Reputation: 308
I am trying to deserialize a JSON String using Jackson 2 with RestAssured (java tool for IT tests).
I have a problem. The String I am trying to deserialize is :
{"Medium":{"uuid":"2","estimatedWaitTime":0,"status":"OPEN_AVAILABLE","name":"Chat","type":"CHAT"}}
There is the object type "Medium" at the begining of the String. This cause Jackson failing during deserialization:
com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "Medium"
I've set the "IGNORE_ON_UNKNOWN_PROPERTIES" to false and then I got no exception during deserialisation. However, all of my properties are 'null' in java.
Response getAvailability -> {"Medium":{"uuid":"2","estimatedWaitTime":0,"status":"OPEN_AVAILABLE","name":"Chat","type":"CHAT"}}
@@@ MEDIUM name -> null
@@@ MEDIUM uuid -> null
@@@ MEDIUM wait time -> null
@@@ MEDIUM wait time -> null
@@@ MEDIUM status -> null
Does anyone can help me ? (note: I can't change my input JSON string).
Upvotes: 0
Views: 1216
Reputation: 71
You need to put annotation @JsonRootName("Medium") on your bean class and configure object mapper to mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE).
Upvotes: 1
Reputation: 608
You need a way to remove the Object name that is the part of the input JSON. Since you cannot change the input string, Use this code to change this input string to a tree and get the value of "Medium" node.
ObjectMapper m = new ObjectMapper();
JsonNode root = m.readTree("{\"Medium\":{\"uuid\":\"2\",\"estimatedWaitTime\":0,\"status\":\"OPEN_AVAILABLE\",\"name\":\"Chat\",\"type\":\"CHAT\"}}");
JsonNode obj = root.get("Medium");
Medium medium = m.readValue(obj.asText, Medium.class);
Upvotes: 0
Reputation: 190
{
"Medium": {
"uuid": "2",
"estimatedWaitTime": 0,
"status": "OPEN_AVAILABLE",
"name": "Chat",
"type": "CHAT"
}
}
as you can see uuid and other params are part of medium object , so class in which it can be deserialized is.
class Medium
{
string name;
// specify other params also.
}
class BaseObject
{
Medium Medium;
}
and then use jackson.deserialize('json', BaseObject.class)
above i had given pseudo code
Upvotes: 2