Getting Php Curl Error - php

Question

I want to grab first 10 results of that any url i passed it to function as parameter ..and i want to make data scraper for some sites
Getting Syntax error when I print the result on screen
syntax error on this line I don't why it giving me syntax error kindly help me....
print_r( $dse->crawl()->parse() );
<?php
class CURL_CRAWLER{
public $url;
public $request_type;
public $data;
public $post_params;
function __construct($url = '' , $request_type = 'GET')
{
$this->url = $url;
$this->request_type = $request_type;
$this->data = '';
$this->post_params = array();
}
/**crawl a document **/
function crawl()
{
$curl = curl_init( $this->url );
curl_setopt($curl, CURLOPT_HEADER, false);
curl_setopt($curl, CURLOPT_TIMEOUT, 60);
curl_setopt($curl, CURLOPT_USERAGENT, 'cURL PHP');
curl_setopt($curl, CURLOPT_RETURNTRANSFER, TRUE);
$this->data = curl_exec($curl);
curl_close($curl);
return $this; //make it a chainable method
}
/** Parse result data **/
function parse(){
$result = array();
$count = 0;
$dom = new DOMDocument;
$dom->preserveWhiteSpace = false;
$dom->loadHTML($this->data);
$xpath = new DOMXPath($dom);
$news = $xpath->query('//td[#bgcolor="#DDDDDD"]/table/tr[position()=2]/td[position()=2]');
foreach( $news as $n){
$result[] = $n->nodeValue;
$count++;
if ($count >9)
break; //we just need 10 results. Index starts from 0
}
return $result;
}
}
error_reporting(0);
$dse = new CURL_CRAWLER('http://www.dsebd.org/display_news.php');
echo "<pre>";
print_r( $dse->crawl()->parse() );
echo "<pre>";
?>

Answer 1

Your syntax error is that you should use explicit "greater than" sign instead of HTML entities > - server doesn't need those, it is not a browser that can render it correctly. Just change:
print_r( $dse->crawl()->parse() );
^^^^ ^^^^
to:
print_r( $dse->crawl()->parse() );