java - 通用解析器返回 HashMap 而不是 Pojo

Question

我有这个代码(A):

JsonFileHandler<Device> jsonFileHandlerDevice;
final List<Device> devicesList = jsonFileHandlerDevice.getList();

和

class JsonFileHandler<T>:

@Override
public List<T> getList() {
    List<T> t = null;
    ObjectMapper mapper = new ObjectMapper();

    if (!file.exists()) {
        return null;
    } else {
        try {
            t = mapper.readValue(file, new TypeReference<List<T>>(){});
        } catch (IOException e) {
            e.printStackTrace();
            throw new RuntimeException(e);
        }
    }
    return t;
}

和这段代码(B):

@Override
public List<Device> getList() {
    List<Device> t = null;
    ObjectMapper mapper = new ObjectMapper();

    if (!file.exists()) {
        return null;
    } else {
        try {
            t = mapper.readValue(file, new TypeReference<List<Device>>(){});
        } catch (IOException e) {
            e.printStackTrace();
            throw new RuntimeException(e);
        }
    }
    return t;
}

和这个 json 文件:

[ {
  "mobileOs" : "ios",
  "osVersion" : 4.2,
  "allocatedPort" : 0,
  "hasSim" : false,
  "uuid" : "uuid2",
  "wazers" : [ {
    "email" : null,
    "emailPassword" : null,
    "first" : null,
    "last" : null,
    "driverPhone" : null,
    "riderPhone" : null,
    "username" : null,
    "password" : null,
    "workEmail" : null,
    "car" : null,
    "model" : null,
    "color" : null,
    "plate" : null,
    "obId" : null
  } ],
  "riders" : [ {
    "email" : null,
    "emailPassword" : null,
    "first" : null,
    "last" : null,
    "driverPhone" : null,
    "riderPhone" : null,
    "username" : null,
    "password" : null,
    "workEmail" : null,
    "car" : null,
    "model" : null,
    "color" : null,
    "plate" : null,
    "obId" : null
  }, {
    "email" : null,
    "emailPassword" : null,
    "first" : null,
    "last" : null,
    "driverPhone" : null,
    "riderPhone" : null,
    "username" : null,
    "password" : null,
    "workEmail" : null,
    "car" : null,
    "model" : null,
    "color" : null,
    "plate" : null,
    "obId" : null
  } ]
} ]

代码仅在执行代码 (B) 时解析 OK

并且当运行代码 (A) 时，我们得到一个 HashMap 而不是 Pojo Device。

Answer 1

我假设您可以通过 Spring 或您用作 IoC 容器的任何方式以某种方式获得类的泛型类型，因此为简单起见，我通过构造函数传递类型。话虽这么说，我目前可以想到的一种解决方案如下:

class JsonFileHandler<T> {

    private File file = new File("/Users/dambros/Desktop/test");
    private final Class<T> type;

    public JsonFileHandler(Class<T> type) {
        this.type = type;
    }

    public List<T> getList() {
        List<T> t;
        ObjectMapper mapper = new ObjectMapper();
        mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

        if (!file.exists()) {
            return null;
        } else {
            try {
                JavaType javaType = mapper.getTypeFactory().constructParametrizedType(ArrayList.class, List.class, type);
                t = mapper.readValue(file, javaType);
            } catch (IOException e) {
                e.printStackTrace();
                throw new RuntimeException(e);
            }
        }
        return t;
    }

}

使用如下所示的调用将返回一个 POJO 列表而不是一个 HashMap:

JsonFileHandler<Device> jsonFileHandlerDevice = new JsonFileHandler<>(Device.class);
final List<Device> devicesList = jsonFileHandlerDevice.getList();

Obs.: 映射器配置只是为了避免必须在 JSON 上写入所有条目并使测试更容易。