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在java中构建URL

试着建立http://IP:4567/foldername/1234?abc=xyz.我不太了解它,但我在下面写了从谷歌搜索的代码:

import java.net.MalformedURLException;
import java.net.URI;
import java.net.URL;

public class MyUrlConstruct {

    public static void main(String a[]){

        try {
            String protocol = "http";
            String host = "IP";
            int port = 4567;
            String path = "foldername/1234";
            URL url = new URL (protocol, host, port, path);
            System.out.println(url.toString()+"?");
        } catch (MalformedURLException ex) {
            ex.printStackTrace();
        }
    }
}
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我能够建立URL http://IP:port/foldername/1234?.我被困在查询部分.请帮助我前进.

java url build

29
推荐指数
5
解决办法
6万
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