如果返回的类型不是已知类型,我该如何使用 intoGroups() ?
How can I use intoGroups() if the returned type is not a known type?
我有一个用例,我通过连接两个表 ITEM
和 ITEM_DESCRIPTION
来构造我的结果。从那里我开始使用几列,然后我想方便地将它们转换成一个对象列表。在我的例子中,这些对象实际上是 DTO 对象,但当然它们也可以是业务对象。
这就是我现在的做法:
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,结果应该是一个项目列表,其中每个项目都有不同语言的项目详细信息:
StoreItemDTO
- Long id
// Maps language-id to item details
- Map<Long, StoreItemDetails> itemDetails
StoreItemDetails
- Long id
- String name
- String description
我找不到 intoGroups()
的版本 return 是一个有用的类型。我可以想象有类似 Map<Long, List<Record>>
的东西,但我做不到。
但是,intoGroups(RecordMapper<? super R, K> keyMapper)
可能 就是我要找的东西。如果映射器还允许我将结果记录实际转换为自定义对象,如 MyCustomPojo
,那么我可以非常方便地检索和转换数据。我不知道这是否可能。类似于:
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;
}
});
可惜编译器只允许
fetch.intoGroups(new RecordMapper<Record, Result<?>>() {
@Override
public Result<?> map(Record record) {
return null;
}
});
经过一番摆弄编译器后,事实证明这是可以做到的。
我不得不“作弊”,将我的结果地图声明为匿名之外的final
,实际上我没有“使用" keyMapper
参数,因为我刚刚返回 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,我仍然愿意接受建议并愿意接受任何可以优雅地缩短此代码的答案 - 如果此时可能的话。 :)
我有一个用例,我通过连接两个表 ITEM
和 ITEM_DESCRIPTION
来构造我的结果。从那里我开始使用几列,然后我想方便地将它们转换成一个对象列表。在我的例子中,这些对象实际上是 DTO 对象,但当然它们也可以是业务对象。
这就是我现在的做法:
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,结果应该是一个项目列表,其中每个项目都有不同语言的项目详细信息:
StoreItemDTO
- Long id
// Maps language-id to item details
- Map<Long, StoreItemDetails> itemDetails
StoreItemDetails
- Long id
- String name
- String description
我找不到 intoGroups()
的版本 return 是一个有用的类型。我可以想象有类似 Map<Long, List<Record>>
的东西,但我做不到。
但是,intoGroups(RecordMapper<? super R, K> keyMapper)
可能 就是我要找的东西。如果映射器还允许我将结果记录实际转换为自定义对象,如 MyCustomPojo
,那么我可以非常方便地检索和转换数据。我不知道这是否可能。类似于:
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;
}
});
可惜编译器只允许
fetch.intoGroups(new RecordMapper<Record, Result<?>>() {
@Override
public Result<?> map(Record record) {
return null;
}
});
经过一番摆弄编译器后,事实证明这是可以做到的。
我不得不“作弊”,将我的结果地图声明为匿名之外的final
,实际上我没有“使用" keyMapper
参数,因为我刚刚返回 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,我仍然愿意接受建议并愿意接受任何可以优雅地缩短此代码的答案 - 如果此时可能的话。 :)