Saving image straight to directory in PHP?

Question

I have an image file. Example: http://images5.fanpop.com/image/photos/31100000/random-random-31108109-500-502.jpg

I want to save the image to a directory called images-folder in my host. What would be the best way to do this using PHP?

http://stackoverflow.com/questions/724391/saving-image-from-php-url-using-php — CharliePrynn, Feb 27 '13 at 16:36
http://www.edmondscommerce.co.uk/php/php-save-images-using-curl/ "`imagecreatefromjpeg($img);` where $img is a url for an external image" The site itself deals with the 'what if your host disabled this' — AmazingDreams, Feb 27 '13 at 16:36
Thank you guy's you all helped allot and I have exactly what I need now because of the help, I am relay grateful :) — , Feb 27 '13 at 17:04

score 14 · Accepted Answer · edited Apr 12 '14 at 18:50

Yes, it is very simple. Here is a little cURL script to do just that:

$image_link = "http://images5.fanpop.com/image/photos/31100000/random-random-31108109-500-502.jpg";//Direct link to image
$split_image = pathinfo($image_link);

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL , $image_link);
curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US) AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.A.B.C Safari/525.13");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
$response= curl_exec ($ch);
curl_close($ch);
$file_name = "images-folder/".$split_image['filename'].".".$split_image['extension'];
$file = fopen($file_name , 'w') or die("X_x");
fwrite($file, $response);
fclose($file);

This should do what you want. It will save the image to the directory and then name the image as random-random-31108109-500-502<-filename .jpg<-extension.

Thanks a lot... Its working.. – Bharat Chodvadiya Sep 11 '15 at 12:38 — Bharat Chodvadiya, Sep 11 '15 at 12:38

score 8 · Answer 2 · edited Apr 12 '14 at 18:51

8

Even simpler without cURL:

<?php
    $link= "http://images5.fanpop.com/image/photos/31100000/random-random-31108109-500-502.jpg";
    $destdir = 'images-folder/';
    $img=file_get_contents($link);
    file_put_contents($destdir.substr($link, strrpos($link,'/')), $img);
?>

edited Apr 12 '14 at 18:51

Peter Mortensen

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answered Feb 27 '13 at 16:39

Ghigo

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using this way will cause an image damage unable to open – atom2ueki Sep 07 '14 at 23:06

score 1 · Answer 3 · answered Feb 27 '13 at 16:38

1

Here is another, somewhat easier to follow, example, straight from the site I commented

$remote_img = 'http://www.somwhere.com/images/image.jpg';
$img = imagecreatefromjpeg($remote_img);
$path = 'images/';
imagejpeg($img, $path);

http://www.edmondscommerce.co.uk/php/php-save-images-using-curl/

answered Feb 27 '13 at 16:38

AmazingDreams

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Explain the downvote, please. – AmazingDreams Feb 27 '13 at 16:45
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I don't see why you got down voted either your solution work's good , I will up vote you because your answer is genuine. – Feb 27 '13 at 17:10
3

You don't need to read the file as an image to download the image. You're allocating all of that memory and creating an editable image buffer with all of the overhead associated with that JUST to save the image. That will not scale. Also you're introducing a security risk by asking PHP to interpret a remote file as a JPEG. libjpeg exploits aren't unknown. These are the reasons I downvoted. – hsanders Mar 20 '13 at 15:46

Saving image straight to directory in PHP?

3 Answers3

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