晚上好大家
我想在Java中使用Socket类来获取一个网页,我已经做到了这一点作为使用Java Socket类
import java.net.*;
import java.io.*;
class htmlPageFetch{
public static void main(String[] args){
try{
Socket s = new Socket("127.0.0.1", 80);
DataInputStream dIn = new DataInputStream(s.getInputStream());
DataOutputStream dOut = new DataOutputStream(s.getOutputStream());
dOut.write("GET /index.php HTTP/1.0\n\n".getBytes());
boolean more_data = true;
String str;
while(more_data){
str = dIn.readLine();
if(str==null)
more_data = false;
System.out.println(str);
}
}catch(IOException e){
}
}
}
获取一个网页,但它只是给空的。
输出
HTTP/1.1 302 Found
Date: Wed, 01 Dec 2010 13:49:02 GMT
Server: Apache/2.2.11 (Unix) DAV/2 mod_ssl/2.2.11 OpenSSL/0.9.8k PHP/5.2.9 mod_apreq2-20051231/2.6.0 mod_perl/2.0.4 Perl/v5.10.0
X-Powered-By: PHP/5.2.9
Location: http://localhost/xampp/
Content-Length: 0
Content-Type: text/html
null
雅拉兹这是真的,但我想这样做使用套接字,因为它是我的任务分析收到的输出 – codeomnitrix 2010-12-01 13:59:28