[英]How to deserialize a JSON to an object ignoring a key using Jackson?
可以说我有一个JSON:
{
"MovieCount": 153,
"MoviesList": [{...},{...},...]
}
我有:
mapper.readValue(moviesListArrayString, new TypeReference<List<Movie>>(){});
但是当moviesListArrayString
为:
[{...},{...},...]
如果我的String
是原始JSON,如何告诉Jackson忽略MovieCount
并反序列化MoviesList
?
我将创建一个具有两个属性的包装器对象:
Number movieCount
List<Movie> moviesList
然后,将值读取为YourWrapperObject.class
并使用moviesList
属性值执行任何操作,而忽略moviecount
属性。
快速,非常丑陋但实用的示例
package test;
import java.util.List;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;
public class Main {
public static void main(String[] args) {
ObjectMapper mapper = new ObjectMapper();
String json = "{\"MovieCount\": 153,\"MoviesList\": [{},{}]}";
try {
List<Movie> movies = ((MovieWrapper) mapper.readValue(json, MovieWrapper.class)).moviesList;
System.out.println(movies.size());
}
catch (Throwable t) {
t.printStackTrace();
}
}
static class MovieWrapper {
@JsonProperty(value = "MovieCount")
int movieCount;
@JsonProperty(value = "MoviesList")
List<Movie> moviesList;
}
static class Movie {
}
}
输出量
2
一种解决方案是将完整的JSON作为JsonNode读取,然后仅反序列化您感兴趣的内容:
final JsonNode node = mapper.readTree(...);
movieList = mapper.readValue(node.get("MoviesList").traverse(), typeRefHere);
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