Use Jackson to deserialize JSON string or object into a String field

Question

I am using Jackson 2 library and I am trying to read a JSON response, which looks like:

{ "value":"Hello" }

When value is empty, JSON response looks like:

{ "value":{} }

My model POJO class looks like this

public class Hello {

    private String value;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }

}

The problem is that when response looks like {value:{}}, Jackson is trying to read an Object, but my model class field is a string, so it throws an Exception:

JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token.

My question is how Jackson can successfully read JSONs who look like:

 {"value":"something"}

and at the same time if response looks like this {"value":{}} (empty response for me), pass null to value field of my Hello model class.

I am using the code below in order to read JSON string:

String myJsonAsString = "{...}";
ObjectMapper mapper = new ObjectMapper();
mapper.readValue(myJsonAsString , Hello.class);

Isn't there any way to fix the part that generates this JSON, instead of working around ths ugly design? — JB Nizet, Mar 11 '17 at 20:08
Nope there isn't unfortunately. API is not mine. I am just the client side. — Konstantinos Raptis, Mar 11 '17 at 20:12
Ouch, proper output would leave out the field or set it to null... — john16384, Mar 11 '17 at 20:24
It's not that odd. MongoDB queries often have parts of the JSON that can be strings or objects. — Doradus, Jul 30 '22 at 00:57

score 14 · Accepted Answer · answered Mar 11 '17 at 20:22

You can use a custom deserializer for this feld. Here is one that returns the string if it's there, or null in any other case:

public class Hello {

    @JsonDeserialize(using = StupidValueDeserializer.class)
    private String value;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }
}

public class StupidValueDeserializer extends JsonDeserializer<String> {
    @Override
    public String deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        JsonToken jsonToken = p.getCurrentToken();
        if (jsonToken == JsonToken.VALUE_STRING) {
            return p.getValueAsString();
        }
        return null;
    }
}

score 2 · Answer 2 · answered Jun 01 '21 at 13:38

Txh for JB Nizet, but if you get type another than String (e.g Object), Jackson deserialiser tried to deserialize inner Object and can throw a latent exception. After that, other fields in json filled as null in Java.

To avoid this you'll ignore children

@Override
public String deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
    JsonToken jsonToken = p.currentToken();
    if (jsonToken == JsonToken.VALUE_STRING) {
        return p.getValueAsString();
    }
    p.skipChildren();
    return "other_string";
}

Use Jackson to deserialize JSON string or object into a String field

2 Answers2

Linked