类似这样的，请求SERVER该怎样实现呢？

[hisben]

谢谢各位给个思路

package com.uumap.mobile.sale.service;


import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.URL;

public class TestMobileGps {
public static void main(String[] args) {
try {
new TestMobileGps().testHtml();
new TestMobileGps().testWord();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
/**
* 文字返回
* @throws Exception
*/
public void testWord() throws Exception {
BufferedReader in = null;

String urlStr = server + "/requestword.action?x=121.43095048643&y=31.219888317383";
URL url = new URL(urlStr);
in = new BufferedReader(
new InputStreamReader(url.openStream()));
StringBuffer sb = new StringBuffer();
for (;;) {
int c = in.read();
if (c == -1)
break;
sb.append((char) c);
}
System.out.println("sb="+sb.toString());


}
/**
* 图片返回
* @throws Exception
*/
public void testPic() throws Exception {
BufferedReader in = null;

String urlStr = server + "/requestpic.action?x=121.43095048643&y=31.219888317383";
URL url = new URL(urlStr);
in = new BufferedReader(
new InputStreamReader(url.openStream()));
StringBuffer sb = new StringBuffer();
for (;;) {
int c = in.read();
if (c == -1)
break;
sb.append((char) c);
}
System.out.println("sb="+sb.toString());


}
/**
* 网页返回
* @throws Exception
*/
public void testHtml() throws Exception {
BufferedReader in = null;

String urlStr = server + "/requesthtml.action?x=121.43095048643&y=31.219888317383&mobileNo=13698569608&requestId=12345";
URL url = new URL(urlStr);
in = new BufferedReader(
new InputStreamReader(url.openStream()));
StringBuffer sb = new StringBuffer();
for (;;) {
int c = in.read();
if (c == -1)
break;
sb.append((char) c);
}
System.out.println("sb="+sb.toString());

}

private String server = "http://www.test1.uumap.net";
}