There is no annotation-only solution for your problem. Somehow you have to convert JSON Object to java.lang.String and you need to specify that conversion.
You can:
- Write custom deserializer which is probably most obvious solution but forbidden in question.
- Register custom
com.fasterxml.jackson.databind.deser.DeserializationProblemHandler and handle com.fasterxml.jackson.databind.exc.MismatchedInputException situation in more sophisticated way.
- Implement
com.fasterxml.jackson.databind.util.Converter interface and convert JsonNode to String. It is semi-annotational way to solve a problem but we do not implement the worst part - deserialisation.
Let's go to point 2. right away.
2. DeserializationProblemHandler
Solution is pretty simple:
ObjectMapper mapper = new ObjectMapper();
mapper.addHandler(new DeserializationProblemHandler() {
@Override
public Object handleUnexpectedToken(DeserializationContext ctxt, JavaType targetType, JsonToken t, JsonParser p, String failureMsg) throws IOException {
if (targetType.getRawClass() == String.class) {
// read as tree and convert to String
return p.readValueAsTree().toString();
}
return super.handleUnexpectedToken(ctxt, targetType, t, p, failureMsg);
}
});
Read a whole piece of JSON as TreeNode and convert it to String using toString method. Helpfully, toString generates valid JSON. Downside, this solution has a global scope for given ObjectMapper instance.
3. Custom Converter
This solution requires to implement com.fasterxml.jackson.databind.util.Converter interface which converts com.fasterxml.jackson.databind.JsonNode to String:
class JsonNode2StringConverter implements Converter<JsonNode, String> {
@Override
public String convert(JsonNode value) {
return value.toString();
}
@Override
public JavaType getInputType(TypeFactory typeFactory) {
return typeFactory.constructType(new TypeReference<JsonNode>() {
});
}
@Override
public JavaType getOutputType(TypeFactory typeFactory) {
return typeFactory.constructType(new TypeReference<String>() {
});
}
}
and now, you can use annotation like below:
@JsonDeserialize(contentConverter = JsonNode2StringConverter.class)
private List<String> companies;
Solutions 2. and 3. solve this problem almost in the same way - read node and convert it back to JSON, but uses different approaches.
If, you want to avoid deserialising and serialising process you can take a look on solution provided in this article: Deserializing JSON property as String with Jackson and take a look at:
