javascript - 使用多种变体过滤 MongoDB 集合

Question

我有一个这样的集合:

[
    {"userId": "0000", "algorithm": "algo1", "status": "Running",   "waitingTime": 0},
    {"userId": "0001", "algorithm": "algo1", "status": "Received",  "waitingTime": 0},
    {"userId": "0000", "algorithm": "algo2", "status": "Completed", "waitingTime": 123},
    {"userId": "0000", "algorithm": "algo2", "status": "Error",     "waitingTime": 134},
    {"userId": "0001", "algorithm": "algo2", "status": "Error",     "waitingTime": 150},
    {"userId": "0001", "algorithm": "algo3", "status": "Completed", "waitingTime": 100},
    {"userId": "0000", "algorithm": "algo3", "status": "Completed", "waitingTime": 120},
    {"userId": "0001", "algorithm": "algo1", "status": "Received",  "waitingTime": 0}
]

我需要找到每个用户的最长等待时间，但仅限于状态“已完成”。每个文档的输出应该有 2 个 props。

• “_id” - 应为原始集合中的“userId”
• “maxWaitingTime” - 应等于“已完成”状态的最长等待时间

对于此示例，输出应如下所示:

[
    {"_id": "0000", "maxWaitingTime": 123},
    {"_id": "0001", "maxWaitingTime": 100}
]

我尝试将一些过滤查询组合在一起，但效果不佳。我可以结合一两个陈述，但对于这一个我很挣扎。

我尝试将 unique 方法与 find 结合起来以获得唯一的 userId，但没有成功:

db.tasks.distinct("userId").find();

我还尝试执行 find 函数内部的逻辑:

printjsononeline(db.tasks.distinct(userId).find({
    userId: "0001"
}).map(doc => doc));

但不幸的是，它不起作用。

Answer 1

Playground

db.collection.aggregate([
  {
    "$match": {//Filter completed
      "status": "Completed"
    }
  },
  {
    $group: {//Group by user id and find max
      "_id": "$userId",
      "waitingTime": {
        "$max": "$waitingTime"
      }
    }
  }
])

注意:Mongo 区分大小写。人们确实使用 waiting_time 而不是 waitingTime。