2

To convert json String to pojo using jackson API can use : 

String jsonInString = "{\"age\":33,\"messages\":[\"msg 1\",\"msg 2\"],\"name\":\"mkyong\"}";
User user1 = mapper.readValue(jsonInString, User.class);

This requires that create class User that matches structure of json String.

Using json-simple API can use instead : 

JSONObject json = (JSONObject)new JSONParser().parse(jsonInString);

Using json-simple do not need to include a pojo that matches json format. Can similar be used in jackson ? json-simple is less verbose as do not have to create the class that matches json structure.

sactiw
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blue-sky
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2 Answers2

3

Jackson can deserialize a json String into a general-purpose Map: 

Map<String, Object> m  = new ObjectMapper().readValue(jsonInString, Map.class);
for (Map.Entry<String, Object> entry : m.entrySet()) {
    System.out.println(entry.getKey() + " -> " + entry.getValue() + "(" + entry.getValue().getClass().getName() + ")");
}

output:

age -> 33(java.lang.Integer)
messages -> [msg 1, msg 2](java.util.ArrayList)
name -> mkyong(java.lang.String)
Sharon Ben Asher
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2

You can use similar API

JsonNode node = mapper.readTree(jsonInString);