jackson Unrecognized field

Question

I use jackson for converting JSON to Object class.

JSON:

{     "aaa":"111",     "bbb":"222",      "ccc":"333"  } 

Object Class:

class Test{     public String aaa;     public String bbb; } 

Code:

ObjectMapper mapper = new ObjectMapper(); Object obj = mapper.readValue(content, valueType); 

My code throws exception like that:

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "cccc" (Class com.isoftstone.banggo.net.result.GetGoodsInfoResult), not marked as ignorable 

And I don't want to add a prop to class Test,I just want jackson convert the exist value whith is also exist in Test.

Answer 1

Jackson provides a few different mechanisms to configure handling of "extra" JSON elements. Following is an example of configuring the ObjectMapper to not FAIL_ON_UNKNOWN_PROPERTIES.

import org.codehaus.jackson.annotate.JsonAutoDetect.Visibility; import org.codehaus.jackson.annotate.JsonMethod; import org.codehaus.jackson.map.DeserializationConfig; import org.codehaus.jackson.map.ObjectMapper;  public class JacksonFoo {   public static void main(String[] args) throws Exception   {     // { "aaa":"111", "bbb":"222", "ccc":"333" }     String jsonInput = "{ \"aaa\":\"111\",                           \"bbb\":\"222\",                           \"ccc\":\"333\" }";      ObjectMapper mapper = new ObjectMapper().setVisibility(JsonMethod.FIELD,                          Visibility.ANY);     mapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES,                      false);      Test test = mapper.readValue(jsonInput, Test.class);   } }  class Test {   String aaa;   String bbb; } 

For other approaches, see http://wiki.fasterxml.com/JacksonHowToIgnoreUnknown

Answer 2

As of Jackson 2.0 the inner enum (DeserializationConfig.Feature) has been moved to a standalone enum (DeserializationFeature):

mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);