Вам нужно написать свой JsonDeserializer и указать его в аннотации @JsonDeserialize над тем сеттером поля с которым у вас проблемы
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.annotation.JsonDeserialize;
import java.io.IOException;
public class TestJackson {
Integer a;
public Integer getA() {
return a;
}
@JsonDeserialize(using = StringIntegerDeserializer.class)
public void setA(Integer a) {
this.a = a;
}
@Override
public String toString() {
return "TestJackson{a=" + a + "}";
}
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
String json = "{\"a\":\"5.666\"}";
System.out.println(mapper.readValue(json, TestJackson.class));
json = "{\"a\":5.666}";
System.out.println(mapper.readValue(json, TestJackson.class));
json = "{\"a\":5}";
System.out.println(mapper.readValue(json, TestJackson.class));
}
public static class StringIntegerDeserializer extends JsonDeserializer<Integer> {
@Override
public Integer deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
String valueAsString = jsonParser.getValueAsString();
int integer = (int) Double.parseDouble(valueAsString);
return integer;
}
}
}
