How to parse json but leave one particular node serialised as a string?

Question

Say I have this JSON example:

{
    "title" : "tttt",
    "custom" : {
        "a" : "aaaaa",
        "b" : "bbbbb",
        "c" : { 
            "d" : "dddd"
        }
    }
}

I want to deserialize this into the following class:

public class Article {
    private String title;
    private String custom;

    public void setTitle(String title) { this.title = title; }
    public String getTitle() { return title; }
    public void setCustom(String custom) { this.custom = custom; }
    public String getCustom() { return custom; }        
}

The result I want is for the title to be deserialized as per normal but the json subtree under "custom" to be not deserialized and set as the raw json string:

ObjectMapper mapper = new ObjectMapper();
Article article = mapper.readValue(content, Article.class);
assertEquals(article.getTitle(), "tttt");
assertEquals(article.getCustom(),
   "{\"a\" : \"aaaaa\","        +
        "\"b\" : \"bbbbb\","    +
        "\"c\" : {"             + 
            "\"d\" : \"dddd\" " +
        "}"                     +
    "}");

One other important note is that I can't change the original JSON under the custom node to use escaped json so it will be treated as a string.

katoquro · Accepted Answer · 2015-02-20 11:07:20Z

4

I've found a little bit easier way. You can just add to your dto next setter:

public void setCustom(JsonNode custom) { this.custom = custom.toString(); }

and for getting as String simple annotate it with @JsonRawValue

answered Feb 20, 2015 at 11:07

katoquro

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1 Comment

dpolivaev Over a year ago

There is a code example at stackoverflow.com/a/11452577/1833472

Michał Ziober · Accepted Answer · 2013-10-18 00:27:13Z

2

You have to write new deserializer for your class. It could look like this:

class ArticleJsonDeserializer extends JsonDeserializer<Article> {

    @Override
    public Article deserialize(JsonParser parser, DeserializationContext context)
            throws IOException, JsonProcessingException {
        RawArticle rawArticle = parser.readValueAs(RawArticle.class);

        return rawArticle.toArticle();
    }

    private static class RawArticle {
        public String title;
        public JsonNode custom;

        Article toArticle() {
            Article article = new Article();
            article.setTitle(title);
            article.setCustom(custom.toString());

            return article;
        }
    }
}

Your Article class should look like this:

@JsonDeserialize(using = ArticleJsonDeserializer.class)
class Article {
   ....
}

Now you can easily deserialize your JSON to Article class.

answered Oct 18, 2013 at 0:27

Michał Ziober

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2 Comments

Shane Rowatt Over a year ago

Thanks a million. I ended up writing a Serializer that extended StdSerializer<Article> and a deserializer that extended StdDeserializer<Article> as this allowed me to wrap the Article in another generic type and still be able to read/write the json.

muttonUp Over a year ago

FWIW: This in effect deserializes to a JsonNode and then serializes back to a string. Which may create json that is not 100% identical to the json originally submitted.

Community · Accepted Answer · 2017-05-23 12:07:46Z

1

You can write a custom Deserializer .

Reference this SO question

edited May 23, 2017 at 12:07

CommunityBot

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answered Oct 17, 2013 at 23:09

ssedano

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