Java MapStruct：如何将所有属性映射到列表的第一个元素？

我需要一个映射来实现这一点：

@Mapping(source = "a", target = "result.transactions[0].a"),
@Mapping(source = "b", target = "result.transactions[0].b"),
@Mapping(source = "c", target = "result.transactions[0].c"),
...
Response dataToResponse(DataModel model);

但是这种语法不起作用（顺便说一句：这适用于Springbean包装器）。 像这样的解决方案只是半熟的解决方案：

@AsList
public <T> List<T> asList( T in ) {
    List<T> result = new ArrayList<T>();
    if ( in!=null ) {
       result.add(in);
   }
   return result;
}

然后

@Mapping([use Transaction from b4], target = "result");

但是如何从上面传递已映射的字段？

---

（我使用的是最新的最终版本1.1.0.final）

显然没有干净的解决方案。因此，我必须通过将以下映射排除到单独的映射器中来解决此问题：

@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);

在主映射器中，我执行单独的映射器，并通过表达式将其转换为列表：

@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")

---

刚刚遇到这个问题。更优雅的解决方案是这样的：

@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);

@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);

@Mapping(expression = "java(player.getAddressBooks().get(0).getPostCode())", target = "postCode")
@Mapping(expression = "java(player.getAddressBooks().get(0).getCity())", target = "city")
@Mapping(expression = "java(player.getAddressBooks().get(0).getStreetNumber())", target = "streetNumber")
@Mapping(expression = "java(player.getAddressBooks().get(0).getStreetName())", target = "streetName")
@Mapping(expression = "java(player.getAddressBooks().get(0).getCountryISO())", target = "countryISO")

• 这与前面的答案相同

剩下要做的是将单个事务映射到一个元素列表：

default List<Transaction> mapTransactionToList(Transaction source) {
     return ImmutableList.of(source);
}

默认列表mapTransactionToList（事务源）{
返回不可变列表（源）；
}

现在，您可以在

@Mapping

定义中将

模型

映射到

列表

，而MapStruct应该知道该做什么

我认为这看起来比基于“表达式”的解决方案更好，更不容易出错

---

请注意，您可以在映射器界面中包含代码。

我认为使用表达式是此处的工作重点：

public interface MyMapper {
    @Mapping(target = "subjectName", expression = "java(source.getCourses().get(0).getCourseName())")
    Target map(Source source);
} 

另请查看此处了解更多信息：

在我的代码中，我是这样做的：

@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);

@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);

@Mapping(expression = "java(player.getAddressBooks().get(0).getPostCode())", target = "postCode")
@Mapping(expression = "java(player.getAddressBooks().get(0).getCity())", target = "city")
@Mapping(expression = "java(player.getAddressBooks().get(0).getStreetNumber())", target = "streetNumber")
@Mapping(expression = "java(player.getAddressBooks().get(0).getStreetName())", target = "streetName")
@Mapping(expression = "java(player.getAddressBooks().get(0).getCountryISO())", target = "countryISO")