I am using the Thumbalizr API for capturing website screenshots.
I a开发者_如何学Pythonm trying to pass a url from a variable into the script below where it currently says "www.google.com"
.
When I try and do $url = "www.google.com";
and pass that to $image->request($url);
it seems to fail and retrieves a screenshot but one that suggests the url was unreachable.
Any what am I doing wrong?
Here is the original php in full working order.
<?php
define ("_THUMBALIZR",1);
require_once("config.inc.php"); // get config and functions
$image=new thumbalizrRequest(); // create object
$image->request("www.google.com"); // send request
if ($image->headers['Status']=="OK" || $image->headers['Status']=="LOCAL") { // if picture is available
$image->output(); //dump binary image data
} else {
print_r($image->headers); // print text result output - you can dump your own "queued" picture here
}
?>
If you show us your current code, IE the one with $url, that may help. In short, $url being used in place of "www.google.com" should not make a difference, as long as $url is being defined correctly and in the current scope (if inside of a function).
Show the current and we can better help you solve this issue.
It needs to be http://www.google.com :)
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