我将业务对象映射到实体,并且存在实体的结构与业务对象不同的情况
我有userCategories
作为字符串存储在业务对象RecipeBo
中,因为BO不必知道有关实体内部结构的任何信息。这些字符串需要映射到Recipe
和RecipeUserCategoryRel
的关系,除此之外,还需要在{{1}中映射userId
的另一个字段RecipeBo
也是。
我的方法(有效)是创建一个包装器并手动创建关系,但这看起来像修补:
RecipeUserCategoryRel
我是否有更好的方法来处理我在BoMapper中所做的事情,例如:使用转换器还是什么?难点在于映射列表的每个元素并添加public class BoMapper
{
private final static ModelMapper modelMapper = new ModelMapper();
static
{
modelMapper.addMappings(new IngredientMap());
}
public static void map(Object from, Object to)
{
modelMapper.map(from, to);
if (from instanceof RecipeBo && to instanceof Recipe)
{
RecipeBo recipeBo = (RecipeBo)from;
List<String> userCategories = recipeBo.getUserCategories();
List<RecipeUserCategoryRel> recipeUserCategoryRels = new ArrayList<>();
for (String userCategory : userCategories)
{
recipeUserCategoryRels.add(new RecipeUserCategoryRel(userCategory, recipeBo.getUserId()));
}
Recipe recipe = (Recipe)to;
recipe.setRecipeUserCategoryRels(recipeUserCategoryRels);
}
}
}
字段。
答案 0 :(得分:2)
这是一个复杂的情况,因为您从其他层次结构获取userId而不是直接从List获取。 ModelMapper会将List映射到List,但如果你没有将ModelMapper配置为LOOSE,它将无法工作。
modelMapper.getConfiguration()
.setMatchingStrategy(MatchingStrategies.LOOSE);
无论如何,如果以这种方式配置ModelMapper(LOOSE模式),它将映射List并放入Class RecipeUserCategoryRel的String属性(在本例中为userCategory,如果它是一个String并且考虑userId不是String )其他人(我不太确定)我认为它会是空的。
好吧,我认为您的问题的解决方案是创建一个转换器并将其添加到您的ModelMapper实例:
代码如下:
ModelMapper mapper = new ModelMapper();
Converter<RecipeBO, Recipe> converter = new Converter<RecipeBO,
Recipe>() {
@Override
public Recipe convert(MappingContext<RecipeBO, Recipe> context) {
RecipeBO source = context.getSource();
Recipe destination = new Recipe();
List<String> userCategoryValues = source.getUserCategories();
List<RecipeUserCategoryRel> userCategoryToMap = new ArrayList<RecipeUserCategoryRel>();
for(final String userCategory : userCategoryValues){
userCategoryToMap.add(new RecipeUserCategoryRel(userCategory,source.getUserId()));
}
destination.setRecipeUserCategoryRels(userCategoryToMap);
//... Map other properties if you need
return destination;
}
};
//Option 1
mapper.createTypeMap(RecipeBO.class, Recipe.class).setConverter(converter);
//If you add as a converter directly also works (I don't know which one is better,
//choose above option (createTypeMap + setConverter) or the next (addConverter)
//Option 2 -> mapper.addConverter(converter);
我已经过测试,它有效!!
如果我有下一个食谱:
RecipeBO recipe = new RecipeBO();
recipe.setUserId("1");
String values[] = new String[] { "abc", "klm", "xyz", "pqr" };
List<String> list = Arrays.asList(values);
recipe.setUserCategories(list);
和RecipeBO:
Recipe recipe = new Recipe();
List<RecipeUserCategoryRel> recipes = new ArrayList<>();
recipes.add(new RecipeUserCategoryRel("abc", "1"));
recipes.add(new RecipeUserCategoryRel("klm", "1"));
recipes.add(new RecipeUserCategoryRel("xyz", "1"));
recipes.add(new RecipeUserCategoryRel("pqr", "1"));
recipe.setRecipeUserCategoryRels(recipes);
当我将RecipeBO映射到食谱时:
Recipe actual = mapper.map(getRecipeBO(), Recipe.class);
我得到下一个输出:
<强>输出:强>
- RecipeUserCategoryRel(userCategory=abc, userId=1)
- RecipeUserCategoryRel(userCategory=klm, userId=1)
- RecipeUserCategoryRel(userCategory=xyz, userId=1)
- RecipeUserCategoryRel(userCategory=pqr, userId=1)