Show Image Using File_Get_Contents

Show image using file_get_contents

Do i need to modify the headers and just echo it or something?

exactly.

Send a header("content-type: image/your_image_type"); and the data afterwards.

file_get_contents displaying image incorrectly

Try this:

<?php
  $img = 'http://images.itracki.com/2011/06/favicon.png';
  $info = getimagesize($img);
  header('Content-type: ' . $info['mime']);
  readfile($img);
?>

You should use readfile() instead of file_get_contents() in this situation because readfile() will send the file contents directly to the browser rather than storing it in memory for a later retrieval, which is what file_get_contents() does.

Also note that the content-type is retrieved dynamically so that you don't have to worry about specifying it incorrectly in the header.

Some images may have an extension that doesn't match its content-type or mime-type, so this corrects for those type of issues.

Fetch image from url using file_get_contents

@h2ooooooo actually bro im not looking for a single image ... i just want to save this 9 images in a folder

<?php
    for ($i = 2 ; $i <= 10; $i++) {
        $imageName = "rate-bar-bg-" . $i . ".jpg";
        $imageContent = file_get_contents("http://www.lafourchette.com/p-3.3.0/default/" . $imageName);
        file_put_contents($imageName, $imageContent);
    }
?>

Remote image display is not working (now working using file_get_contents and curl)

You are trying to send remotely inlined data... It's a weird idea, and i'm not sure the browser understand what are you trying to achieve...

imageserver.php

<?php
$url = 'http://google.com/images/srpr/logo11w.png';   
function data_uri($file) 
{  
    $contents = file_get_contents($file);

    $finfo = new finfo(FILEINFO_MIME);
    $mime = $finfo->buffer($contents);

    return array($mime, $contents);
}
list($mime, $content) = data_uri($url);
header('Content-type: '.$mime);
die($content);

?>

file_get_contents or readfile for displaying filesystem image

If you don't need to manipulate the images (resizing, adding watermarks,...) readfile() will be the best choice as it writes the file directly to the output buffer. file_get_contents() will read the file into memory instead - requiring a lot of memory with large files.

Get Image With file_get_contents it return Not Found Error

Try this.

There is problem with URL Special character.then you have to decode some special character from url basename.

$imgfile = 'http://www.lagrolla.com.au/image/m fr 137 group.jpg';
$destinationPath = '/path/to/folder/';
$filename = basename($imgpath);
$imgpath = str_replace($filename,'',$imgpath).rawurldecode($filename);
copy($imgfile,$destination_path.$filename);

Image is not downloading using file_get_contents

What if you change the fopen() to:

$file = fopen($destination, "wb+"); // binary writing

Let me know.

EDIT

Here's the code that will work:

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
ini_set('log_errors', 0);

$opts = array(
  'http'=>array(
    'method'=>"GET",
    'header'=>"Accept-language: en\r\n" .
        "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.6; rv:23.0) Gecko/20100101 Firefox/23.0\r\n" . 
    "Referer: http://www.funnyjunk.com\r\n"
  )
);

$image_url = 'http://static.fjcdn.com/pictures/The+smarterest.+Wow+a+description+I+could+say+anything+I_298f02_4766792.jpg';

$doc_root= $_SERVER['DOCUMENT_ROOT'];
$image_upload_path = $doc_root . '/';
$image_name2 = 'img-' . time() . '.jpg';

$destination  = $image_upload_path . $image_name2;

$data = file_get_contents($image_url, false, stream_context_create($opts));
//$file = fopen($destination, "w+");
//fputs($file, $data);
//fclose($file);
header("Content-type: image/jpeg");
echo $data;
?>

stream_context_create() with the referrer to the original site did the trick.

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