Java中的HTTP客户端连接
问题描述:
我正在尝试使用套接字进行HTTPClient连接,但无法弄清楚。当我运行代码时,我收到以下消息;Java中的HTTP客户端连接
HTTP/1.1 400 Bad Request
Date: Sun, 22 Apr 2012 13:17:12 GMT
Server: Apache/2.2.16 (Debian)
Vary: Accept-Encoding
Content-Length: 317
Connection: close
Content-Type: text/html; charset=iso-8859-1
<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<html><head>
<title>400 Bad Request</title>
</head><body>
<h1>Bad Request</h1>
<p>Your browser sent a request that this server could not understand.<br />
</p>
<hr>
<address>Apache/2.2.16 (Debian) Server at pvs.ifi.uni-heidelberg.de Port 80</address>
</body></html>
的代码是在这里:
import java.net.*;
import java.io.*;
public class a1 {
public static void main (String[] args) throws IOException {
Socket s = null;
try {
String host = "host1";
String file = "file1";
int port = 80;
在这里,我创建套接字:
s = new Socket(host, port);
OutputStream out = s.getOutputStream();
PrintWriter outw = new PrintWriter(out, false);
outw.print("GET " + file + " HTTP/1.1\r\n");
outw.print("Accept: text/plain, text/html, text/*\r\n");
outw.print("\r\n");
outw.flush();
InputStream in = s.getInputStream();
InputStreamReader inr = new InputStreamReader(in);
BufferedReader br = new BufferedReader(inr);
String line;
while ((line = br.readLine()) != null) {
System.out.println(line);
}
}
catch (UnknownHostException e) {}
catch (IOException e) {}
if (s != null) {
try {
s.close();
}
catch (IOException ioEx) {}
}
}
}
任何帮助将不胜感激。谢谢。
答
如果你真的想编写自己的HTTP客户端(难度比它看起来,但很教育性),你缺少的HTTP 1.1的要求Host
头:
outw.print("Host: " + host + ":" + port + "\r\n");
见RFC 2616, section 14.23. Host:
客户端必须在所有HTTP/1.1请求消息中包含一个主机头字段。
所以你的要求不好,它错过要求Host
头。
试试这个:http://stackoverflow.com/questions/1359689/how-to-send-http-request-in-java/1359700#1359700 – duffymo 2012-04-22 13:26:04
或者你可以使用Apache Commons http。 – bmargulies 2012-04-22 13:26:41
如果你真的想编写自己的HTTP客户端(如果这不仅仅是为了好玩,那么不要看上面的注释),你应该首先尝试HTTP 1.0并在请求的路径之前添加一个斜杠:' “GET /”+文件+“HTTP/1.0 \ r \ n”'。 – 2012-04-22 13:32:23