Question

我正在尝试转换来自JSON的imbdapi的输出。我一直在研究如何做到这一点，但似乎无法弄清楚GSON。谢谢，对不起，我自己无法弄清楚，我的意思似乎并不像是在要求喂食。如果有人能提供一些帮助，我将非常感激。

public class Test {

    public static String getMovieInfo(String movie) {
        BufferedReader rd;
        OutputStreamWriter wr;
        //Scanner s = new Scanner(System.in);
        //System.out.println("Enter input:");
        //movie = s.nextLine();
        //movie = movie.replaceAll(" ", "%20");
        if (movie != null)
        {
            try {
                URL url = new URL("http://www.imdbapi.com/?i=&t=" + movie);
                URLConnection conn = url.openConnection();
                conn.setDoOutput(true);
                wr = new OutputStreamWriter(conn.getOutputStream());
                wr.flush();

                // Get the response
                rd = new BufferedReader(
                        new InputStreamReader(conn.getInputStream()));
                String line = rd.readLine();
                if (line != null) {
                   return line;
                } else {

                    return "Sorry! That's not a valid URL.";
                }
            } catch (UnknownHostException codeyellow) {
                System.err.println("Caught UnknownHostException: " + codeyellow.getMessage());
            }
            catch (IOException e)
            {
                System.out.println("Caught IOException:" + e.getMessage());
            }

        }
        else
        {
            return "passed parameter is null!";
        }

        return "an error occured, see console!";
    }
}

Answer 1

好的！我自己想出了这个，Quick-JSON的文档有点粗糙，但我想通了。 https://code.google.com/p/quick-json/如果有其他人想要它，感谢布兰登给我看。谢谢！

将JSON字符串转换为Java

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