Java 使用多个列进行分组有问题吗？

我有这个方法：

public List<IncomeChannelCategoryMap> allIncomeChannels(final List<String> list) {
    final CriteriaQuery<IncomeChannelCategoryMap> criteriaQuery = builder.createQuery(IncomeChannelCategoryMap.class);
    final Root<IncomeChannelMapEntity> root = criteriaQuery.from(IncomeChannelMapEntity.class);

    final List<Selection<?>> selections = new ArrayList<>();
    selections.add(root.get(IncomeChannelMapEntity_.incomeChannel).get(IncomeChannelEntity_.code));
    selections.add(root.get(IncomeChannelMapEntity_.logicalUnitCode));
    selections.add(root.get(IncomeChannelMapEntity_.logicalUnitIdent));
    selections.add(root.get(IncomeChannelMapEntity_.keyword));
    criteriaQuery.multiselect(selections);
    Predicate codePredicate = root.get(IncomeChannelMapEntity_.incomeChannel).get(IncomeChannelEntity_.code).in(list);
    criteriaQuery.where(codePredicate);
    return entityManager.createQuery(criteriaQuery).getResultList();

}

从

一切都很好，但有一个问题：

例如，直销可以有另一个logicalUnitCode，所以现在我只得到一个，我想为logicalUnitCode取得成功，就像我为logicalUnitCode取得成功一样

有什么建议吗？ 因此，我想要实现的是：

      {
        "incomeChannelCode": "DIRECT_SALES",
        "logicalUnitCode": "R_CATEGORY","R_TYPE",
        "logicalUnitIdent": "7,8"
      }

---

以下是您的更新代码：

@Override
public List<IncomeChannelCategoryMap> allIncomeChannels(final EntityRequest<IncomeChannel> request) throws ApiException {
    List<String> lists = request.getEntity().getIncomeChannels();
    List<IncomeChannelCategoryMap> channels = incomeChannelMapDAO.allIncomeChannels(lists);
    return new ArrayList<>(channels.stream().collect(Collectors.toMap(IncomeChannelCategoryMap::getIncomeChannelCode,
            Function.identity(), (i1, i2) -> {
                i1.setLogicalUnitIdent(i1.getLogicalUnitIdent() + ", " + i2.getLogicalUnitIdent());
                if (!i1.getLogicalUnitCode().contains(i2.getLogicalUnitCode())) {
                    i1.setLogicalUnitCode(i1.getLogicalUnitCode() + ", " + i2.getLogicalUnitCode());
                }
                return i1;
            })).values());
}

@覆盖
公共列表allIncomeChannels（最终EntityRequest请求）引发异常{
List lists=request.getEntity（）.getIncomeChannels（）；
列表通道=incomechannellmapdao.allIncomeChannels（列表）；
返回新的ArrayList（channels.stream（）.collect（Collectors.toMap）（IncomeChannelCategoryMap:：getIncomeChannelCode、，
Function.identity（），（i1，i2）->{
i1.SetLogicalInitiadent（i1.GetLogicalInitiadent（）+”，“+i2.GetLogicalInitiadent（））；
如果（！i1.getLogicalUnitCode（）。包含（i2.getLogicalUnitCode（）））{
i1.setLogicalUnitCode（i1.getLogicalUnitCode（）+”，“+i2.getLogicalUnitCode（））；
}
返回i1；
})).values（））；
}

---

就像

logicalinitident

一样，现在

logicalUnitCode

也将被分组。这里我假设你不想在这里重复。我所说的重复是指，如果两个结果的

logicalUnitCode

都是

“R\u CATEGORY”

，那么您希望它作为输出一次。如果一个是

“R\u CATEGORY”

，另一个是

“R\u TYPE”

，则希望将它们分组为

“R\u CATEGORY，R\u TYPE”

，作为输出。如果我的假设是正确的，那么这就是您所需的答案。

您好，无。那

R\u类型

是从哪里来的？我在原始输出中看不到它。

[
{
    "incomeChannelCode": "DIRECT_SALES",
    "logicalUnitCode": "R_CATEGORY",
    "logicalUnitIdent": "7"
  },
  {
    "incomeChannelCode": "DIRECT_SALES",
    "logicalUnitCode": "R_CATEGORY",
    "logicalUnitIdent": "8"
  }
]

      {
        "incomeChannelCode": "DIRECT_SALES",
        "logicalUnitCode": "R_CATEGORY","R_TYPE",
        "logicalUnitIdent": "7,8"
      }

@Override
public List<IncomeChannelCategoryMap> allIncomeChannels(final EntityRequest<IncomeChannel> request) throws ApiException {
    List<String> lists = request.getEntity().getIncomeChannels();
    List<IncomeChannelCategoryMap> channels = incomeChannelMapDAO.allIncomeChannels(lists);
    return new ArrayList<>(channels.stream().collect(Collectors.toMap(IncomeChannelCategoryMap::getIncomeChannelCode,
            Function.identity(), (i1, i2) -> {
                i1.setLogicalUnitIdent(i1.getLogicalUnitIdent() + ", " + i2.getLogicalUnitIdent());
                if (!i1.getLogicalUnitCode().contains(i2.getLogicalUnitCode())) {
                    i1.setLogicalUnitCode(i1.getLogicalUnitCode() + ", " + i2.getLogicalUnitCode());
                }
                return i1;
            })).values());
}