无法使用 Jackson 从对象值（无委托或基于属性的创建者）反序列化

Question

我正在尝试使用 Jackson 反序列化以下 JSON 负载：

{"code":null,"reason":"subscription yet available","message":"{ Message:\"subscription yet available\", SubscriptionUID:\"46b62920-c519-4555-8973-3b28a7a29463\" }"}

但我得到了这个 JsonMappingException:

Cannot construct instance of `com.ids.utilities.DeserializeSubscription` (no Creators, like default construct, exist): cannot deserialize from Object value (no delegate- or property-based Creator)
 at [Source: (String)"{"code":null,"reason":"subscription yet available","message":"{ Message:\"subscription yet available\", SubscriptionUID:\"46b62920-c519-4555-8973-3b28a7a29463\" }"}"; line: 1, column: 2]

我创建了两个 classes。第一个class：

import lombok.Data;

@Data
public class DeserializeSubscription {

    private String code;
    private String reason;
    private MessageSubscription message;


    public DeserializeSubscription(String code, String reason, MessageSubscription message) {
        super();
        this.code = code;
        this.reason = reason;
        this.message = message;
    }

和第二个class

import lombok.Data;

@Data
public class MessageSubscription {

    private String message;
    private String subscriptionUID;


    public MessageSubscription(String message, String subscriptionUID) {
        super();
        this.message = message;
        this.subscriptionUID = subscriptionUID;
    }

主要class：

                 try 
                 {

                    ObjectMapper mapper = new ObjectMapper();
                    mapper.enable(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT);
                    DeserializeSubscription desSub=null;

                    desSub=mapper.readValue(e.getResponseBody(), DeserializeSubscription.class);

                    System.out.println(desSub.getMessage().getSubscriptionUID());
                 }
                 catch (JsonParseException e1) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                 }
                 catch (JsonMappingException e1) {
                     System.out.println(e1.getMessage());
                        e.printStackTrace();
                 }
                 catch (IOException e1) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                 }

我找到了这个解决方案，但没有成功 https://facingissuesonit.com/2019/07/17/com-fasterxml-jackson-databind-exc-invaliddefinitionexception-cannot-construct-instance-of-xyz-no-creators-like-default-construct-exist-cannot-deserialize-from-object-value-no-delega/

我在我的应用程序中使用的 jackson maven

<!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-databind</artifactId>
        <version>2.10.2</version>
    </dependency>

Answer 1

消息很清楚：(no Creators, like default construct, exist)

您需要向 class 或 NoArgsConstructor 注释添加无参数构造函数：

@Data
public class DeserializeSubscription {
  public DeserializeSubscription (){}

或

@NoArgsConstructor
@Data
public class DeserializeSubscription {

Answer 2

你要考虑几个情况：

message JSON 中的字段是原始的 String。在 POJO 级别，它是一个 MessageSubscription 对象。
message JSON 中的值包含未加引号的属性名称，这是非法的，但 Jackson 也会处理它们。
如果构造函数不适合 JSON 我们需要使用注释来配置它。

要处理不带引号的名称，我们需要启用 ALLOW_UNQUOTED_FIELD_NAMES 功能。为了处理 JSON 有效负载和 POJO 之间的不匹配，我们需要为 MessageSubscription class.

实现自定义反序列化器

自定义解串器可能如下所示：

class MessageSubscriptionJsonDeserializer extends JsonDeserializer<MessageSubscription> {
    @Override
    public MessageSubscription deserialize(JsonParser p, DeserializationContext ctxt) throws IOException {
        final String value = p.getValueAsString();
        final Map<String, String> map = deserializeAsMap(value, (ObjectMapper) p.getCodec(), ctxt);

        return new MessageSubscription(map.get("Message"), map.get("SubscriptionUID"));
    }

    private Map<String, String> deserializeAsMap(String value, ObjectMapper mapper, DeserializationContext ctxt) throws IOException {
        final MapType mapType = ctxt.getTypeFactory().constructMapType(Map.class, String.class, String.class);
        return mapper.readValue(value, mapType);
    }
}

现在，我们需要自定义DeserializeSubscription的构造函数：

@Data
class DeserializeSubscription {

    private String code;
    private String reason;
    private MessageSubscription message;

    @JsonCreator
    public DeserializeSubscription(
            @JsonProperty("code") String code,
            @JsonProperty("reason") String reason,
            @JsonProperty("message") @JsonDeserialize(using = MessageSubscriptionJsonDeserializer.class) MessageSubscription message) {
        super();
        this.code = code;
        this.reason = reason;
        this.message = message;
    }
}

使用示例：

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.annotation.JsonDeserialize;
import com.fasterxml.jackson.databind.type.MapType;
import lombok.Data;

import java.io.File;
import java.io.IOException;
import java.util.Map;

public class JsonPathApp {

    public static void main(String[] args) throws Exception {
        File jsonFile = new File("./resource/test.json").getAbsoluteFile();

        ObjectMapper mapper = new ObjectMapper();
        mapper.enable(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT);
        mapper.enable(JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES);

        DeserializeSubscription value = mapper.readValue(jsonFile, DeserializeSubscription.class);
        System.out.println(value);
    }
}

对于提供的 JSON 负载，上面的示例打印：

DeserializeSubscription(code=null, reason=subscription yet available, message=MessageSubscription(message=subscription yet available, subscriptionUID=46b62920-c519-4555-8973-3b28a7a29463))

Answer 3

这可能是由于使用了不受支持的数据类型，例如无符号整数。

我在反序列化具有 ULong 字段的 JSON 对象时收到此错误。并通过将字段类型更改为普通有符号（长）整数来解决它。

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