Jackson 自定义解串器无法在 Spring 引导中工作

Question

我为我的实体创建了一个自定义反序列化器，但它一直抛出异常：

我有两个类：AppUser 和 AppUserAvatar

AppUser.java

@Entity
@Table(name = "user")
public class AppUser implements Serializable {

    @Transient
    private static final long serialVersionUID = -3536455219051825651L;

    @JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
    @Column(name = "password", nullable = false, length = 256)
    private String password;

    @JsonIgnore
    @Column(name = "is_active", nullable = false)
    private boolean active;

    @JsonIgnore
    @OneToMany(mappedBy = "appUser", targetEntity = AppUserAvatar.class, fetch = FetchType.LAZY)
    private List<AppUserAvatar> appUserAvatars;

    //// Getters and Setters and toString() ////
}

AppUserAvatar.java

@Entity
@Table(name = "user_avatar")
public class AppUserAvatar extends BaseEntityD implements Serializable {

    @Transient
    private static final long serialVersionUID = 8992425872747011681L;

    @Column(name = "avatar", nullable = false)
    @Digits(integer = 20, fraction = 0)
    @NotEmpty
    private Long avatar;

    @JsonDeserialize(using = AppUserDeserializer.class)
    @JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "user_id", nullable = false)
    private AppUser appUser;

    //// Getters and Setters and toString() ////
}

AppUserDeserializer.java 包裹 com.nk.accountservice.deserializer;

import com.edoctar.accountservice.config.exception.InputNotFoundException;
import com.edoctar.accountservice.domain.candidate.AppUser;
import com.edoctar.accountservice.service.candidate.AppUserService;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.JsonNode;
import org.springframework.beans.factory.annotation.Autowired;

import java.io.IOException;
import java.io.Serializable;

public class AppUserDeserializer extends JsonDeserializer implements Serializable {

    private static final long serialVersionUID = -9012464195937554378L;

    private AppUserService appUserService;

    @Autowired
    public void setAppUserService(AppUserService appUserService) {
        this.appUserService = appUserService;
    }

    @Override
    public Object deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {
        JsonNode node = jsonParser.getCodec().readTree(jsonParser);
        Long userId = node.asLong();
        System.out.println(node);
        System.out.println(node.asLong());
        AppUser appUser = appUserService.findById(userId);
        System.out.println("appuser: " + appUser);
        if (appUser == null) try {
            throw new InputNotFoundException("User not found!");
        } catch (InputNotFoundException e) {
            e.printStackTrace();
            return null;
        }

        return appUser;
    }
}

样本 xhr 男孩是：

{
  "appUser": 1,
  "avatar": 1
}

每次提交请求都抛出异常

Resolved [org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: (was java.lang.NullPointerException); nested exception is com.fasterxml.jackson.databind.JsonMappingException: (was java.lang.NullPointerException) (through reference chain: com.edoctar.accountservice.domain.candidate.AppUserAvatar["appUser"])]

我发现 appUserService.findById() 方法没有被调用。我真的很困惑。我不知道我哪里出错了。对于任何解决方案都将非常有用。谢谢。

Answer 1

更新的答案： 您不能使用自动装配的属性，因为您不在 Spring 上下文中。您正在将 class AppUserDeserializer 作为注释

中的参考传递

@JsonDeserialize(using = AppUserDeserializer.class)

这种情况是FasterJackson库创建了AppUserDeserializer的实例，所以不考虑Autowired注解。

你可以用一个小技巧来解决你的问题。在 AppUserService:

中添加对 spring 创建的实例的静态引用

 @Service
 public AppUserService {

   public static AppUserService instance;

   public AppUserService() {
     // Modify the constructor setting a static variable holding a
     // reference to the instance created by spring
     AppUserService.instance = this;
   }

   ...
 }

在 AppUserDeserializer 中使用该引用：

public class AppUserDeserializer extends JsonDeserializer implements Serializable {

    private AppUserService appUserService;

    public AppUserDeserializer() {
      // Set appUserService to the instance created by spring
      this.appUserService = AppUserService.instance;
    }

    ...

}

原始答案：要正确初始化 Autowired 属性，您必须注释 class AppUserDeserializer，否则 appUserService 为空，如果您没有使用 set 方法显式初始化它。

尝试用 @Component 注释 AppUserDeserializer:

@Component   // Add this annotation
public class AppUserDeserializer extends JsonDeserializer implements Serializable {
   ...
}

Answer 2

尝试更改这行代码：

private boolean active;

到

private Boolean active;

布尔基元无法处理空值，可能会导致 NPE。

Answer 3

您可以继续尝试正确注入 AppUserService，但根据我的说法，这不是最干净的解决方案。一般来说，我不喜欢使用 @Entity 作为通信模型或视图模型的想法。通过这种方式，您将实体耦合到视图模型的 producer/consumer 。您基本上是在将模型部分短路。

我会做的是在反序列化阶段将json的内容映射到不同的class，稍后使用这个来构造相应的实体。

Jackson 自定义解串器无法在 Spring 引导中工作

Jackson Custom Deserializer Not Working In Spring Boot

java

jackson

json-deserialization

spring-boot