Modelmapper 无法识别 UUID

Question

嘿，我刚开始使用 Modelmapper 将 jOOQ 记录映射到 POJO。

这是我尝试转换其记录的表的架构 (Postgresql)

CREATE TABLE IF NOT EXISTS actor(
    actor_id UUID DEFAULT uuid_generate_v4(),
    first_name VARCHAR(256) NOT NULL,
    last_name VARCHAR(256) NOT NULL,
    PRIMARY KEY(actor_id)
);

这是 POJO 的样子：

@JsonDeserialize(builder = Actor.Builder.class)
public class Actor {
    private final UUID actorId;
    private final String firstName;
    private final String lastName;

    private Actor(final Builder builder) {
        actorId = builder.actorId;
        firstName = builder.firstName;
        lastName = builder.lastName;
    }

    public static Builder newBuilder() {
        return new Builder();
    }

    public UUID getActorId() {
        return actorId;
    }

    public String getFirstName() {
        return firstName;
    }

    public String getLastName() {
        return lastName;
    }

    @JsonIgnoreProperties(ignoreUnknown = true)
    public static final class Builder {
        private UUID actorId;
        private String firstName;
        private String lastName;

        private Builder() {
        }

        public Builder withActorId(final UUID val) {
            actorId = val;
            return this;
        }

        public Builder withFirstName(final String val) {
            firstName = val;
            return this;
        }

        public Builder withLastName(final String val) {
            lastName = val;
            return this;
        }

        public Actor build() {
            return new Actor(this);
        }
    }
}

我正在我的应用程序中创建一个ModelMapper bean 并向它注册一个UUID 转换器。

   @Bean
    public ModelMapper modelMapper() {
        final ModelMapper mapper = new ModelMapper();
        Provider<UUID> uuidProvider = new AbstractProvider<UUID>() {
            @Override
            public UUID get() {
                return UUID.randomUUID();
            }
        };

        final Converter<String, UUID> uuidConverter = new AbstractConverter<>() {
            @Override
            protected UUID convert(final String source) {
                return UUID.fromString(source);
            }
        };
        mapper.createTypeMap(String.class, UUID.class);
        mapper.addConverter(uuidConverter);
        mapper.getTypeMap(String.class, UUID.class).setProvider(uuidProvider);


        mapper.getConfiguration()
                .setSourceNameTokenizer(NameTokenizers.UNDERSCORE)
                .addValueReader(new RecordValueReader())
                .setDestinationNameTransformer(NameTransformers.builder("with"))
                .setDestinationNamingConvention(NamingConventions.builder("with"));
        mapper.validate();
        return mapper;
    }

然后我使用模型映射器将 ActorRecord 从 jOOQ 自动生成的代码映射到 POJO

public Optional<Actor> getActor(final UUID actorId) {
    return Optional.ofNullable(dsl.selectFrom(ACTOR)
            .where(ACTOR.ACTOR_ID.eq(actorId))
            .fetchOne())
            .map(e -> modelMapper.map(e, Actor.Builder.class).build());
}

除了 UUID 始终为 null 外，此方法有效。例如：

{"actor_id":null,"first_name":"John","last_name":"Doe"}

但是，当我在Builder 中更改以下内容时：

    public Builder withActorId(final String val) {
        actorId = UUID.fromString(val);
        return this;
    }

有效！不幸的是，这不适用于重载方法：

    public Builder withActorId(final String val) {
        actorId = UUID.fromString(val);
        return this;
    }

    public Builder withActorId(final UUID val) {
        actorId = val;
        return this;
    }

因为这也返回null。

您可以从自动生成的 jOOQ 代码中看到它应该处理 UUID：

/**
 * The column <code>public.actor.actor_id</code>.
 */
public final TableField<ActorRecord, UUID> ACTOR_ID = createField(DSL.name("actor_id"), org.jooq.impl.SQLDataType.UUID.nullable(false).defaultValue(org.jooq.impl.DSL.field("uuid_generate_v4()", org.jooq.impl.SQLDataType.UUID)), this, "");

我不确定我到底错过了什么。我不想为我的每个实体创建一个自定义转换器，因为我有很多实体，它们都包含（至少 1 个）UUID。理想情况下，我想配置ModelMapper 以了解UUID，并且只要它看到它就可以处理它。谢谢！

注意：我也尝试过使用 Lombok @Data 对象，但它也不起作用。

@JsonDeserialize(builder = Actor.ActorBuilder.class)
@Data
public class Actor {
    private UUID actorId;
    private String firstName;
    private String lastName;

    @JsonPOJOBuilder(withPrefix = "with")
    public static class ActorBuilder {

    }
}

Answer 1

UUID.fromString(val) 是不允许的。我昨天有同样的问题。尝试把 ModelMapper 配置的 UUID 转换为 String。