具有多个参数的 MapStruct QualifiedByName

Question

我遇到过这样一种情况，我的映射方法有 3 个参数，而这三个参数都用于派生目标类型的属性之一。

我在接口中创建了一个默认映射方法，保留了派生属性的逻辑，现在为了调用这个方法，我可以在@Mapping 注释中使用expression = "java( /*method call here*/ )"。

有没有办法用任何 mapstruct 注释来做到这一点，比如@qualifiedByName，我尝试评论具有表达式属性的注释并使用了qualifiedByName，但它不起作用：

@Mapper
public interface OneMapper {

    @Mapping(target="id", source="one.id")
    //@Mapping(target="qualified",expression = "java( checkQualified (one, projId, code) )")
    @Mapping(target="qualified",qualifiedByName="checkQualifiedNamed")
    OneDto createOne (One one, Integer projId, Integer val, String code);

    @Named("checkQualifiedNamed")
    default Boolean checkQualified (One one, Integer projId, Integer val, String code) {
        if(one.getProjectId() == projId && one.getVal() == val && one.getCode().equalsIgnoreCase(code)) {
            return Boolean.TRUE;
        }
        return Boolean.FALSE;                   
    }
}

Answer 1

目前 MapStruct 不支持具有多个源属性的映射方法。

但是，在您的情况下，您可以使用 1.2.0 中的 @Context。据我了解，projId 和 code 只是作为映射的助手，它们不用于映射目标属性。

所以你可以做类似的事情（理论上应该可行）：

@Mapper
public interface OneMapper {

    @Mapping(target="id", source="one.id")
    @Mapping(target="qualified", qualifiedByName="checkQualifiedNamed")
    OneDto createOne (One one, @Context Integer projId, @Context String code);

    @Named("checkQualifiedNamed")
    default Boolean checkQualified (One one, @Context Integer projId, @Context String code) {
        if(one.getProjectId() == projId && one.getCode().equalsIgnoreCase(code)) {
            return Boolean.TRUE;
        }
        return Boolean.FALSE;                   
    }
}

另一种选择是将所有这些属性提取到一个单独的类中并传递（这将允许同一类型的多个参数）。

这个类看起来像：

public class Filter {

    private final Integer projId;
    private final Integer val;
    private final String code;

    public Filter (Integer projId, Integer val, String code) {
        this.projId = projId;
        this.val = val;
        this.code = code;
    }

    //getters
}

您的映射器将如下所示：

@Mapper
public interface OneMapper {

    @Mapping(target="id", source="one.id")
    @Mapping(target="qualified", qualifiedByName="checkQualifiedNamed")
    OneDto createOne (One one, @Context Filter filter);

    @Named("checkQualifiedNamed")
    default Boolean checkQualified (One one, @Context Filter filter) {
        if(one.getProjectId() == filter.getProjId() && one.getVal() == filter.getVal() && one.getCode().equalsIgnoreCase(filter.getCode())) {
            return Boolean.TRUE;
        }
        return Boolean.FALSE;                   
    }
}

然后您可以像这样调用映射器：mapper.createOne(one, new Filter(projId, val, code));

Answer 2

从 1.2 版开始支持： http://mapstruct.org/documentation/stable/reference/html/#mappings-with-several-source-parameters

例如这样：

@Mapping(source = "person.description", target = "description")
@Mapping(source = "address.houseNo", target = "houseNumber")
DeliveryAddressDto personAndAddressToDeliveryAddressDto(Person person, Address address);

更新

由于 Mapstruct 允许将多个源参数映射到单个目标，我建议从映射器中提取 checkQualified 方法，而不是预先计算结果并使用 checkQualified 方法的结果调用映射器。 Mapstruct 是一个 mapping 库，不擅长执行任意逻辑。这并非不可能，但就个人而言，我看不出它在您的特定情况下增加的价值。

提取逻辑后，您的映射器可能如下所示：

@Mapper
public interface OneMapper {
    OneDto toOneDto(One one, Boolean qualified);
}

映射器可以这样使用：

One one = new One(1, 10, 100, "one");
boolean qualified = checkQualified(one, 10, 100, "one");
boolean notQualified = checkQualified(one, 10, 100, "two");
OneDto oneDto = mapper.toOneDto(one, isQualified);

有关完整示例，请参阅：https://github.com/phazebroek/so-mapstruct/blob/master/src/main/java/nl/phazebroek/so/MapStructDemo.java

Answer 3

如果您需要基于来自同一源对象的多个源字段计算单个目标字段，您可以将完整源对象传递给自定义映射器函数，而不是单个字段：

示例实体：

@Entity
@Data
public class User {
    private String firstName;
    private String lastName;
}

DTO 示例：

public class UserDto {
    private String fullName;
}

...和映射器...而不是传递单个源（名字）：

@Mapper
public abstract class UserMapper {

    @Mapping(source = "firstName", target = "fullName", qualifiedByName = "nameTofullName")
    public abstract UserDto userEntityToUserDto(UserEntity userEntity);


    @Named("nameToFullName")
    public String nameToFullName(String firstName) {
        return String.format("%s HOW DO I GET THE LAST NAME HERE?", firstName);
    }

...传递完整的实体对象（userEntity）作为源：

@Mapper
public abstract class UserMapper {

    @Mapping(source = "userEntity", target = "fullName", qualifiedByName = "nameToFullName")
    public abstract UserDto userEntityToUserDto(UserEntity userEntity);


    @Named("nameToFullName")
    public String nameToOwner(UserEntity userEntity) {
        return String.format("%s %s", userEntity.getFirstName(), userEntity.getLastName());
    }

Answer 4

您可以创建一个默认方法，该方法在内部调用带有附加上下文参数的mapstruct方法。这样，您可以获得'qualifiedByName'部分中的所有参数

@Mapper
public interface OneMapper {

    default OneDto createOne(One one, Integer projId, Integer val, String code) {
        return createOneWithContext(one,porjId,val,code
                                    one,porjId,val,code //as context params
        );
    }

    @Mapping(target="id", source="one.id")
    @Mapping(target="qualified",source="one",qualifiedByName="checkQualifiedNamed")
    OneDto createOneWithContext (One one, Integer projId, Integer val, String code
                     @Context One oneAsContext, 
                     @Context Integer projIdAsContext, 
                     @Context Integer valAsContext, 
                     @Context String codeAsContext
    
);

    @Named("checkQualifiedNamed")
    default Boolean checkQualified (One one, @Context Integer projId, @Context Integer val, @Context String code) {
        if(one.getProjectId() == projId && one.getVal() == val && one.getCode().equalsIgnoreCase(code)) {
            return Boolean.TRUE;
        }
    return Boolean.FALSE;                   
    }
}



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