如何对值进行分组和连接？

Question

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();    }响应是这样的：[{    "incomeChannelCode": "DIRECT_SALES",    "logicalUnitCode": "R_CATEGORY",    "logicalUnitIdent": "7"  },  {    "incomeChannelCode": "DIRECT_SALES",    "logicalUnitCode": "R_CATEGORY",    "logicalUnitIdent": "8"  }]我想要实现的是：  {    "incomeChannelCode": "DIRECT_SALES",    "logicalUnitCode": "R_CATEGORY",    "logicalUnitIdent": "7,8"  }有什么建议我怎样才能实现这个目标？我尝试过这个，我在一些例子中发现了这一点：builder.function("group_concat", String.class, root.get(IncomeChannelMapEntity_.logicalUnitIdent));但这是行不通的。还有其他建议吗？@Data@Builder@NoArgsConstructor@AllArgsConstructorpublic class IncomeChannelCategoryMap implements Serializable {    private static final long serialVersionUID = 1L;    private String incomeChannelCode;    private String logicalUnitCode;    private String logicalUnitIdent;    private String keyword;}

Answer 1

尝试这个：

ArrayList<IncomeChannelCategoryMap> list = entityManager.createQuery(criteriaQuery).getResultList();

List<IncomeChannelCategoryMap> finalList = new ArrayList<>(list.stream().collect(

                 Collectors.toMap(IncomeChannelCategoryMap::getIncomeChannelCode, Function.identity(), (IncomeChannelCategoryMap i1, IncomeChannelCategoryMap i2) -> {

                     i1.setLogicalUnitIdent(i1.getLogicalUnitIdent()+","+i2.getLogicalUnitIdent());

                     return i1;

                 })).values());

 return finalList;

注意：请相应地添加您的 getter 方法，我只是假设您有这些方法名称。

Answer 2

您应该使用本机查询方法，正如我在上一个问题中建议的那样：

public List<IncomeChannelCategoryMap> allIncomeChannels(final List<String> list) {

    List<Object[]> resultList = entityManager.createNativeQuery(

            "select income_channel_code, logicalunit_code, string_agg(logicalunitident,',') idents, keyword from r_income_channel_map where income_channel_code in (:codes) group by logicalunit_code, income_channel_code, keyword")

            .setParameter("codes", list).getResultList();

    return resultList.stream().map(IncomeChannelCategoryMap::new).collect(Collectors.toList());

}

您需要将此构造函数添加到您的IncomeChannelCategoryMap类中：

IncomeChannelCategoryMap(Object[] objects) {

    this.incomeChannelCode = (String) objects[0];

    this.logicalUnitCode = (String) objects[1];

    this.logicalUnitIdent = (String) objects[2];

    this.keyword = (String) objects[3];

}