请教，php如何获取远程JSon内容 并post一些参数
目的： php请求远程php页面（页面是json内容），提交一些参数（例如name，pwd字段），将json返回
希望大家给个例子
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- PHP code
$data = file_get_contents($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);
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- 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); }
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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));
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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