用Java构建URL

问题描述

尝试构建 http://IP:4567/foldername/1234?abc = xyz .我对此了解不多，但是我从Google搜索中写下了以下代码:

Trying to build http://IP:4567/foldername/1234?abc=xyz. I don't know much about it but I wrote below code from searching from google:

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();
        }
    }
}

我能够构建URL http://IP:port/foldername/1234?.我被困在查询部分.请帮助我前进.

I am able to build URL http://IP:port/foldername/1234?. I am stuck at query part. Please help me to move forward.

推荐答案

在一般的非Java术语中，URL是URI的一种特殊类型.您可以使用 URI 类(比古老的URL类(自Java 1.0开始就存在)更现代，可以更可靠地创建URI，您可以使用 toURL URI方法:

In general non-Java terms, a URL is a specialized type of URI. You can use the URI class (which is more modern than the venerable URL class, which has been around since Java 1.0) to create a URI more reliably, and you can convert it to a URL with the toURL method of URI:

String protocol = "http";
String host = "example.com";
int port = 4567;
String path = "/foldername/1234";
String auth = null;
String fragment = null;
URI uri = new URI(protocol, auth, host, port, path, query, fragment);
URL url = uri.toURL();

请注意，路径必须以斜杠开头.

Note that the path needs to start with a slash.

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