jackson json 转换Bean, Bean 里没有对应的值 jackson Unrecognized field

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.

 

 

 

 

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