fsockopen不能正确获取网页，返回403错误

Question

又是appid 又是 login_sig...我们没办法给你做测试的，除非有人以前做过并且成功了

Answer 1

直接浏览器访问，得
ptui_checkVC('0','!YZG','\x00\x00\x00\x00\x16\x0b\xcd\xd4');
显然是一个 jsonp 应用
echo file_get_contents('http://check.ptlogin2.qq.com/check?regmaster=&uin=369872340&appid=549000912&js_ver=10049&js_type=1&login_sig=4aNczek7pQ2h04JL8nS0xfDqPQsJ-oF5czyfk1lOAY47-HswQg3rmElsCZ6QIAyO&u1=http%3A%2F%2Fqzs.qq.com%2Fqzone%2Fv5%2Floginsucc.html%3Fpara%3Dizone&r=0.42962526359176423');
得
ptui_checkVC('1','FT8RVzgWagPfnzHuLXG9LOxegxgkseUV','\x00\x00\x00\x00\x16\x0b\xcd\xd4');....

Answer 2

为什么file_get_contents可以正确获得而fsockopen不行呢？能不能用fsockopen搞定呢？我需要fsockopen的方案

Answer 3

$host = 'check.ptlogin2.qq.com';
$param = '/check?regmaster=&uin=369872340&appid=549000912&js_ver=10049&js_type=1&login_sig=4aNczek7pQ2h04JL8nS0xfDqPQsJ-oF5czyfk1lOAY47-HswQg3rmElsCZ6QIAyO&u1=http%3A%2F%2Fqzs.qq.com%2Fqzone%2Fv5%2Floginsucc.html%3Fpara%3Dizone&r=0.42962526359176423';$out = array(
  "GET $param HTTP/1.0",
  "Host: $host",
  'Connection: Close',
  '', ''
);
$fp = fsockopen($host, 80, $errer, $errno, 1);
fwrite($fp, join("\r\n", $out) );
while(! feof($fp)){
  echo fread($fp, 1024);
}
HTTP/1.1 200 OK
Server: tencent http server
Pragma: No-cache
P3P: CP="CAO PSA OUR"
Set-Cookie: confirmuin=0; PATH=/; DOMAIN=ptlogin2.qq.com;
Set-Cookie: ptvfsession=ec74c1fdab7c67a28e60e26b39ca2cd789648df0341f311ebd4a970791b4e4af72576feb6fc4abf6e403d3a52ff4ccd5; PATH=/; DOMAIN=ptlogin2.qq.com;
Set-Cookie: ptisp=ctc; PATH=/; DOMAIN=qq.com;
Connection: close
Content-Type: application/x-javascript; charset=utf-8ptui_checkVC('0','!WKV','\x00\x00\x00\x00\x16\x0b\xcd\xd4');

Answer 4

<Directory />
     Options FollowSymLinks
     AllowOverride None
     Order deny,allow
     Deny from all
</Directory>你这东西是如何配置的

Answer 5

太感谢版主们了，谢谢啊！特别是xuzuning 问题已经解决了。原来HOST和Host是两码事。看了xuzuning 的代码，把我原来的HOST改为Host就能正常访问了。特此说明，给将来遇到类似问题的童鞋参考参考