Java Jackson未处理的异常?
我是android编程新手,我一直在关注这一点 创建GCM服务器程序。然而,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课程:Java Jackson未处理的异常?,java,android,jackson,google-cloud-messaging,unhandled-exception,Java,Android,Jackson,Google Cloud Messaging,Unhandled Exception,我是android编程新手,我一直在关注这一点 创建GCM服务器程序。然而,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课程: import java.io.BufferedReader; import java.io.DataOutputStream; import java.io.IOException; import java.io.InputStreamReader; import java.net.HttpURLConnection; import java.
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,但是我得到了错误:
未处理的异常:com.fasterxml.jackson.databind.JsonMappingException
在线mapper.writeValue(wr,content)代码>
导致此异常的原因是什么?如何修复它?是一个通用数据绑定包,可用于流式API(jackson core)实现。这就是为什么需要添加并捕获3个异常。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();
}
希望它对您有用。如在它周围添加一个尝试捕获?当我这样做的时候,我得到了一个不兼容类型错误,上面写着:Required:java.lang.Throwable-Found:com.fasterxml.jackson.databind.JsonMappingException谢谢你的帮助,但是我仍然得到了“JsonMappingException”的一个不兼容类型错误。你添加了jackson内核吗?对不起,我是新手,我只知道如何添加JAR,但jackson core看起来像一个项目。不,你也可以为jackson core
安装JAR。这会解决你的问题!!