杰克逊：使用带有非默认构造函数的构建器

Question

我有以下课程：

@JsonDeserialize(builder = Transaction.Builder.class)
public final class Transaction {

    @JacksonXmlText
    private final TransactionType transactionType;
    @JacksonXmlProperty(isAttribute = true)
    private final boolean transactionAllowed;

    private Transaction(Builder builder) {
        transactionType = builder.transactionType;
        transactionAllowed = builder.transactionAllowed;
    }

    public static final class Builder {
        private final TransactionType transactionType;
        private boolean transactionAllowed;

        public Builder(TransactionType transactionType) {
            this.transactionType = transactionType;
        }

        public Builder withTransactionAllowed() {
            transactionAllowed = true;
            return this;
        }

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

TransactionType 是一个简单的枚举：

public enum TransactionType {
    PU,
    CV;
}

当我创建一个新的 Transaction 实例并使用 Jackson 映射器对其进行序列化时，我得到以下 xml：

<transaction transactionAllowed="true">PU</transaction>

问题是我无法反序列化它。我得到以下异常：

com.fasterxml.jackson.databind.JsonMappingException:
No suitable constructor found for type [simple type, class Transaction$Builder]

如果我将 @JsonCreator 放在 Builder 构造函数上，如下所示：

@JsonCreator
public Builder(TransactionType transactionType) {
    this.transactionType = transactionType;
}

我得到以下异常：

com.fasterxml.jackson.databind.exc.InvalidFormatException: Can not construct
instance of Transaction from String value 'transactionAllowed': value not one of
declared Enum instance names: [PU, CV]

如果我将 @JsonProperty 放在构造函数参数上，如下所示：

@JsonCreator
public Builder(@JsonProperty TransactionType transactionType) {
    this.transactionType = transactionType;
}

我收到另一个错误：

com.fasterxml.jackson.databind.JsonMappingException: Could not find creator
property with name '' (in class Transaction$Builder)

任何想法如何解决这个问题？

Answer 1

毕竟我成功编写了这个构建器：

public static final class Builder {
    @JacksonXmlText
    private final TransactionType transactionType;
    private boolean transactionAllowed;

    @JsonCreator
    public Builder(@JsonProperty(" ") TransactionType transactionType) {
        this.transactionType = transactionType;
    }

    public Builder withTransactionAllowed() {
        transactionAllowed = true;
        return this;
    }

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

注意@JsonProperty 的值不是空的，它实际上可以是除空字符串之外的任何值。我不相信这是有史以来最好的解决方案，但它确实有效。但我不会接受我自己的答案，因为可能有更好的解决方案。