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php怎样挪用api接口【php题目】

作者:搜搜PHP网发布时间:2019-11-26分类:PHP问题浏览:101


导读:经由过程php模仿post要求即可挪用。引荐:php服务器php模仿POST提交的2种要领1、经由过程curl函数PHP中运用cURL完成Get和Post要...

经由过程php模仿post要求即可挪用。

引荐:php服务器

php模仿POST提交的2种要领

1、经由过程curl函数

PHP中运用cURL完成Get和Post要求的要领

$post_data = array();  
$post_data['clientname'] = "test08";  
$post_data['clientpasswd'] = "test08";  
$post_data['submit'] = "submit";  
$url='http://xxx.xxx.xxx.xx/xx/xxx/top.php';  
$o="";  
foreach ($post_data as $k=>$v)  
{  
    $o.= "$k=".urlencode($v)."&";  
}  
$post_data=substr($o,0,-1);  
$ch = curl_init();  
curl_setopt($ch, CURLOPT_POST, 1);  
curl_setopt($ch, CURLOPT_HEADER, 0);  
curl_setopt($ch, CURLOPT_URL,$url);  
//为了支撑cookie  
curl_setopt($ch, CURLOPT_COOKIEJAR, 'cookie.txt');  
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data);  
$result = curl_exec($ch);

2、经由过程fsockopen.

fsockopen — 翻开一个网络衔接或许一个Unix套接字衔接

$URL=‘http://xxx.xxx.xxx.xx/xx/xxx/top.php';  
$post_data['clientname'] = "test08";  
$post_data['clientpasswd'] = "test08";  
$post_data['submit'] = "ログイン";  
$referrer="";  
// parsing the given URL  
$URL_Info=parse_url($URL);  
// Building referrer  
if($referrer=="") // if not given use this script as referrer  
$referrer=$_SERVER["SCRIPT_URI"];  
   
// making string from $data  
foreach($post_data as $key=>$value)  
$values[]="$key=".urlencode($value);  
   
$data_string=implode("&",$values);  
// Find out which port is needed - if not given use standard (=80)  
if(!isset($URL_Info["port"]))  
$URL_Info["port"]=80;  
// building POST-request:  
$request.="POST ".$URL_Info["path"]." HTTP/1.1\n";  
$request.="Host: ".$URL_Info["host"]."\n";  
$request.="Referer: $referrer\n";  
$request.="Content-type: application/x-www-form-urlencoded\n";  
$request.="Content-length: ".strlen($data_string)."\n";  
$request.="Connection: close\n";  
$request.="\n";  
$request.=$data_string."\n";  
$fp = fsockopen($URL_Info["host"],$URL_Info["port"]);  
fputs($fp, $request);  
while(!feof($fp)) {  
    $result .= fgets($fp, 128);  
}  
fclose($fp);

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标签:phpapi接口