//print only image
} else {
//print image and text too
}
So this is simple to understand what i want to achieve. So i get links like theese:
http://rockdizfile.com/atfmzkm7236t
http://rockdizfile.com/xuj5oincoqmy
http://rockdizfile.com/pg8wg9ej3pou
So theese links are from one cloud storage site I want to make a php script that automates their downloading.
So I can't find which is the script or the thing these links download button starts and how can I start that so i can download it with php on my server?
Basically my idea is to download a lot of files but don't wanna do it manually so need automatic way of doing it. As far as I know I make a request which is the following 2 urls:
http://rockdizfile.com/pg8wg9ej3pou
http://wi32.rockdizfile.com/d/wsli6rbhfp4r2ge4t7cqeeztijrprelfiw4afvqg5iwspmvqabpkmgiz/Desislava%20feat.%20Mandi%20&%20Ustata%20-%20Pusni%20go%20pak%20(CDRIP).mp3
So the first url is executing the next one but here comes the tricky part as far as I tested that last string Desislava%20feat.%20Mandi%20&%20Ustata%20-%20Pusni%20go%20pak%20(CDRIP).mp3 is the file name we get when downloading so if you change it with for example somefile.mp3 it will download somefile.mp3 but with the same file content as http://wi32.rockdizfile.com/d/wsli6rbhfp4r2ge4t7cqeeztijrprelfiw4afvqg5iwspmvqabpkmgiz/Desislava%20feat.%20Mandi%20&%20Ustata%20-%20Pusni%20go%20pak%20(CDRIP).mp3 so the data is hidden in this hash wsli6rbhfp4r2ge4t7cqeeztijrprelfiw4afvqg5iwspmvqabpkmgiz or i think so. And now is the tricky part how to get this hash? we have almost everything we have the code for the url atfmzkm7236t the hash wsli6rbhfp4r2ge4t7cqeeztijrprelfiw4afvqg5iwspmvqabpkmgiz and the filename Desislava%20feat.%20Mandi%20&%20Ustata%20-%20Pusni%20go%20pak%20(CDRIP).mp3 There must be a way to download from this site without clicking so please help me kinda a hack this :)
you can use PHP's header function to force a file to download
header('Content-disposition: attachment; filename=index.php');
readfile('Link');
You should know that this will not give you the ability to download PHP files from external websites.
You can only use this if you got the direct link to a file
It's impossibly to tell you without the source code
e.g. sha1("Test Message") gives you 35ee8386410d41d14b3f779fc95f4695f4851682 but sha256("Vote this up") gives you 65e03c456bcc3d71dde6b28d441f5a933f6f0eaf6222e578612f2982759378ed
totally different... unless you're hidden function add's "65e03c456bcc3d71dde6b28dxxxxxxxxxxxxxxxxxxxxxxxxxx" (where xxxxxxxxxxxxxxxxxxxxxxxxxx is a bunch of numbers I can't be arsed to work out) to each hash...
then sha1("Test Message") gives you 65e03c456bcc3d71dde6b28d441f5a933f6f0eaf6222e578612f2982759378ed
The file is embedded into the swf player.
alert(jwplayer('mp3player').config.file);
Something like:
<?PHP echo file_get_contents($_GET["url"]); ?>
<script>
document.location=jwplayer('mp3player').config.file;
</script>
Though I've actually just noticed they change 5 digits of the URL on each page request, and the script above uses 2 page requests. One to get the URL and HTML source and another to try and download the file, meaning the URL has changed before the second request has started.
What I want to do is following:
User takes screenshot with the application like jing. ok!
Pastes link that Jing returned back. ok!
Server processes the link that user entered, and extracts images source url. But, I have no idea how server will get "clean" image source URL. For example, this is the link that Jing returned after sharing screenshot http://screencast.com/t/zxBzNNkcg but real url of image looks like http://content.screencast.com/users/TT13/folders/Jing/media/c25ec5c6-bc6a-413c-a50b-ada95fac4ed2/2012-07-25_0221.png
Server returns back image source URL. no idea!
Is there any possible way to get image url with Javascript or PHP?
you could use Simple PHP DOM Parser to retrieve the image from the url without considering the url for as long as it contains and image inside, like so:
foreach($html->find('div[class=div-that-contain-the image]') as $div) {
foreach($div->find('img') as $img){
echo "<img src='" . $img->src . "'/>";
}
}
That is my solution.
You can retrieve the page containing the image using a DOM library like Query Path.
Using that you can extract the URL to the image.
So in your step 3:
Get source of shared screenshot page (maybe use file_get_contents)
Extract screenshot's image src, using Query Path.
Return image src URL to user
Yes. If you right click on the image and go Copy Image Location, you'll see it's http://content.screencast.com/users/TT13/folders/Jing/media/c25ec5c6-bc6a-413c-a50b-ada95fac4ed2/2012-07-25_0221.png
If you were to do it programmatically, you would use cURL and simplehtmldom.sourceforge.net to parse the outputed HTML for the actual link.
Javascript answer:
Will the output image always have the same class, embeddedObject?
If so, how about something like:
myVar = document.getElementsByClassName('embeddedObject');
myVar[0].getAttribute("src");
The myVar[0] reference of course assumes that there is only ever one image on the page with the embeddedObject class, otherwise you'd need to sort, or know which index to reference each time.
Also, this sadly doesn't seem to be supported in IE8 (which browser do you need to support?):
http://caniuse.com/getelementsbyclassname