请问，php怎么获取远程JSon内容并post一些参数

$data = file_get_contents($url);//目的页面内容获取
$t = json_decode($data,1);//转换为PHP数组
//处理...
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $urlo);//数据发送地址
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data);//发送的数据数组
curl_exec($ch);

------解决方案--------------------

PHP code


function requrest($url,$posts)
{
      if(is_array($posts) && !empty($posts))
      {
            foreach($posts as $key=>$value)
            {
                $post[] = $key.'='.urlencode($value);
            }
            $posts = implode('&',$post);
       }

      $curl = curl_init();

      $options = array( 
                     CURLOPT_URL=>$url,
                     CURLOPT_CONNECTTIMEOUT => 2,
                     CURLOPT_TIMEOUT => 10,
                     CURLOPT_RETURNTRANSFER => true,
                     CURLOPT_POST => 1,
                     CURLOPT_POSTFIELDS=>$posts,
                     CURLOPT_USERAGENT=>'Mozilla/5.0 (Windows NT 5.1; rv:5.0) Gecko/20100101 Firefox/5.0'
                    );          

                   curl_setopt_array($curl,$options);
                   
                   $retval = curl_exec($curl);

                   return $retval;
}

$posts = array('name'=>'root',
               'pwd'=>'123456'
               );

$retval = request($url,$posts);

if($retval !== false)
{
    $Arr = json_decode($retval,true);
   
}

------解决方案--------------------

如果你的服务器是linux或unix，也可以用工具来实现:

PHP code

system("curl -d 'name=xx&password=xxx' 'http://www.xx.com/xx.php' > ./tmp");

while(!file_exists('./tmp')){
  $jsoncode = file_get_contents('./tmp');
  sleep(1);
}

var_dump(json_decode($jsoncode));

------解决方案--------------------

1.可以用file_get_contents
参考视频：
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2009/0416/810.html


2.也可以用curl
参考视频：
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2010/0621/4795.html
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2010/0628/4848.html