java - jackson 未处理的异常？

Question

我是 android 编程的新手，我正在关注这个 tutorial创建 GCM 服务器程序。但是，我遇到了一个令人沮丧的错误，非常感谢任何帮助。

这是我的 POST2GCM 类:

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import com.fasterxml.jackson.databind.ObjectMapper;


public class POST2GCM extends Content {

private static final long serialVersionUID = 1L;

public static void post(String apiKey, Content content){

        try{

        // 1. URL
        URL url = new URL("https://android.googleapis.com/gcm/send");

        // 2. Open connection
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();

        // 3. Specify POST method
        conn.setRequestMethod("POST");

        // 4. Set the headers
        conn.setRequestProperty("Content-Type", "application/json");
        conn.setRequestProperty("Authorization", "key="+apiKey);

        conn.setDoOutput(true);

            // 5. Add JSON data into POST request body

            //`5.1 Use Jackson object mapper to convert Contnet object into JSON
            ObjectMapper mapper = new ObjectMapper();

            // 5.2 Get connection output stream
            DataOutputStream wr = new DataOutputStream(conn.getOutputStream());

            // 5.3 Copy Content "JSON" into
            mapper.writeValue(wr, content);

            // 5.4 Send the request
            wr.flush();

            // 5.5 close
            wr.close();

            // 6. Get the response
            int responseCode = conn.getResponseCode();
            System.out.println("\nSending 'POST' request to URL : " + url);
            System.out.println("Response Code : " + responseCode);

            BufferedReader in = new BufferedReader(
                    new InputStreamReader(conn.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();

            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();

            // 7. Print result
            System.out.println(response.toString());

            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
}

我已经包含了“jackson-databind-2.5.1.jar”文件，但我收到了错误:

Unhandled Exception: com.fasterxml.jackson.databind.JsonMappingException

在线mapper.writeValue(wr, content);

是什么导致了这个异常，我该如何解决？

Answer 1

jackson-databind是一个通用数据绑定(bind)包，适用于流式 API (jackson-core) 实现。这就是为什么你需要添加 jackson-core并捕获 3 个异常。 writeValue方法抛出 IOException、JsonGenerationException 和 JsonMappingException。

try {
    mapper.writeValue(wr, content);
} catch (JsonMappingException e) {
    e.printStackTrace();
} catch (JsonGenerationException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
}

希望对你有用。