无法使用 Jackson 从对象值(没有基于委托或基于属性的 Creator)反序列化

Cannot deserialize from Object value (no delegate- or property-based Creator) using Jackson

提问人:Scripta14 提问时间:3/17/2020 最后编辑:Michał ZioberScripta14 更新时间:10/26/2021 访问量:44287

问:

我正在尝试使用以下有效载荷进行反序列化:JSONJackson

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

但我得到这个: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]

我创建了两个类。第一堂课:

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;
    }

和二等舱

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;
    }

在主类中:

                 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();
                 }

我已经找到了这个解决方案,但我没有 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>
Java JSON Jackson 化 JSON 反序列化

评论


答:

17赞 Jens 3/17/2020 #1

信息非常明确:(no Creators, like default construct, exist)

您需要向类或注释添加一个无参数构造函数:NoArgsConstructor

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

@NoArgsConstructor
@Data
public class DeserializeSubscription {

评论

1赞 Scripta14 3/17/2020
对不起,我之前尝试过第一个解决方案,并使用第二个解决方案获得了相同的结果
0赞 Jens 3/17/2020
@Scripta14* 与第二个解决方案*的结果相同,结果是什么?
0赞 Scripta14 3/17/2020
同样的例外: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]
6赞 Michał Ziober 3/18/2020 #2

您必须考虑以下几种情况:

  • message字段 in 是原始的。在水平上,它是一个对象。JSONStringPOJOMessageSubscription
  • messagevalue in 包含未加引号的属性名称,这是非法的,但也可以处理它们。JSONJackson
  • 如果构造函数不适合,我们需要使用注释来配置它。JSON

要处理未加引号的名称,我们需要启用ALLOW_UNQUOTED_FIELD_NAMES功能。为了处理有效负载和之间的不匹配,我们需要为类实现自定义反序列化器。JSONPOJOMessageSubscription

自定义反串行程序可能如下所示:

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 的构造函数:DeserializeSubscription

@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;
    }
}

如何使用它的例子:

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);
    }
}

对于上面示例打印的提供的有效负载:JSON

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

评论

0赞 Scripta14 3/18/2020
非常感谢您的解决方案。.不幸的是,我复制了您的代码,但出现此错误:。我已经使用了我的json字符串Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Unexpected end-of-input within/between Object entries at [Source: (String)"{ Message:"; line: 1, column: 3] (through reference chain: test.exatest.DeserializeSubscription["message"])
0赞 Michał Ziober 3/18/2020
@Scripta14,可能你已经破坏了有效载荷。另外,您是否启用了功能?我已经针对您的有效载荷对其进行了测试,并且运行良好。你使用哪个版本?JSONJsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMESJackson
0赞 Scripta14 3/18/2020
上面我编写了我在应用程序中使用的 Jackson maven 版本。我复制了你的代码。唯一的区别是,我使用了我在这篇文章上面写的 json 字符串。
0赞 Michał Ziober 3/18/2020
@Scripta14,你逃过有效载荷了吗?有效载荷:JSON"{\"code\":null,\"reason\":\"subscription yet available\",\"message\":\"{ Message:\\\"subscription yet available\\\", SubscriptionUID:\\\"46b62920-c519-4555-8973-3b28a7a29463\\\" }\"}"
1赞 Scripta14 3/18/2020
多谢。。。你的 json 有效负载运行良好,它缺少一些逃生
0赞 david.barkhuizen 10/26/2021 #3

这可能是由于使用了不支持的数据类型,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正常有符号(长)整数来解决它。