使用 gson 从 URL 动态加载 JSON 元数据

Question

如何将生成的输出"put" 看起来有效 JSON 转换为实际的 JSON 对象？

根据this answer，gson 要求定义class。肯定有一种动态方式来加载或定义 class？还是通用类型？

在 XML 中，模式或 DTD 将可用。我可以使用 gson 为 JSON、 动态加载或找到类似的东西吗？

代码：

package net.bounceme.noagenda;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.MalformedURLException;
import java.net.URL;
import java.util.ArrayList;
import java.util.List;
import static java.lang.System.out;

public class NoAgenda {

    public static void main(String[] args) throws MalformedURLException, IOException {
        List<URL> urls = new ArrayList<>();
        new NoAgenda().iterateURLs(urls);
    }

    private void iterateURLs(List<URL> urls) throws MalformedURLException, IOException {
        urls.add(new URL("https://www.flickr.com/photos/"));
        urls.add(new URL("http://www.javascriptkit.com/dhtmltutors/javascriptkit.json"));
        urls.add(new URL("http://api.wunderground.com/api/54f05b23fd8fd4b0/geolookup/conditions/forecast/q/US/CO/Denver.json"));
        for (URL url : urls) {
            connect(url);
        }
    }

    private void connect(URL url) throws IOException {
        out.println(url);
        String line = null;
        StringBuilder sb = new StringBuilder();
        BufferedReader in = new BufferedReader(
                new InputStreamReader(url.openStream()));
        while ((line = in.readLine()) != null) {
            sb.append(line + "\n");
        }
        in.close();
        out.println(sb);

        //      HOW DO I TURN THIS INTO AN ACTUAL JSON??
    }
}

维基百科says:

Schema and metadata

JSON Schema

JSON Schema[20] specifies a JSON-based format to define the structure of JSON data for validation, documentation, and interaction control. A JSON Schema provides a contract for the JSON data required by a given application, and how that data can be modified.

我正在使用的任意三个 URL：

https://www.flickr.com/photos/

http://www.javascriptkit.com/dhtmltutors/javascriptkit.json

http://api.wunderground.com/api/54f05b23fd8fd4b0/geolookup/conditions/forecast/q/US/CO/Denver.json

Answer 1

您可以使用以下内容

StringBuilder sb = new StringBuilder();

String line;
br = new BufferedReader(new InputStreamReader(is));
while ((line = br.readLine()) != null) {
    sb.append(line);
}
JSONObject json = new JSONObject(sb.toString());