当我通过一个套接字发送一个正常的HTTP请求时,服务器不响应一个OK响应。我复制了FireFox的HTTP标头。下面是代码:通过套接字手动发送HTTP请求
Socket s = new Socket(InetAddress.getByName("stackoverflow.com"), 80);
PrintWriter pw = new PrintWriter(s.getOutputStream());
pw.print("GET/HTTP/1.1");
pw.print("Host: stackoverflow.com");
pw.flush();
BufferedReader br = new BufferedReader(new InputStreamReader(s.getInputStream()));
String t;
while((t = br.readLine()) != null) System.out.println(t);
br.close();
然而,这里是我收到的回应:
HTTP/1.0 408 Request Time-out
Cache-Control: no-cache
Connection: close
Content-Type: text/html
<html><body><h1>408 Request Time-out</h1>
Your browser didn't send a complete request in time.
</body></html>
我知道,我可以用URL.openStream()
做到这一点,但为什么服务器不识别的HTTP请求当我手动发送它?
我想你把所有的报头之后发送一个额外的换行; 'pw.println();',并且使用'println()'作为头文件? – Torious
@Torious是的,这就是问题所在。谢谢:) –
对于HTTP,换行符必须是\ r \ n的格式。 – EJP