我遇到过一种情况,我的映射方法有3个参数,所有这三个参数都用于导出目标类型的一个属性。
我在接口中创建了一个默认的映射方法,保留了导出属性的逻辑,现在为了调用这个方法,我可以在expression = "java( /*method call here*/ )"
注释中使用@Mapping
。
有没有办法用@qualifiedByName
之类的任何一个maptruct注释来做这个,我试着用注释具有表达式属性的注释并使用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;
}
}
答案 0 :(得分:4)
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
。根据我的理解,projId
和code
只是映射的帮助器,它们不用于映射目标属性。
所以你可以做一些事情(它应该在理论上起作用):
@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它(这将允许多个相同类型的参数)。
该课程如下:
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,您可以调用映射器,如:mapper.createOne(one, new Filter(projId, val, code));
答案 1 :(得分:1)
从1.2版开始支持: http://mapstruct.org/documentation/stable/reference/html/#mappings-with-several-source-parameters
例如这样的例子:
@Mappings({
@Mapping(source = "person.description", target = "description"),
@Mapping(source = "address.houseNo", target = "houseNumber")
})
DeliveryAddressDto personAndAddressToDeliveryAddressDto(Person person, Address address);
答案 2 :(得分:0)
您可以创建一个默认方法,该方法使用其他上下文参数在内部调用mapstruct方法。这样,您就可以在'qualifiedByName'部分获取所有参数
@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;
}
}
```