請教,php如何擷取遠程JSon內容 並post一些參數
請教,php如何擷取遠程JSon內容 並post一些參數
目的: php請求遠程php頁面(頁面是json內容),提交一些參數(例如name,pwd欄位),將json返回
希望大家給個例子
------解決方案--------------------
PHP code
$data = file_get_contents($url);//目的頁面內容擷取$t = json_decode($data,1);//轉換為PHP數組//處理...$ch = curl_init();curl_setopt($ch, CURLOPT_URL, $urlo);//資料發送地址curl_setopt($ch, CURLOPT_POST, 1);curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data);//發送的資料數組curl_exec($ch);
------解決方案--------------------
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); }
------解決方案--------------------
如果你的伺服器是linux或unix,也可以用工具來實現:
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));
------解決方案--------------------
1.可以用file_get_contents
參考視頻:
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2009/0416/810.html
2.也可以用curl
參考視頻:
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2010/0621/4795.html
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/2010/0628/4848.html