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

王朝php·作者佚名  2007-01-09
窄屏简体版  字體: |||超大  

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/1.0\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

<?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);

?>

 
 
 
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