Create Image from Url Any File Type

Create Image From Url Any File Type

Maybe you want this:

$jpeg_image = imagecreatefromfile( 'photo.jpeg' );
$gif_image = imagecreatefromfile( 'clipart.gif' );
$png_image = imagecreatefromfile( 'transparent_checkerboard.PnG' );
$another_jpeg = imagecreatefromfile( 'picture.JPG' );
// This requires you to remove or rewrite file_exists check:
$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg' );
// SEE BELOW HO TO DO IT WHEN http:// ARGS IS NEEDED:
$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg?foo=hello&bar=world' );

Here's how it's done:

function imagecreatefromfile( $filename ) {
    if (!file_exists($filename)) {
        throw new InvalidArgumentException('File "'.$filename.'" not found.');
    }
    switch ( strtolower( pathinfo( $filename, PATHINFO_EXTENSION ))) {
        case 'jpeg':
        case 'jpg':
            return imagecreatefromjpeg($filename);
        break;

        case 'png':
            return imagecreatefrompng($filename);
        break;

        case 'gif':
            return imagecreatefromgif($filename);
        break;

        default:
            throw new InvalidArgumentException('File "'.$filename.'" is not valid jpg, png or gif image.');
        break;
    }
}

With some small modifications to switch same function is ready for web url's:

    /* if (!file_exists($filename)) {
        throw new InvalidArgumentException('File "'.$filename.'" not found.');
    } <== This needs addiotional checks if using non local picture */
    switch ( strtolower( array_pop( explode('.', substr($filename, 0, strpos($filename, '?'))))) ) {
        case 'jpeg':

After that you can use it with http://www.tld/image.jpg:

$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg' );
$gif_image = imagecreatefromfile( 'http://www.example.com/art.gif?param=23&another=yes' );

Some proofs:

As you can read from official PHP manual function.imagecreatefromjpeg.php GD allows loading images from URLs that is supported by function.fopen.php, so there is no need to fetch image first and save it to file, and open that file.

How do you find the filetype of an image in a url with nonobvious filetype in Python

Building on the responses to this question, you could try:

import requests
from PIL import Image       # pillow package
from io import BytesIO

url = "your link"

image = Image.open( BytesIO( requests.get( url ).content))
file_type = image.format

This calls for downloading the entire file, though. If you're looking to do this in bulk, you might want to explore the option in the comment above that mentions "magic bytes"...

Edit:
You can also try to get the image type from the headers of the response to your url:

headers = requests.get(url).headers
file_type =headers.get('Content-Type', "nope/nope").split("/")[1]
# Will print 'nope' if 'Content-Type' header isn't found
print(file_type)
# Will print 'gif' or 'jpeg' for your listed urls

Edit 2:
If you're really only concerned with the file type of the link and not the file itself, you could use the head method instead of the get method of the requests module. It's faster:

headers = requests.head(url).headers
file_type =headers.get('Content-Type', "nope/nope").split("/")[1]

save any image type from a url as jpg

I guess this is not possible (taking in mind all of the existing image filetypes out there) but for the most basic ones there should be libraries for converting.

Also for the thing you tried - just renaming the file extension to .jpg doesn't actually make it a jpg (IrfanView, for example, warns you upon opening such file and asks you if you want to change the extension back to correct one)

Linking images regardless of file type

You can do it in CSS like this:

background-image: url('users/avatars/$username.jpg'), url('users/avatars/$username.png');

Getting a file type from URL

Assuming your Content-Type HTTP headers are accurate, you can avoid downloading the whole file just to check the type by creating a HEAD request. Assuming you don't also need the whole file for something else, this could be a much-quicker operation, especially for large files.

Working Example:

var xhr = new XMLHttpRequest();xhr.open('HEAD', 'https://crossorigin.me/http://placehold.it/350x150', true);
xhr.onload = function() {    var contentType = xhr.getResponseHeader('Content-Type');    console.log(contentType);};
xhr.send();

How to save an image from an URL which does not contain any image file extensions (e.g. png) in Python?

you can get do the following, after getting the response:

_, extension = response.headers["content-type"].split("/")
fname = f"sample_image.{extension}"
file = open(fname, "wb")
file.write(response.content)
file.close()

How can I detect the file type of image at a URL?

Well, https://stackoverflow.com/content/img/so/logo is a 404. If it were not, then you could use

#!/usr/bin/perl

use strict;
use warnings;

use LWP::Simple;

my ($content_type) = head "https://stackoverflow.com/content/img/so/logo.png";

print "$content_type\n" if defined $content_type;

__END__

As Kent Fredric points out, what the web server tells you about content type need not match the actual content sent by the web server. Keep in mind that File::MMagic can also be fooled.

#!/usr/bin/perl
use strict;
use warnings;

use File::MMagic;
use LWP::UserAgent;

my $mm = File::MMagic->new;

my $ua = LWP::UserAgent->new(
    max_size => 1_000 * 1_024,
);

my $res = $ua->get('https://stackoverflow.com/content/img/so/logo.png');

if ( $res->code eq '200' ) {
    print $mm->checktype_contents( $res->content );
}
else {
    print $res->status_line, "\n";
}
__END__

Determine file extension for image urls

If you can trust that the URLs are not malformed, how about this:

FilenameUtils.getExtension(URI.create(url).getPath())

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