java - JsonIgnoreProperties 不适用于 JsonCreator 构造函数

Question

我有以下实体，我将其用作对 Controller 的请求之一的目标 POJO:

Entity
@Table(name="user_account_entity")
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonSerialize(using = UserAccountSerializer.class)
public class UserAccountEntity implements UserDetails {
    //...
    private String username;

    private String password;

    @PrimaryKeyJoinColumn
    @OneToOne(mappedBy= "userAccount", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private UserEntity user;

    @PrimaryKeyJoinColumn
    @OneToOne(mappedBy= "userAccount", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private UserAccountActivationCodeEntity activationCode;

    @JsonCreator
    public UserAccountEntity(@JsonProperty(value="username", required=true) final String username, @JsonProperty(value="password", required=true) final String password) {
      //....
    }

    public UserAccountEntity() {}

    //.....
}

当我在请求中放入意外字段时，它会抛出 MismatchedInputException 并失败并显示以下消息:

com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `com.myproject.project.core.entity.userAccountActivationCode.UserAccountActivationCodeEntity` (although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value ('9WL4J')
 at [Source: (PushbackInputStream); line: 4, column: 20] (through reference chain: com.myproject.project.core.entity.userAccount.UserAccountEntity["activationCode"])

在我的 Controller 中:

@InitBinder
public void binder(WebDataBinder binder) {
    binder.addValidators(new CompoundValidator(new Validator[] {
            new UserAccountValidator(),
            new UserAccountActivationCodeDTOValidator() }));
}

我向其发出请求的端点是:

@Override
public UserAccountEntity login(@Valid @RequestBody UserAccountEntity account,
        HttpServletResponse response) throws MyBadCredentialsException, InactiveAccountException {
    return userAccountService.authenticateUserAndSetResponsenHeader(
            account.getUsername(), account.getPassword(), response);
}

更新 1

UserAccountSerializer 的代码:

public class UserAccountSerializer extends StdSerializer<UserAccountEntity> {

    public UserAccountSerializer() {
        this(null);
    }

    protected UserAccountSerializer(Class<UserAccountEntity> t) {
        super(t);
    }

    @Override
    public void serialize(UserAccountEntity value, JsonGenerator gen,
            SerializerProvider provider) throws IOException {
        gen.writeStartObject();
        gen.writeStringField("id", value.getId());
        gen.writeStringField("username", value.getUsername());
        gen.writeEndObject();
    }

}

Answer 1

错误被触发是因为你的 json 中有:

"activationCode" : "9WL4J"

但是 Jackson 不知道如何将字符串“9WL4J”映射到对象 UserAccountActivationCodeEntity

我猜字符串“9WL4J”是 UserAccountActivationCodeEntity 的主键 ID 的值，在这种情况下，您应该在 json 中包含:

"activationCode" : {"id" : "9WL4J"}

如果不是这种情况，请使用自定义反序列化器告诉 Jackson 如何将字符串映射到对象。您可以在实体上使用 @JsonDeserialize。