I want to get current URL inside a PNG generated by PHP (header('Content-Type: image/png');)
But when i use http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];,
The URL output is the file's location (http://test.com/png.php) but not the page which get the img with <img> tag.
What can i do to get current URL?
Thanks!
Sounds like you're looking for $_SERVER['HTTP_REFERER'], not REQUEST_URI.
I'm not sure I understood you, but I'll give it a go. If you serve an image via PHP, it's URL will always point to a PHP script. If you want it to look like it's really a .png image, you have 2 choices:
1) Create a rewrite rule which will redirect some .png URL to your PHP script and then serve that URL to users.
2) Stick with the current URL, but also add an (non-standard!) Content-Disposition header:
header('Content-Disposition: attachment; filename="my_image.png"');
This way the URL will still point to a PHP script, but when a user tries to save the image, the suggested name won't be the PHP script, but the name specified in Content-Disposition.
Related
I am at a situation, where I need to download files from the URL, it is easy with the direct file URLs like https://somedomain.com/some-path/somefile.exe
file_put_contents( $save_file_loc, file_get_contents($url_to_download);
But what to do when you have delayed force download from the URL which actually prints HTML and how to differentiate those URL?
Example URL: https://filehippo.com/download_mozilla-firefox-64/post_download/
EDIT: On above url the file download starts using JS, as I tested with blocking JS and download did not start.
Thanks in advance for your help.
Read the html of the URL using file_get_contents
Find the URL of the file within the HTML. You'll have to visit the page and view source to locate the URL. In your example of https://filehippo.com/download_mozilla-firefox-64/post_download/ it's found in between data-qa-download-url="https://dl5.filehippo.com/367/fb9/ef3863463463b174ae36c8bf09a90145/Firefox_Installer.exe?Expires=1594425587&Signature=18ab87cedcf3464363469231db54575665668c4f6&url=https://filehippo.com/download_mozilla-firefox-64/&Filename=Firefox_Installer.exe"
As you may have noticed, the page may have pre-approved the request so it's not guaranteed to work if the host has checks using cookies or other methods.
Create a regex based on the above to extract the URL using preg_match
Then file_get_contents the URL of the file to download it.
I want to show an img like this:
http://picviewer.umov.me/Pic/GetImage?id=92013681&token=ee103380297bbb2df0d8855949d791df
How should i use php to show the img with dynamic parameters?
You need to set the proper content type headers and then stream the binary data.
$imagepath = '/path/to/file';
header("Content-type: image/jpeg");
echo file_get_contents( $imagepath );
First i'll explain what does that URL do: it's configured to point to a script, or class->function, that manages the searching in the database, for the real path, the path you want to hide for the user, then returns the image to the request.
That said, in two steps, what you have to do is the following.
Check how to customize your url, this could be a start:
How to create friendly URL in php?
create your custom url pointing to a script or class/function like this:
Return a PHP page as an image
Thats, all... Assuming that you already have the images/database covered of course. if not, well you'll have to make the necessary database and tables...
I am trying to read multiple image files from a folder (.htaccess protected) and display in a HTML page using php readfile().
The problem is I can see only the first image is read and the next is not shown in the browser. The code is as below
<?php
$image1 = 'files/com_download\256\50\www\res\icon\android\icon-36-ldpi.png';
$image2 = 'files/com_download\256\50\www\res\icon\android\icon-48-mdpi.png';
$imginfo = getimagesize($image1);
header("Content-type: ".$imginfo['mime']);
readfile($image1);
$imginfo = getimagesize($image2);
header("Content-type: ".$imginfo['mime']);
readfile($image2);
?>
I could see the first image 'icon-36-ldpi.png' successfully read and displayed in the browser and the second image is not read and not displayed in the browser.
Am I missing something? Any advice please.
Sorry if I am doing stupid but the requirement is to read multiple image files and render in the browser like a grid view. I cannot use img tag because of security reasons.
You can't dump both images out at once. Why not make two images in your html so the browser makes two calls to your script. Then use a GET param to pass the filename you want to display.
---Edit---
Important Security Note
There is an attack vector which you open up when doing soething like this. Someone could easily view your source html and change the parameter to get your image script to output any file they want. They could even use "../../" to go up directories and search for well known files that exist. e.g. "../../../wp_config.php". Now the attacker has your wordpress database credentials. The correct way to prevent against this is to always validate the input parameter properly. For example, only output if the file name ends with ".jpg"
I want code that loads an image to a PHP server and then send it to browser.
For example I want sample.php to send an image to browser once it is requested.
in other words, I want to create a PHP file that acts like a proxy for an image.
why are you doing this?
why don't deliver the image directly?
if you are trying to display a random image you may as well just redirect to the image using
header("Location: address-of-image");
for delivering the file to your clients from your server and not from its original location you can just do. however your php.ini settings need to allow external file opens
readfile("http://www.example.com/image.jpg")
correct headers are not required if you are going to display the image in an img tag,
altough i would recommend it. you should check the filetype of the image or in most cases just set an octet-stream header so the browser doesnt assume an incorrect type like text or something and tries to display binary data.
to do so just do
header("Content-type: application/octet-stream")
one more thing to consider may be setting correct headers for caching...
You need to use
$image = fopen("image.png");
Modify the headers(not sure exacly if it's correct)
headers("Content-type: image/png");
And then send the image
echo fread($image, file_size("image.png"));
I need to provide a static url to a client, eg. http://domain.com/video.mp4. However this URL needs to provide a random video from a selection of 5 videos each time it is accessed.
Is this possible using PHP and mod_rewrite? Or some other way?
Thanks
You may be able to provide a url to a PHP script (http://domain.com/video.php), which then sets its content-type header to the type of a video and then randomly reads out a file using readfile.
I don't think you need to use mod_rewrite in this case.
header('Content-type: video/mp4'); // I don't know the correct MIME type
$files = array('vid1.mp4', 'vid2.mp4', 'vid3.mp4');
readfile($files[array_rand($files)]);
Does your client require the url to end in .mp4?
If not you could indeed use a mod_rewrite by selecting a random number from, say, 1 to 5