将内部JSON解析为Java POJO

Question

我有以下JSON，我想将其转换为Java POJO。

我使用com.fasterxml.jackson.databind.ObjectMapper作为对象。

    {  
   "deviceType":"firefox",
   "devices":{  
      "device#1234":{  
         "userName":"pcp",
         "rowId":12345
      }
   },
   "serviceId":12345,
   "message":{  
      "alert":"Hi, I am alert"
   }
}

Java POJO Class

public class TestDevices extends App {
    private String message;
    private JSONObject devices;
    private List<String> deviceKeys;
    private String deviceType;
    private String campaignId;

    public TestDevices() {
        // To make reflection happy
    }
    public TestDevices(long serviceId) {
        super(serviceId);
    }

    public TestDevices(long serviceId, String message, JSONObject devices, List<String> deviceKeys, String deviceType) {
        super(serviceId);
        this.message = message;
        this.devices = devices;
        this.deviceKeys = deviceKeys;
        this.deviceType = deviceType;
    }

    public String getCampaignId() {
        return campaignId;
    }
    public void setCampaignId(String campaignId) {
        this.campaignId = campaignId;
    }

    public String getMessage() {
        return message;
    }
    public void setMessage(String message) {
        this.message = message;
    }
    public JSONObject getDevices() {
        return devices;
    }
    public void setDevices(JSONObject devices) {
        this.devices = devices;
    }
    public List<String> getDeviceKeys() {
        return deviceKeys;
    }
    public void setDeviceKeys(List<String> deviceKeys) {
        this.deviceKeys = deviceKeys;
    }
    public String getDeviceType() {
        return deviceType;
    }
    public void setDeviceType(String deviceType) {
        this.deviceType = deviceType;
    }

    @Override
    public String toString() {
        return "Device [message=" + message + ", devices=" + devices + ", deviceKeys=" + deviceKeys + ", deviceType="
                + deviceType + ", serviceId=" + serviceId + "]";
    }
}

我的杰克逊召集人

public static TestDevices buildMockObject(String payload)
            throws JsonParseException, JsonMappingException, IOException {
        ObjectMapper mapper = new ObjectMapper();
        mapper.configure(Feature.AUTO_CLOSE_SOURCE, true);
        TestDevices devices = mapper.readValue(payload, TestDevices.class);
        return devices;
    }

但我收到此错误：

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "device#1234" (class org.json.JSONObject), not marked as ignorable (0 known properties: ])
 at [Source: {"deviceType":"firefox","devices":{"device#1234":{"userName":"pcp","rowId":12345}},"serviceId":12345,"message":{"alert":"Hi, I am alert"}}; line: 1, column: 51] (through reference chain: com.shephertz.karma.TestDevices["devices"]->org.json.JSONObject["device#1234"])

注意： 我上面给出的json中的devices和message个键包含具有动态键的json对象。

由于错误说它无法将JSON字段解析为我的java类。 我做错了什么以及如何解决上述错误？