使用FasterXML将字符串格式的日期转换为Joda DateTime

Question

public static void main( String[] args )
    throws JsonParseException, JsonMappingException, IOException {

    final String INPUT_JSON = "{\"start\" : \"11/23/2016\"}";
    // final String INPUT_JSON = "{\"start\" : \"1972-12-28T12:00:01.000Z\"}";
    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.registerModule( new JodaModule() );
    DateFormat dateFormat = new SimpleDateFormat( "MM/dd/yyyy" );
    objectMapper.setDateFormat( dateFormat );
    Bean bean = objectMapper.readValue( INPUT_JSON, Bean.class );
    DateTime start = bean.getStart();
    System.out.println( start );
}

public class Bean {
   public DateTime start;

   public Bean() {
    // TODO Auto-generated constructor stub
   }

   public DateTime getStart() {
    return start;
  }

  public void setStart( DateTime start ) {
    this.start = start;
  }
}

Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Invalid format: "11/23/2016" is malformed at "/23/2016" (through reference chain: com.firstfuel.analytics.bpu.translator.Bean["start"])
at com.fasterxml.jackson.databind.JsonMappingException.wrapWithPath(JsonMappingException.java:232)
at com.fasterxml.jackson.databind.JsonMappingException.wrapWithPath(JsonMappingException.java:197)
at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.wrapAndThrow(BeanDeserializerBase.java:1420)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:244)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:118)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:2986)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2091)
at com.firstfuel.analytics.bpu.translator.VTModelPostProcessRequestTranslator.main(VTModelPostProcessRequestTranslator.java:125)

Caused by: java.lang.IllegalArgumentException: Invalid format: "11/23/2016" is malformed at "/23/2016"
at org.joda.time.format.DateTimeFormatter.parseMillis(DateTimeFormatter.java:754)
at org.joda.time.convert.StringConverter.getInstantMillis(StringConverter.java:65)
at org.joda.time.base.BaseDateTime.<init>(BaseDateTime.java:150)
at org.joda.time.DateTime.<init>(DateTime.java:265)
at com.fasterxml.jackson.datatype.joda.deser.DateTimeDeserializer.deserialize(DateTimeDeserializer.java:48)
at com.fasterxml.jackson.datatype.joda.deser.DateTimeDeserializer.deserialize(DateTimeDeserializer.java:20)
at com.fasterxml.jackson.databind.deser.SettableBeanProperty.deserialize(SettableBeanProperty.java:525)
at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:99)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:242)
... 4 more

Answer 1

我知道这个问题已经得到了回答，但我想与您分享一种解决这个问题的替代方法，而不需要这么多行代码。只需将@JSONFormat注释添加到pojo属性。

@JsonFormat( pattern="MM/dd/yyyy")
public DateTime start;

并序列化时间部分

@JsonFormat(pattern = "dd-MM-yyyy hh:mm:ss")
public DateTime field = new DateTime();

您还需要将JodaModule注册到jackson以利用它。

final ObjectMapper result = new ObjectMapper();  
result.registerModule(new JodaModule());