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
发布时间:2020-03-16 19:28:48 所属栏目:PHP教程 来源:互联网
导读: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
可以使用以下函数来实现: <?php 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; } ?> -------------------------------------------------------------------------------------------------- 也可以这样 <?php $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); ?> (编辑:焦作站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |