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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的_PHP教程

时间:2021-07-01 10:21:17 帮助过:3人阅读

可以使用以下函数来实现:

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);
?>


www.bkjia.comtruehttp://www.bkjia.com/PHPjc/317860.htmlTechArticle可以使用以下函数来实现: ?php functionposttohost($url,$data){ $url=parse_url($url); if(!$url)return"couldn'tparseurl"; if(!isset($url['port'])){$url['port']="";} if(!is...

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