如何解析或拆分Java中的URL地址？

Question

如果我有 url 地址。

https://graph.facebook.com/me/home?limit=25&since=1374196005

我可以獲取（或拆分）參數（避免硬編碼）嗎？

像這樣

https /// graph.facebook.com /// me/home /// {limit=25, sincse=1374196005}

Answer 1

使用 Android 的Uri類。 http://developer.android.com/reference/android/net/Uri.html

Uri uri = Uri.parse("https://graph.facebook.com/me/home?limit=25&since=1374196005");
String protocol = uri.getScheme();
String server = uri.getAuthority();
String path = uri.getPath();
Set<String> args = uri.getQueryParameterNames();
String limit = uri.getQueryParameter("limit");

Answer 2

對於純 Java ，我認為這段代碼應該可以工作：

import java.net.URL;
import java.net.URLDecoder;
import java.util.HashMap;
import java.util.Map;

public class UrlTest {
    public static void main(String[] args) {
        try {
            String s = "https://graph.facebook.com/me/home?limit=25&since=1374196005";
            URL url = new URL(s);
            String query = url.getQuery();
            Map<String, String> data = new HashMap<String, String>();
            for (String q : query.split("&")) {
                String[] qa = q.split("=");
                String name = URLDecoder.decode(qa[0]);
                String value = "";
                if (qa.length == 2) {
                    value = URLDecoder.decode(qa[1]);
                }

                data.put(name, value);
            }
            System.out.println(data);
        } catch (Exception e) {
            e.printStackTrace();
        }

    }

}

Answer 3

kotlin

https://graph.facebook.com/me/home?limit=25&since=1374196005

import java.net.URI
import android.net.Uri

val uri = URI(viewEvent.qr)
val guid = Uri.parse(uri.fragment).getQueryParameter("c")

Answer 4

內置java.net.URL應該服務。

URL url = new URL("https://graph.facebook.com/me/home?limit=25&since=1374196005");

String[] parts = {
   url.getProtocol(), 
   url.getUserInfo(), 
   url.getHost(), 
   url.getPort(), 
   url.getPath(),
   url.getQuery(), 
   url.getRef() 
};