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How to post data from a php file to another address without using forms and hidden variables_PHP Tutorial
- WBOYOriginal
- 2016-07-13 17:27:00941browse
可以使用以下函数来实现：
function posttohost($url, $data) {
$url = parse_url($url);
if (!$url) return "couldnt parse url";
if (!isset($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/1.0
", $url[path], $url[query] ? "?" : "", $url[query]));
fputs($fp, "Host: $url[host]
");
fputs($fp, "Content-type: application/x-www-form-urlencoded
");
fputs($fp, "Content-length: " . strlen($encoded) . "
");
fputs($fp, "Connection: close ");
fputs($fp, "$encoded ");
$line = fgets($fp,1024);
if (!eregi("^HTTP/1.. 200", $line)) return;
$results = ""; $inheader = 1;
while(!feof($fp)) {
$line = fgets($fp,1024);
if ($inheader && ($line == " " || $line == " ")) {
$inheader = 0;
}
elseif (!$inheader) {
$results .= $line;
}
}
fclose($fp);
return $results;
}
?>
也可以这样
$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);
?>