如何从一个php文件向另一个地址post数据，不用表单和隐藏的变量

LAMP-兄弟连

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/1.0\n", $url['path'], $url['query'] ? "?" : "", $url['query']));
fputs($fp, "Host: $url[host]\n");
fputs($fp, "Content-type: application/x-www-form-urlencoded\n");
fputs($fp, "Content-length: " . strlen($encoded) . "\n");
fputs($fp, "Connection: close\n\n");

fputs($fp, "$encoded\n");

$line = fgets($fp,1024);
if (!eregi("^HTTP/1\.. 200", $line)) return;

$results = ""; $inheader = 1;
while(!feof($fp)) {
$line = fgets($fp,1024);
if ($inheader && ($line == "\n" || $line == "\r\n")) {
$inheader = 0;
}
elseif (!$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);
?>

Digico

用～GET喽

james.liu

好好好～～～～～～～～～～～～～～～～～～难

===false

panzt

hubinhust

恩，
可以打开一个socket。直接写如Post Data

mz198424

phpfan

原帖由 hubinhust 于 2008-12-20 20:50 发表
恩，
可以打开一个socket。直接写如Post Data

正解