如何在单个字段上创建一个对象作为GroupOperation id的Spring Data MongoDB GroupOperation？

Question

这是我想要根据一个命名对象创建的group操作。

private static GroupOperation createStatsGroupOperationFromNomenclature(Nomenclature nomenclature) {
    Fields groupFields = Fields.fields("departmentCode");
    nomenclature.getDepartmentCode().ifPresent(code -> groupFields.and("subDepartmentCode"));
    nomenclature.getSubDepartmentCode().ifPresent(code -> groupFields.and("categoryCode"));
    nomenclature.getCategoryCode().ifPresent(code -> groupFields.and("subCategoryCode"));
    return group(groupFields)
        .count().as("nbProducts")
        .sum("$proposedMatchesAmount").as("nbProposedMatches")
        .sum("$reviewedMatchesAmount").as("nbReviewedMatches");
}

对于前面的函数，如果我在departmentCode参数中提供了一个subDepartmentCode和一个mongo，下面是要执行的mongo查询：

{
  _id: {
    "departmentCode": "$departmentCode",
    "subDepartmentCode": "$subDepartmentCode"
  },
  "nbProduct": {
    $sum: 1
  },
  "proposedMatchesAmount": {
    $sum: "$proposedMatchesAmount"
  },
  "reviewedMatchesAmount": {
    $sum: "$reviewedMatchesAmount"
  }
}

此查询的结果将在以下对象中解析：

@Builder
@Value
public class ProductsStatsDocument {
  @Id
  Nomenclature nomenclature;
  Integer nbProducts;
  Integer nbProposedMatches;
  Integer nbReviewedMatches;
}

当我仅在departmentCode参数中提供命名时，问题就会被追加。然后，构建的组操作具有以下mongo查询语言的等价物：

{
  _id: "$departmentCode",
  "nbProduct": {
    $sum: 1
  },
  "proposedMatchesAmount": {
    $sum: "$proposedMatchesAmount"
  },
  "reviewedMatchesAmount": {
    $sum: "$reviewedMatchesAmount"
  }
}

并且这个查询的结果不能被解析成前一个ProductsStatsDocument，因为结果_id字段id现在是一个字符串，而不是一个命名对象。

即使只有一个字段，也可以强制group方法使用一个对象作为结果_id字段吗？或者，有没有其他方法来构建这样的mongo群组操作？

=================================================================

找到这个问题的“原因”。下面是spring data中的一段代码，它将GroupOperation转换为bson对象：

    } else if (this.idFields.exposesSingleNonSyntheticFieldOnly()) {
      FieldReference reference = context.getReference((Field)this.idFields.iterator().next());
      operationObject.put("_id", reference.toString());
    } else {

下面是exposesSingleNonSyntheticFieldOnly方法：

  boolean exposesSingleNonSyntheticFieldOnly() {
    return this.originalFields.size() == 1;
  }

正如您所看到的，只要只有一个字段可以分组，它就被用作_id结果值。

Answer 1

最后，目前可行的解决方案是创建一个管理文档转换_id部件的自定义AggregationOperation：

public class ProductsStatsGroupOperation implements AggregationOperation {

  private static GroupOperation getBaseGroupOperation() {
    return group()
        .count().as("nbProducts")
        .sum("$proposedMatchesAmount").as("nbProposedMatches")
        .sum("$reviewedMatchesAmount").as("nbReviewedMatches");
  }

  private final Nomenclature nomenclature;

  public ProductsStatsGroupOperation(Nomenclature nomenclature) {
    this.nomenclature = nomenclature;
  }

  @Override
  public Document toDocument(AggregationOperationContext context) {
    Document groupOperation = getBaseGroupOperation().toDocument(context);

    Document operationId = new Document();
    for (Field field : getFieldsToGroupOn()) {
      FieldReference reference = context.getReference(field);
      operationId.put(field.getName(), reference.toString());
    }
    ((Document)groupOperation.get("$group")).put("_id", operationId);

    return groupOperation;
  }

  private Fields getFieldsToGroupOn() {
    Fields groupFields = Fields.fields("departmentCode");
    if (nomenclature.getDepartmentCode().isPresent()) {
      groupFields = groupFields.and("subDepartmentCode");
    }
    if (nomenclature.getSubDepartmentCode().isPresent()) {
      groupFields = groupFields.and("categoryCode");
    }
    if (nomenclature.getCategoryCode().isPresent()) {
      groupFields = groupFields.and("subCategoryCode");
    }
    return groupFields;
  }
}

这个解决方案有一个不好的地方:被覆盖的方法toDocument似乎已被弃用。