Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
445 views
in Technique[技术] by (71.8m points)

java - Send HTTP request manually via socket

When I send a normal HTTP request via a socket, the server does not respond with an OK response. I copied the HTTP header from Firefox. Here is the code:

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();

However, here is the response I received:

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>

I know that I can do this by using URL.openStream(), but why doesn't the server identify the HTTP request when I send it manually?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Two things:

  1. You should use println instead of print to print your entries to separate lines.
  2. HTTP request should end in a blank line (link). So add pw.println("");

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...