How to parse a JSON string into JsonNode in Jackson?

Question

It should be so simple, but I just cannot find it after being trying for an hour.

I need to get a JSON string, for example, {"k1":v1,"k2":v2}, parsed as a JsonNode.

JsonFactory factory = new JsonFactory();
JsonParser jp = factory.createJsonParser("{\"k1\":\"v1\"}");
JsonNode actualObj = jp.readValueAsTree();

gives

java.lang.IllegalStateException: No ObjectCodec defined for the parser, can not deserialize JSON into JsonNode tree

#embarrasing -- nope. If simple things aren't simple, then the API designer has failed, not you. — jameshfisher
– jameshfisher, Commented Aug 15, 2014 at 17:05
@StaxMan's answer in code: JsonFactory factory = new MappingJsonFactory(); — Juan Rojas
– Juan Rojas, Commented May 8, 2020 at 6:55

John Gietzen · Accepted Answer · 2015-04-03 20:37:03Z

440

A slight variation on Richards answer but readTree can take a string so you can simplify it to:

ObjectMapper mapper = new ObjectMapper();
JsonNode actualObj = mapper.readTree("{\"k1\":\"v1\"}");

edited Apr 3, 2015 at 20:37

John Gietzen

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answered Mar 7, 2012 at 10:19

slashnick

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2 Comments

Matthew Herbst Over a year ago

For anyone who needs an ObjectNode rather than a JsonNode use mapper.valueToTree("{\"k1\":\"v1\"}")

minexew Over a year ago

@MatthewHerbst In 2.5.1, this creates a new text node with the string "{\"k1\":\"v1\"}" rather than parsing it as JSON.

Richard Fearn · Accepted Answer · 2018-05-28 15:39:32Z

75

You need to use an ObjectMapper:

ObjectMapper mapper = new ObjectMapper();
JsonFactory factory = mapper.getJsonFactory(); // since 2.1 use mapper.getFactory() instead
JsonParser jp = factory.createJsonParser("{\"k1\":\"v1\"}");
JsonNode actualObj = mapper.readTree(jp);

Further documentation about creating parsers can be found here.

edited May 28, 2018 at 15:39

answered Sep 6, 2010 at 20:20

Richard Fearn

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2 Comments

benathon Over a year ago

Is it possible to take the JsonNode, modify it a bit, and then call mapper.readValue(node, class); and get out a class?

StaxMan Over a year ago

Yes. There's even an alias for that, mapper.treeToValue(). But readValue() also works.

qerub · Accepted Answer · 2012-07-20 14:50:13Z

37

A third variant:

ObjectMapper mapper = new ObjectMapper();
JsonNode actualObj = mapper.readValue("{\"k1\":\"v1\"}", JsonNode.class);

answered Jul 20, 2012 at 14:50

qerub

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3 Comments

David Over a year ago

What did passing in JsonNode.class actually get you here?

qerub Over a year ago

@David: Nothing, so slashnick's variant is obviously better if you just want a JsonNode. However, you usually don't want a JsonNode, so I wanted to show how to tell Jackson what kind of value of you want.

Cory Klein Over a year ago

For me, passing JsonNode was apparently necessary to prevent Jackson from deserializing it as something else - which would have failed.

JJD · Accepted Answer · 2013-10-08 10:27:18Z

5

Richard's answer is correct. Alternatively you can also create a MappingJsonFactory (in org.codehaus.jackson.map) which knows where to find ObjectMapper. The error you got was because the regular JsonFactory (from core package) has no dependency to ObjectMapper (which is in the mapper package).

But usually you just use ObjectMapper and do not worry about JsonParser or other low level components -- they will just be needed if you want to data-bind parts of stream, or do low-level handling.

edited Oct 8, 2013 at 10:27

JJD

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answered Sep 7, 2010 at 2:04

StaxMan

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Comments

pdxleif · Accepted Answer · 2014-10-01 23:01:22Z

5

import com.github.fge.jackson.JsonLoader;
JsonLoader.fromString("{\"k1\":\"v1\"}")
== JsonNode = {"k1":"v1"}

answered Oct 1, 2014 at 23:01

pdxleif

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Comments

Manuel Ortiz · Accepted Answer · 2020-06-24 18:03:28Z

4

New approach to old question. A solution that works from java 9+

ObjectNode agencyNode = new ObjectMapper().valueToTree(Map.of("key", "value"));

is more readable and maintainable for complex objects. Ej

Map<String, Object> agencyMap = Map.of(
        "name", "Agencia Prueba",
        "phone1", "1198788373",
        "address", "Larrea 45 e/ calligaris y paris",
        "number", 267,
        "enable", true,
        "location", Map.of("id", 54),
        "responsible", Set.of(Map.of("id", 405)),
        "sellers", List.of(Map.of("id", 605))
);
ObjectNode agencyNode = new ObjectMapper().valueToTree(agencyMap);

answered Jun 24, 2020 at 18:03

Manuel Ortiz

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Comments

AbstractVoid · Accepted Answer · 2024-01-29 13:03:52Z

0

In kotlin:

import com.fasterxml.jackson.databind.JsonNode
import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.kotlin.readValue

val data = ObjectMapper().readValue<JsonNode>("{\"field\":\"value\"}")

answered Jan 29, 2024 at 13:03

AbstractVoid

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