杰克逊：如何忽略数据类型不匹配的领域

Question

我在gent一个JSON响应在春天休息模板，并试图将其强制转换为自定义对象。

@JsonIgnoreProperties(ignoreUnknown = true) 
public class DataItems { 

private String deviceId; 
    private String dataSourceName; 
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss.ms+0000") 
    private Date timestamp; 
    private String mName; 
    private String mVersion; 
    private double value; 


    /** 
    * constructors and getters and setters 
    */ 

现在，有时候，这个双重值我得到作为字符串的垃圾值。

那么，如何在投射时忽略这些记录。以下是我的铸造代码。

ObjectMapper mapper = new ObjectMapper(); 
DataItems dataItems = mapper.readValue(/*String*/json),DataItems.class); 

样品JSON：

[{ 
     "value": 26.394289, 
     "deviceId": "97572545-3768-6070-0B09-1A566D8F6BC1", 
     "dataSourceName": "AQM", 
     "timestamp": "2016-11-04T05:24:55.000+0000", 
     "mName": "SENS_TEMPERATURE", 
     "mVersion": "1.0" 
     }, 
     { 
     "value": 26.396435, 
     "deviceId": "97572545-3768-6070-0B09-1A566D8F6BC1", 
     "dataSourceName": "AQM", 
     "timestamp": "2016-11-04T05:25:58.000+0000", 
     "mName": "SENS_TEMPERATURE", 
     "mVersion": "1.0" 
     }, 
     { 
     "value": "{\"timestamp\": 1478220851, \"metric\": \"SENS_TAMPER\", \"value\": \"1\", \"source\": \"AQM\"}", 
     "deviceId": "97572545-3768-6070-0B09-1A566D8F6BC1", 
     "dataSourceName": "AQM", 
     "timestamp": "2016-11-04T00:54:11.000+0000", 
     "mName": "SENS_UNREGISTERED", 
     "mVersion": "1.0" 
     }] 

Answer 1

自定义解串器可以解决这个问题。如果double值无法反序列化，则自定义解串器将设置默认值“0.0”。

更改POJO： -

@JsonDeserialize(using = DoubleDeserializer.class) 
private Double value; 

DoubleDeserializer - 自定义解串器： -

public class DoubleDeserializer extends JsonDeserializer<Double> { 

    @Override 
    public Double deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) 
      throws IOException, JsonProcessingException { 

     String doubleValue = jsonParser.readValueAs(String.class); 

     try { 
      return Double.valueOf(doubleValue); 
     } catch (NumberFormatException ne) { 
      System.out.println("Default value is set as the value in json is not compatible with double value"); 
      return new Double(0);   
     } 

    } 
} 

Answer 2

您可以使用类型JsonNode，而不是double：

@JsonIgnoreProperties(ignoreUnknown = true) 
public class DataItems { 
    private String deviceId; 
    private String dataSourceName; 
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss.ms+0000") 
    private Date  timestamp; 
    private String mName; 
    private String mVersion; 

    private JsonNode value; 

    [...] // getters, setters 
} 

现在，你可以尝试get a double from JsonNode-Object：

double value = data.getValue().asDouble(0.0); // 0.0 is default 

使用这种方式，您将不松其他非双数据：

{ 
    "value": "{\"timestamp\": 1478220851, \"metric\": \"SENS_TAMPER\", \"value\": \"1\", \"source\": \"AQM\"}", 
    [...] 
    }