


How to post data from a php file to another address without using forms and hidden variables
Dec 23, 2016 pm 04:30 PM可以使用以下函數(shù)來實現(xiàn):??
function?posttohost($url,?$data)?{?
$url?=?parse_url($url);?
if?(!$url)?return?"couldn't?parse?url";?
if?(!isset($url['port']))?{?$url['port']?=?"";?}?
if?(!isset($url['query']))?{?$url['query']?=?"";?}??
$encoded?=?"";??
while?(list($k,$v)?=?each($data))?{?
$encoded?.=?($encoded???"&"?:?"");?
$encoded?.=?rawurlencode($k)."=".rawurlencode($v);?
}??
$fp?=?fsockopen($url['host'],?$url['port']???$url['port']?:?80);?
if?(!$fp)?return?"Failed?to?open?socket?to?$url[host]";??
fputs($fp,?sprintf("POST?%s%s%s?HTTP/1.0n",?$url['path'],?$url['query']???"?"?:?"",?$url['query']));?
fputs($fp,?"Host:?$url[host]n");?
fputs($fp,?"Content-type:?application/x-www-form-urlencodedn");?
fputs($fp,?"Content-length:?"?.?strlen($encoded)?.?"n");?
fputs($fp,?"Connection:?closenn");??
fputs($fp,?"$encodedn");??
$line?=?fgets($fp,1024);?
if?(!eregi("^HTTP/1..?200",?$line))?return;??
$results?=?"";?$inheader?=?1;?
while(!feof($fp))?{?
$line?=?fgets($fp,1024);?
if?($inheader?&&?($line?==?"n"?||?$line?==?"rn"))?{?
$inheader?=?0;?
}?
elseif?(!$inheader)?{?
$results?.=?$line;?
}?
}?
fclose($fp);??
return?$results;?
}?
?>?
--------------------------------------------------------------------------------------------------?
也可以這樣??
$URL="www.mysite.com/test.php";?
$ch?=?curl_init();?
curl_setopt($ch,?CURLOPT_URL,"https://$URL");?
curl_setopt($ch,?CURLOPT_POST,?1);?
curl_setopt($ch,?CURLOPT_POSTFIELDS,?"Data1=blah&Data2=blah");?
curl_exec?($ch);?
curl_close?($ch);?
?>?
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