如何使用 Jackson 反序列化对象数组

Question

Jackson data binding documentation 表示 Jackson 支持反序列化“所有受支持类型的数组”，但我无法弄清楚具体的语法。

对于单个对象，我会这样做：

//json input
{
    "id" : "junk",
    "stuff" : "things"
}

//Java
MyClass instance = objectMapper.readValue(json, MyClass.class);

现在对于一个数组我想这样做：

//json input
[{
    "id" : "junk",
    "stuff" : "things"
},
{
    "id" : "spam",
    "stuff" : "eggs"
}]

//Java
List<MyClass> entries = ?

有人知道是否有一个神奇的缺失命令吗？如果不是，那么解决办法是什么？

Answer 1

首先创建一个映射器：

import com.fasterxml.jackson.databind.ObjectMapper;// in play 2.3
ObjectMapper mapper = new ObjectMapper();

作为数组：

MyClass[] myObjects = mapper.readValue(json, MyClass[].class);

作为列表：

List<MyClass> myObjects = mapper.readValue(jsonInput, new TypeReference<List<MyClass>>(){});

另一种指定List类型的方法：

List<MyClass> myObjects = mapper.readValue(jsonInput, mapper.getTypeFactory().constructCollectionType(List.class, MyClass.class));

Answer 2

来自Eugene Tskhovrebov

List<MyClass> myObjects = Arrays.asList(mapper.readValue(json, MyClass[].class))

这个解决方案似乎对我来说是最好的。

Answer 3

对于通用实现：

public static <T> List<T> parseJsonArray(String json,
                                         Class<T> classOnWhichArrayIsDefined) 
                                         throws IOException, ClassNotFoundException {
   ObjectMapper mapper = new ObjectMapper();
   Class<T[]> arrayClass = (Class<T[]>) Class.forName("[L" + classOnWhichArrayIsDefined.getName() + ";");
   T[] objects = mapper.readValue(json, arrayClass);
   return Arrays.asList(objects);
}

Answer 4

首先创建一个线程安全的 ObjectReader 实例。

ObjectMapper objectMapper = new ObjectMapper();
ObjectReader objectReader = objectMapper.reader().forType(new TypeReference<List<MyClass>>(){});

然后使用它：

List<MyClass> result = objectReader.readValue(inputStream);

Answer 5

try {
    ObjectMapper mapper = new ObjectMapper();
    JsonFactory f = new JsonFactory();
    List<User> lstUser = null;
    JsonParser jp = f.createJsonParser(new File("C:\\maven\\user.json"));
    TypeReference<List<User>> tRef = new TypeReference<List<User>>() {};
    lstUser = mapper.readValue(jp, tRef);
    for (User user : lstUser) {
        System.out.println(user.toString());
    }

} catch (JsonGenerationException e) {
    e.printStackTrace();
} catch (JsonMappingException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
}

Answer 6

我无法使用 this answer，因为我的 linter 不允许未经检查的强制转换。

这是您可以使用的替代方法。我觉得它实际上是一个更清洁的解决方案。

public <T> List<T> parseJsonArray(String json, Class<T> clazz) throws JsonProcessingException {
  var tree = objectMapper.readTree(json);
  var list = new ArrayList<T>();
  for (JsonNode jsonNode : tree) {
    list.add(objectMapper.treeToValue(jsonNode, clazz));
  }
  return list;
}

Answer 7

这是一个可以转换 json2object 或 Object2json 的实用程序， 无论你的 pojo（实体 T）

import java.io.IOException;
import java.io.StringWriter;
import java.util.List;

import com.fasterxml.jackson.core.JsonGenerationException;
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;

/**
 * 
 * @author TIAGO.MEDICI
 * 
 */
public class JsonUtils {

    public static boolean isJSONValid(String jsonInString) {
        try {
            final ObjectMapper mapper = new ObjectMapper();
            mapper.readTree(jsonInString);
            return true;
        } catch (IOException e) {
            return false;
        }
    }

    public static String serializeAsJsonString(Object object) throws JsonGenerationException, JsonMappingException, IOException {
        ObjectMapper objMapper = new ObjectMapper();
        objMapper.enable(SerializationFeature.INDENT_OUTPUT);
        objMapper.disable(SerializationFeature.FAIL_ON_EMPTY_BEANS);
        StringWriter sw = new StringWriter();
        objMapper.writeValue(sw, object);
        return sw.toString();
    }

    public static String serializeAsJsonString(Object object, boolean indent) throws JsonGenerationException, JsonMappingException, IOException {
        ObjectMapper objMapper = new ObjectMapper();
        if (indent == true) {
            objMapper.enable(SerializationFeature.INDENT_OUTPUT);
            objMapper.disable(SerializationFeature.FAIL_ON_EMPTY_BEANS);
        }

        StringWriter stringWriter = new StringWriter();
        objMapper.writeValue(stringWriter, object);
        return stringWriter.toString();
    }

    public static <T> T jsonStringToObject(String content, Class<T> clazz) throws JsonParseException, JsonMappingException, IOException {
        T obj = null;
        ObjectMapper objMapper = new ObjectMapper();
        obj = objMapper.readValue(content, clazz);
        return obj;
    }

    @SuppressWarnings("rawtypes")
    public static <T> T jsonStringToObjectArray(String content) throws JsonParseException, JsonMappingException, IOException {
        T obj = null;
        ObjectMapper mapper = new ObjectMapper();
        obj = mapper.readValue(content, new TypeReference<List>() {
        });
        return obj;
    }

    public static <T> T jsonStringToObjectArray(String content, Class<T> clazz) throws JsonParseException, JsonMappingException, IOException {
        T obj = null;
        ObjectMapper mapper = new ObjectMapper();
        mapper = new ObjectMapper().configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
        obj = mapper.readValue(content, mapper.getTypeFactory().constructCollectionType(List.class, clazz));
        return obj;
    }

Answer 8

试试这个

List<MyClass> list = mapper.readerForListOf(MyClass.class).readValue(json)

Answer 9

你也可以创建一个扩展ArrayList的类：

public static class MyList extends ArrayList<Myclass> {}

然后像这样使用它：

List<MyClass> list = objectMapper.readValue(json, MyList.class);