JSON field - Parse String to Int

Question

Have a relatively simple (I think) issue, but being a novice to JSON cannot seem to find a neat solution.

I have an Entity object with Id field as type Integer. However, the incoming Json data to be mapped has the id as a string.

With this a straightforward map does not seem to be possible. Is there a way to change the string data in the JSON to an integer, before mapping?

Example Json data

{"Id": "021", "userAge": 99}

Example Entity

@Entity
public class User{

    @id
    int userId;
    int userAge;
}

Many Thanks.

Sotirios Delimanolis · Accepted Answer · 2015-02-23 17:02:10Z

8

You don't need to.

Jackson is smart enough to convert a JSON string to a numerical value if the target field is numerical.

It's not obvious what the leading 0 is meant to represent, but Jackson will simply ignore it.

_{Also, if your field name is different in Java, you'll need @JsonProperty("theJsonName").}

public class Jackson {
    public static void main(String[] args) throws Exception {
        String json = "{\"userId\": \"021\", \"userAge\": 99}";
        ObjectMapper mapper = new ObjectMapper();
        User user = mapper.readValue(json, User.class);
        System.out.println(user.userId);
    }
}

class User {
    int userId;
    int userAge;
    public void setUserId(int userId) {
        this.userId = userId;
    }
    public void setUserAge(int userAge) {
        this.userAge = userAge;
    }
}

prints

edited Feb 23, 2015 at 17:02

answered Feb 23, 2015 at 16:56

Sotirios Delimanolis

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Comments

Jivings · Accepted Answer · 2015-02-23 16:52:14Z

1

You can write a custom jackson deserializer to cope with this behaviour. There's an good blog post on this topic here.

public class ItemDeserializer extends JsonDeserializer<Item> {

    @Override
    public Item deserialize(JsonParser jp, DeserializationContext ctxt) 
      throws IOException, JsonProcessingException {
        JsonNode node = jp.getCodec().readTree(jp);
        int id = Integer.parseInt(node.get("userId").asText());
        int userAge = (Integer) ((IntNode) node.get("userAge")).numberValue();

        return new Item(id, userAge);
    }
}

answered Feb 23, 2015 at 16:52

Jivings

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