如何在Mongo聚合查询中组合结果

Question

我不熟悉Mongo中的聚合查询，并且一直在努力产生我想要的输出。我有以下聚合查询：

db.events.aggregate([
  { $match: { requestState: "APPROVED" } },
  { $unwind: { path: "$payload.description" } },
  { $group: { _id: { instr: "$payload.description", bu: "$createdByUser", count: { $sum: 1 } } } }
]);

返回以下结果：

{ "_id" : { "instr" : "ABC-123", "bu" : "BU2", "count" : 1 } }
{ "_id" : { "instr" : "ABC-123", "bu" : "BU1", "count" : 1 } }
{ "_id" : { "instr" : "DEF-456", "bu" : "BU1", "count" : 1 } }

如何修改聚合查询，以便只返回2个文档而不是3个文档？将两个“ABC-123”结果合并为一个结果，并使用带有“bu”和“count”字段的新计数数组，即。

{ "_id" : { "instr" : "ABC-123", "counts": [ { "bu" : "BU1", "count" : 1 }, { "bu" : "BU2", "count" : 1 } ] } }

非常感谢

Answer 1

您可以通过\u id.instr将另一个阶段仅添加到$group，并将另一个阶段添加到所需的输出形状的$project

db.events.aggregate([
  {
    $match: { requestState: "APPROVED" }
  },
  {
    $unwind: { path: "$payload.description" }
  },
  {
    $group: {
      _id: { instr: "$payload.description", bu: "$createdByUser", count: { $sum: 1 } }
    }
  },
  {
    $group: {
      _id: { instr: "$_id.instr" },
      counts: { $push: { bu: "$_id.bu", count: "$_id.count" } }
    }
  },
  {
    $project: {
      _id: { instr: "$_id.instr", counts: "$counts" }
    }
  }
]);