Try to echo the $target_url var. What's on it?



Try to echo the $args var. What's on it?



How do you pass the $args?





São Paulo/Brazil



-----Mensagem original-----
De: Miller, Terion [] 
Enviada em: sexta-feira, 8 de maio de 2009 10:27
Para: Jônatas Zechim; Miller, Terion; PHP-General List
Assunto: Re: RES: [PHP] CURL error help





On 5/8/09 8:04 AM, "Jônatas Zechim" <> wrote:


Try to change this:






curl_setopt($ch,CURLOPT_URL, $target_url);



São Paulo/Brazil


Thanks for the suggestion:

Tried it and get this error now:


Warning: file_get_contents($target_url) [function.file-get-contents]: failed
to open stream: No such file or directory in
on line 39


On line #39 from my dom.php file is this :

// get html dom form file

function file_get_html() {

   $dom = new simple_html_dom;

   $args = func_get_args();

   $dom->load(call_user_func_array('file_get_contents', $args), true);

   return $dom;



I think somehow the array is not being read or it's trying to pass the full
array of url's as one url? Therefore saying it does not exist? My best
guess, am I making the array wrong?


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