Question

我有一个需要映射的mongo ObjectId列表，将一些响应（可能存在或可能不存在）分配给该id，然后根据响应生成可用性平均值。例如，我有一个引用Users的ObjectId列表：

["5d978d372f263f41cc624727", "5d978d372f263f41cc624728", "5d978d372f263f41cc624729"]

还有Events的汇总列表，其中包含一个employeeResponses子文档，如果子文档已响应，则记录其_id，response和任何事件{{1 }}：

notes

我需要做的是在[ { ... employeeResponses: [ { _id: "5d978d372f263f41cc624727", response: "I want to work.", notes: "" }, { _id: "5d978d372f263f41cc624728", response: "Available to work.", notes: "" } ] }, { ... employeeResponses: [ { _id: "5d978d372f263f41cc624727", response: "I want to work.", notes: "" }, { _id: "5d978d372f263f41cc624728", response: "Prefer not to work.", notes: "" } ] }, { ... employeeResponses: [ { _id: "5d978d372f263f41cc624727", response: "Not available to work.", notes: "" }, { _id: "5d978d372f263f41cc624729", response: "Not available to work.", notes: "" }, ] } ]数组中将他们的响应（或没有响应）分配给他们的ID：

responses

然后，根据他们的响应，生成成员平均可用性（可用：[ { "_id": "5d978d372f263f41cc624727", "responses": ["I want to work.", "I want to work.", "Not available to work."] }, { "_id": "5d978d372f263f41cc624728", "responses": ["Available to work.", "Prefer not to work.", "No response."] } { "_id": "5d978d372f263f41cc624729", "responses": ["No response.", "No response.", "Not available to work."] } ]； 不可用：["I want to work.", "Available to work"]）：

["Prefer not to work.", "Not available to work.", "No response."]

在@mickl的帮助下，我可以aggregate the responses for 1 ObjectId，但是我一直在努力将其扩展为ObjectId列表，而没有映射ID并按用户进行汇总-这似乎非常低效且昂贵。 / p>

Answer 1

您可以尝试以下方法：

原始示例：https://mongoplayground.net/p/_pDaErcuvkP

https://mongoplayground.net/p/oURNxgNiprl中的调整查询：

    db.collection.aggregate([
  {
    $unwind: {
      path: "$employeeResponses",          
    }
  },
  {
    "$group": {
      _id: "$employeeResponses._id",
      "availability": {
        "$sum": {
          "$cond": [
            {
              "$in": [
                "$employeeResponses.response",
                [
                  "Available to work.",
                  "I want to work."
                ]
              ]
            },
            1,
            0
          ]
        }
      }
    }
  }
])

汇总子文档，按ID列表过滤，按ID分组值

1 个答案: