求php，远程抓取代码，谢谢

Question

抓取网站http://www.131458.com
通过 POST查询提交到http://www.131458.com/handler/load.aspx/Load
传递参数nick=z224324
得到图2的结果，请问curl怎么抓取图二的结果呀？
图一：

图二：

Answer 1

 $data="{'nick':'z224324'}";
$ch = curl_init();
curl_setopt($ch, CURLOPT_REFERER, 'http://www.131458.com/');
curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/31.0.1650.63 Safari/537.36');
curl_setopt($ch, CURLOPT_URL, 'http://www.131458.com/handler/load.aspx/Load');
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type:application/json;charset=utf-8'));
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
$content = curl_exec($ch);

$d=json_decode($content);

echo $d->d;