我有一个用例,其中通过联接两个表ITEM
和ITEM_DESCRIPTION
构造结果。从那里开始,我准备了几列,然后将它们方便地转换为对象列表。在我的情况下,这些对象实际上是DTO对象,但是当然它们也可以是业务对象。
这就是我现在做的方式:
public Map<Long, List<StoreItemDTO>> getItems(Long storeId) {
LOGGER.debug("getItems");
// Get all item_ids for the store
SelectHavingStep<Record1<Long>> where = this.ctx
.select(STORE_ITEM.ID)
.from(STORE_ITEM)
.where(STORE_ITEM.STORE_ID.eq(storeId))
// GROUP BY store_item.id
.groupBy(STORE_ITEM.ID);
// Get all store_item_details according to the fetched item_ids
TableLike<?> storeItemDetails = this.ctx
.select(
STORE_ITEM_DETAILS.ID,
STORE_ITEM_DETAILS.STORE_ITEM_ID,
STORE_ITEM_DETAILS.NAME,
STORE_ITEM_DETAILS.DESCRIPTION,
STORE_ITEM_DETAILS.STORE_LANGUAGE_ID
)
.from(STORE_ITEM_DETAILS)
.where(STORE_ITEM_DETAILS.STORE_ITEM_ID.in(where))
.asTable("storeItemDetails");
// Join the result and use
Field<Long> itemIdField = STORE_ITEM.ID.as("item_id");
Result<?> fetch = this.ctx
.select(
STORE_ITEM.ID.as("item_id"),
itemIdField,
storeItemDetails.field(STORE_ITEM_DETAILS.ID),
storeItemDetails.field(STORE_ITEM_DETAILS.NAME),
storeItemDetails.field(STORE_ITEM_DETAILS.DESCRIPTION),
storeItemDetails.field(STORE_ITEM_DETAILS.STORE_LANGUAGE_ID)
)
.from(STORE_ITEM)
.join(storeItemDetails)
.on(storeItemDetails.field(STORE_ITEM_DETAILS.STORE_ITEM_ID).eq(STORE_ITEM.ID))
.fetch();
Map<Long, ?> groups = fetch.intoGroups(STORE_ITEM.ID);
return null;
}
如您所见,结果应该是项目列表,其中每个项目都有一个使用不同语言的明细:
StoreItemDTO
- Long id
// Maps language-id to item details
- Map<Long, StoreItemDetails> itemDetails
StoreItemDetails
- Long id
- String name
- String description
我找不到会返回有用类型的
intoGroups()
版本。我可以想象有类似Map<Long, List<Record>>
的东西,但我却无法做到。但是,有一个
intoGroups(RecordMapper<? super R, K> keyMapper)
可能正是我想要的。如果映射器还允许我实际将结果记录转换为自定义对象,例如MyCustomPojo
,那么我可以非常方便地检索和转换数据。我不知道这是否可行。就像是:public static class MyCustomPojo {
public Long itemId;
// etc.
}
// ..
Map<Long, List<MyCustomPojo>> result = fetch.intoGroups(STORE_ITEM.ID, new RecordMapper<Record, List<MyCustomPojo>>() {
@Override
public List<MyCustomPojo> map(List<Record> record) {
// 'record' is grouped by STORE_ITEM.ID
// Now map each 'record' into every item of each group ..
return resultList;
}
});
但是不幸的是编译器只允许
fetch.intoGroups(new RecordMapper<Record, Result<?>>() {
@Override
public Result<?> map(Record record) {
return null;
}
});
最佳答案
经过一番摆弄之后,编译器被证明是可以做到的。
我不得不通过在匿名之外将生成的地图声明为final
来“作弊”,实际上我并没有“使用” keyMapper
参数,因为我只是返回了null
。
这是我想出的:
public Map<Long, StoreItemDTO> getItems(Long storeId) {
// Get all item_ids for the store
SelectHavingStep<Record1<Long>> where = this.ctx
.select(STORE_ITEM.ID)
.from(STORE_ITEM)
.where(STORE_ITEM.STORE_ID.eq(storeId))
.groupBy(STORE_ITEM.ID);
// Get all store_item_details according to the fetched item_ids
TableLike<?> storeItemDetails = this.ctx
.select(
STORE_ITEM_DETAILS.ID,
STORE_ITEM_DETAILS.STORE_ITEM_ID,
STORE_ITEM_DETAILS.NAME,
STORE_ITEM_DETAILS.DESCRIPTION,
STORE_ITEM_DETAILS.STORE_LANGUAGE_ID
)
.from(STORE_ITEM_DETAILS)
.where(STORE_ITEM_DETAILS.STORE_ITEM_ID.in(where))
.asTable("storeItemDetails");
// Join the result and use
final Field<Long> itemIdField = STORE_ITEM.ID.as("item_id");
Result<?> fetch = fetch = this.ctx
.select(
itemIdField,
storeItemDetails.field(STORE_ITEM_DETAILS.ID),
storeItemDetails.field(STORE_ITEM_DETAILS.NAME),
storeItemDetails.field(STORE_ITEM_DETAILS.DESCRIPTION),
storeItemDetails.field(STORE_ITEM_DETAILS.STORE_LANGUAGE_ID)
)
.from(STORE_ITEM)
.join(storeItemDetails)
.on(storeItemDetails.field(STORE_ITEM_DETAILS.STORE_ITEM_ID).eq(STORE_ITEM.ID))
.fetch();
final Map<Long, StoreItemDTO> itemIdToItemMap = new HashMap<>();
fetch.intoGroups(
record -> {
Long itemDetailsId = record.getValue(STORE_ITEM_DETAILS.ID);
// ... sake of compactness
StoreItemDetailsDTO storeItemDetailsDto = new StoreItemDetailsDTO();
storeItemDetailsDto.setId(itemDetailsId);
// ... sake of compactness
Long itemId = record.getValue(itemIdField);
StoreItemDTO storeItemDto = new StoreItemDTO();
storeItemDto.setId(itemId);
storeItemDto.getItemDetailsTranslations().put(languageId, storeItemDetailsDto);
StoreItemDTO itemDetailsList = itemIdToItemMap.get(itemId);
if(itemDetailsList == null) {
itemDetailsList = new StoreItemDTO();
itemIdToItemMap.put(itemId, itemDetailsList);
}
itemDetailsList.getItemDetailsTranslations().put(languageId, storeItemDetailsDto);
return null;
});
return itemIdToItemMap;
}
由于我不确定这是否是最优雅的解决方案,因此我仍然愿意征求建议,并愿意接受任何可以优雅缩短此代码的答案-如果目前可以的话。 :)