Bev*_*vor 9 java mapping mapstruct
我需要一个映射来实现此目的:
@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);
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但这种语法不起作用(顺便说一句:这适用于 Spring Bean 包装器)。这样的解决方案只是一个半生不熟的解决方案:
@AsList
public <T> List<T> asList( T in ) {
List<T> result = new ArrayList<T>();
if ( in!=null ) {
result.add(in);
}
return result;
}
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这仅适用于一个属性,因为它总是为每个属性创建一个新列表。我不需要将每个属性映射到新列表的第一个元素。该列表必须重复使用,但我不知道这是如何工作的。实现这一目标的正确方法是什么?我想到了这样的事情:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);
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进而...
@Mapping([use Transaction from b4], target = "result");
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但是我怎样才能从上面传递已经映射的字段呢?(我使用的是最新的最终版本1.1.0.Final)
我认为使用表达式是这里的解决方法:
public interface MyMapper {
@Mapping(target = "subjectName", expression = "java(source.getCourses().get(0).getCourseName())")
Target map(Source source);
}
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另请在此处查看更多信息: https ://github.com/mapstruct/mapstruct/issues/1321
在我的代码中我是这样做的:
@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")
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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);
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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")
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