Question

When I add an parameter to the CURL init function:

curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);

It does print the site content I’m sending a request to, the problem is that I want to manually parse the returned content ($response = curl_exec($ch);) but the problem is that, the site is displaying the page content and I want to keep having the site content on my $response variable so I could parse it, but at the same, stop it from displaying it.

Is that feasible?

The curl code:

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $action);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6");
curl_setopt($ch, CURLOPT_TIMEOUT, 60);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_COOKIEJAR, '');
curl_setopt($ch, CURLOPT_COOKIEFILE, '');
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $fields);
$response = curl_exec($ch);
curl_close($ch);

Answer 1

You should set the CURLOPT_RETURNTRANSFER flag to true. From PHP.NET manual

CURLOPT_RETURNTRANSFER TRUE

To return the transfer as a string of the return value of curl_exec() instead of outputting it out directly.

curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);