I have some mappers like the following, converting a legacy time to a new type. The conversion requires some additional metadata. @Mapper(uses = NestedMapper.class) public interface SomeTypeMapper { @Mapping/* copy fields from legacy */ ...
mapper iamcrypticcoder 2,829 askedJun 6, 2020 at 11:18 3votes 2answers 2kviews Mapstruct generated classes are not found if the application is run from IntelliJ IDEA I have a Java 11 based application built with Gradle 4.8.1 which uses Mapstruct 1.3.0.Final. I use IntelliJ IDEA Ultimate...
Check out build output in "out/production/classes/generated" folder. If it's empty, try to rebuild your project. Make sure changes are applied by manually deleting contents of build output folders and modifying your mapper class before the rebuild. $ rm -rf out build .gradle Verify your G...
Fix the problem directly, I don't know enough the code of MapStruct, but would it be possible not to read the.classfile for the argument name, but only the current@Mappercode? This way the argument name in the actual mapper will have to match the one in the config, and we would ign...
This @Mapper uses another @Mapper. You can put as many classes/mappers here as you'd like - we only have one. Because we've added this flag, when generating the mapper implementation for the DoctorMapper interface, MapStruct will also convert the Patient model into a PatientDto - since ...
The project uses mapstruct to copy the object attributes. Multiple entity classes of the project have the same attributes, such as createTime . Whe...
@Mapper(componentModel = "spring", uses = { MapperUtils.class }) public interface CustomerTagApiMapper { CustomerTagAPI toCustomerTagApi(CustomerTag customerTag); } @Mapper(componentModel = "spring", uses = { CustomerTagApiMapper.class, MapperUtils.class }) ...
By doing so, you will have all the classes related to MapStruct under the same mapstruct package. All told, the current package will be com.mapstruct.demo.mapstruct.mappers. In our example, the following mapper will be enough: @Mapper( componentModel = "spring" ) public interface MapStruct...
Ideally, I'd like to define mappers so I range re-use the range mapping in multiple places: class RangeMapper { public <T extends Comparable> Range<T> toRange(T low, T high) { return Range.closedOpen(low, high); } } @Mapper(uses = RangeMapper.class) interface ModelMapper { @...