ObjectMapper restrictions

Question

Is ObjectMapper supposed to work with a class which only has private members, no constructor and no getter/setters ?

I tried this, but it does not solve the problem.

mapper.setVisibility(JsonMethod.FIELD, JsonAutoDetect.Visibility.ANY);

A class *always* has a constructor. Do you mean that it has only the default constructor? — chrylis -cautiouslyoptimistic-, Jan 08 '17 at 04:35
Possible duplicate of [how to specify jackson to only use fields - preferably globally](http://stackoverflow.com/questions/7105745/how-to-specify-jackson-to-only-use-fields-preferably-globally) — Jason Hoetger, Jan 08 '17 at 04:37

PJ Fanning · Answer 1 · 2017-01-08T04:49:49.650

0

http://www.baeldung.com/jackson-field-serializable-deserializable-or-not

static class MyDtoAccessLevel {
    private String stringValue = "hidden";
}

public static void main(String[] args) throws JsonProcessingException {
    ObjectMapper mapper = new ObjectMapper();
    mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
    MyDtoAccessLevel dtoObject = new MyDtoAccessLevel();
    System.out.println(mapper.writeValueAsString(dtoObject));
    //prints {"stringValue":"hidden"}
}

edited Jan 08 '17 at 04:49

answered Jan 08 '17 at 04:31

PJ Fanning

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Thanks a lot for writing. But I am using org.codehaus.jackson.map.ObjectMapper, which is sort of older version of com.fasterxml.jackson.core. I do not see an option to set PropertyAccessor in older version. :( Thoughts ? – Andy897 Jan 08 '17 at 05:10
And I have no clues as to who down voted your answer. – Andy897 Jan 08 '17 at 05:10
My code sample is for Jackson 2.x. Jackson 1.x is no longer maintained. Would it be feasible to switch? – PJ Fanning Jan 08 '17 at 05:12
No, it is a big big code base, where 100s of dependencies are using 1.x :( – Andy897 Jan 08 '17 at 05:14
You can have both jackson 1.x and 2.x jars on your classpath because they use different package names. – PJ Fanning Jan 08 '17 at 05:30

ObjectMapper restrictions

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