Deserializaing EnumMap using jackson

Question

I've added the @JsonAnySetter annotation to addProperty method

Map<MyEnum, String> mymap = new EnumMap<>(MyEnum.class);

@JsonAnySetter
public void addProperty(MyEnum key, Object value) {
    mymap.put(key, value);
}

but while deserializing I'm getting the below error

com.fasterxml.jackson.databind.JsonMappingException: Invalid 'any-setter' annotation on method 'addProperty()': first argument not of type String or Object, but java.lang.Enum

If I change the map to simple HashMap with key of type String then it works fine.

Could someone please let me know what nneds to be done for deserializing to EnumMap

Could you please show an example json string you are trying to deserialze? — CausingUnderflowsEverywhere, Jan 19 '20 at 04:37

score 1 · Answer 1 · answered Jan 19 '20 at 04:48

1

The documentation states that:

First argument is the name of property ,so it must be String.

You can convert it to Enum in your set method.

answered Jan 19 '20 at 04:48

Chetan Komakula

369
2
5

Vivek · Accepted Answer · 2020-01-19T07:39:31.990

The @JsonAnySetter annotation seems to work only with the String or Object key types.

The workaround I found is to overload the addProperty method with the key type as String, then convert the String key to Enum type and call the addProperty method with Enum key.

Use the @JsonAnySetter annotation on the overloaded method.

Map<MyEnum, String> mymap = new EnumMap<>(MyEnum.class);

public void addProperty(MyEnum key, Object value) {
    mymap.put(key, value);
}

@JsonAnySetter
private void addProperty(String key, Object value) {
    addProperty(MyEnum.valueOf(key), value);
}

Note that I've made the overloaded addProperty method private as I don't want to expose this method version.

@JsonAnySetter annotation seems to work with private methods.

Deserializaing EnumMap using jackson

2 Answers2