doutangu4671
doutangu4671
2013-09-20 06:14

图像不是使用file_get_contents下载的

已采纳

I am using the below mentioned code to download image.

$image_url='http://www.abcd.com/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);
$file = fopen($destination, "w+");
fputs($file, $data);
fclose($file);

The above mentioned code is working fine in development environment to download image from any website. The code also works in live site to download images from other website except only one site.

The image seems to be downloaded but whenever i am trying to open the image using any image viewer, an error message is being displayed which is given below.

Could not load image 'img-65456465.jpg'. Error interpreting JPEG image file (Not a JPEG file: starts with 0x3c 0x21)

I checked the error log as well but found no issue. I am not understanding why its image download functionality is not working with that particular website only.

Note: 1) I checked that the original image type hosted in that website is JPEG only.

EDIT: When i am trying to view the downloaded image using browser, i found the below mentioned error.

The image 'http://www.my-live-site.com/folder/img-65456465.jpg' can not be displayed because it contains error.

As per suggestion, i am providing the original website link. Website: http://www.funnyjunk.com/ Image: any image link from this website. For example http://static.fjcdn.com/pictures/The+smarterest.+Wow+a+description+I+could+say+anything+I_298f02_4766792.jpg

Any help on this will be appreciated.

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4条回答

  • dqwn64004 dqwn64004 8年前

    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
    " .
            "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.6; rv:23.0) Gecko/20100101 Firefox/23.0
    " . 
        "Referer: http://www.funnyjunk.com
    "
      )
    );
    
    $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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  • douzhan5058 douzhan5058 8年前

    Hey just open your url

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

    and refresh this page. you will redirected to someother page i think this is the problem image has been protected.

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  • drmcm84800 drmcm84800 8年前

    The link you provided isn't an image when you download it. When you open the image URL in your browser, you're redirected to the HTML page which contains the image. This is confirmed when I run the file command on this image, it tells me it is a HTML text document.

    piNAS:~#: file The+smarterest.+Wow+a+description+I+could+say+anything+I_298f02_4766792.jpg
    The+smarterest.+Wow+a+description+I+could+say+anything+I_298f02_4766792.jpg: HTML document text
    

    It's probably set that images can't be directly linked to, and then it redirects you to the HTML page.

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  • duanchifo2866 duanchifo2866 8年前

    your code is fine i ve tested with one new url working fine try this code

    <?php
    $image_url='http://blog.svnlabs.com/wp-content/uploads/2012/06/PHP-DOWNLOAD-FILES-BY-URL.png';
    
    $doc_root= $_SERVER['DOCUMENT_ROOT'];
    echo $doc_root;
    $image_upload_path = $doc_root . '/';
    $image_name2 = 'img-' . time() . '.jpg';
    
    $destination  = $image_upload_path . $image_name2;
    
    $data = file_get_contents($image_url);
    $file = fopen($destination, "w+");
    fputs($file, $data);
    fclose($file);
    echo $file;
    ?>
    
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