wkhtmltoimage and PHP debug

Question

I'm trying to use wkhtmltoimage in PHP to grab a screen. It's not working, but I don't know how to debug it. I contacted my server admins and they said that exec was allowed. Here's my directory structure:

-index.php
-lib/
  -screenshot.php
  -wkhtmltoimage/
     -wkhtmltoimage-i386
     -wkhtmltoimage-amd64

The code in screenshot.php is the following:

function screenshot($url, $save_path, $zoom = 0.5, $width = 500)
{
    $zoom = escapeshellarg($zoom);
    $width = escapeshellarg($width);
    $url = escapeshellarg($url);
    $save_path = escapeshellarg($shell_path);

    print_r(scandir("lib/wkhtmltoimage"));

    $output = array();

    echo exec("./lib/wkhtmltoimage/wkhtmltoimage-i386 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("lib/wkhtmltoimage/wkhtmltoimage-i386 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("./wkhtmltoimage/wkhtmltoimage-i386 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("wkhtmltoimage/wkhtmltoimage-i386 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("./lib/wkhtmltoimage/wkhtmltoimage-amd64 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("lib/wkhtmltoimage/wkhtmltoimage-amd64 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("./wkhtmltoimage/wkhtmltoimage-amd64 --enable-plugins --zoom $zoom --width $width $url test.png", $output);
    echo exec("wkhtmltoimage/wkhtmltoimage-amd64 --enable-plugins --zoom $zoom --width $width $url test.png", $output);

    print_r($output);
    die();
}

index.php calls the function screenshot on url "http://www.nasa.gov/".

Here's the output I get:

Array
(
    [0] => .
    [1] => ..
    [2] => wkhtmltoimage-amd64
    [3] => wkhtmltoimage-i386
)
Array
(
)

Visibly, I'm not even sure where the error comes from. It could be that the path is wrong, or that the input is wrong somehow, but I don't know how to get any output on these factors. Even the function doesn't seem to return any line of output through my array.

How can I debug this? Where do I start? Are there PHP functions for checking directories, or file execution?

Not sure if this will help, but I had to redirect stderror to stdout: $blah = shell_exec("$cmd 2>&1"); That might help you out. — Mark Steudel
– Mark Steudel, Commented Jun 16, 2011 at 17:24

Lazlo · Accepted Answer · 2011-06-20 03:14:33Z

3

Stupid mistake, I had to use the root path. (e.g. /home/...)

Does anyone also feel like half of the time in web development is wasted in solving path problems? Grrr.

answered Jun 20, 2011 at 3:14

Lazlo

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