PHP downloading Image from a URL

Question

I have a TXT files with URLs to images on every new file. I am reading the files and fetching each new URL and fecthcing the image and storing it locally as below.

$filename = 'images.txt';
displayTXTList($filename);

function displayTXTList($fileName) {
    if(file_exists($fileName)) {
        $file = fopen($fileName, 'r');
        while (!feof($file)) {
            $url = fgets($file);
            $image = basename($url);
            downloadImages($image);
            downloadImages(str_replace(".jpg", 'l.jpg ', $image));
            downloadImages(str_replace(".jpg", 's.jpg ', $image));
        }
        fclose($file);
    }
}

function downloadImages($image) {
    $ch = curl_init('http://i.imgur.com/'.$image);
    $fp = fopen('images/'.$image, 'wb');
    curl_setopt($ch, CURLOPT_FILE, $fp);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_exec($ch);
    curl_close($ch);
    fclose($fp);
    //file_put_contents('images/'.$i, file_get_contents('http://i.imgur.com/'.$i));
}

I am getting the following error

Warning: fclose() expects parameter 1 to be resource, boolean given in C:\wamp\www\temp\nomad\index.php on line 45
Warning: fopen(images/BrRdRl.jpg ) [function.fopen]: failed to open stream: Invalid argument in C:\wamp\www\temp\nomad\index.php on line 40
Warning: curl_setopt(): supplied argument is not a valid File-Handle resource in C:\wamp\www\temp\nomad\index.php on line 41

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When i shorten the list of images to 2-3. I see that only the last image is being downloaded.

sachleen · Accepted Answer · 2013-01-15 08:24:57Z

2

You have a space in the file name:

downloadImages(str_replace(".jpg", 'l.jpg ', $image));
downloadImages(str_replace(".jpg", 's.jpg ', $image));
                                         ^ here

There shouldn't be a space after jpg

answered Jan 15, 2013 at 8:24

sachleen

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1 Comment

Harsha M V Over a year ago

i changed that still i have a problem. When i shorten the list of images to 2-3. I see that only the last image is being downloaded.

Hanky Panky · Accepted Answer · 2013-01-15 08:24:02Z

1

$fp = fopen('images/'.$image, 'wb');

This call is resulting in failure and resource handle is not being generated

answered Jan 15, 2013 at 8:24

Hanky Panky

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2 Comments

Harsha M V Over a year ago

how to solve it ? i can see that from the line that is pointing to this code

Hanky Panky Over a year ago

@HarshaMV Above that call, temporarily echo this echo 'i.imgur.com/'.$image; (with http) This will explain whats wrong in the url

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