Question

我有一个事件集合，user_id和created_at（ISO日期）。我需要每天对对象的数量求和，并且每天只计算一次user_id的事件。

我得到了下一个查询，但它计算了每天同一用户的所有事件：

results = Event.collection.aggregate([
  {"$match" =>  {
    "created_at" => {
      "$gte" => start_time, "$lte" => end_time
    },
    "event_type" => "click"
  }},
  {"$group" => {
    "_id" => {
      "$concat" => [
        {"$substr" => [{"$year" => "$created_at"}, 0, 4]},
        '-',
        {"$cond" => [
           {"$lt" => [{"$month" => "$created_at"}, 9]},
           {"$concat" => ["0", {"$substr" => [{"$month" => "$created_at"}, 0, 2]}]},
           {"$substr" => [{"$month" => "$created_at"}, 0, 2]}
        ]},
        '-',
        {"$cond" => [
           {"$lt" => [{"$dayOfMonth" => "$created_at"}, 9]},
           {"$concat" => ["0", {"$substr" => [{"$dayOfMonth" => "$created_at"}, 0, 2]}]},
           {"$substr" => [{"$dayOfMonth" => "$created_at"}, 0, 2]}
        ]}
      ]
    },
    "count" => {"$sum" => 1}
  }}
])

我试着将它与答案结合起来： Mongoid Grouping by date

但它对我不起作用。

如何区分user_id AKA会忽略同一天同一天的多条记录？

由于

Answer 1

我得到了这个，我只是在代码后面计算users_unique数组长度。

db.events.aggregate(
  {$match: {event_type: 'click', created_at: {
    $gte: start_time,
    $lt: end_time
  }}},
  {$project: {
    user_id: $user_id,
    year: {$year: $created_at}, 
    month: {$month: $created_at},
    day: {$dayOfMonth: $created_at}
  }},
  {$group: {
       _id: {year: $year, month: $month, day: $day},
       unique_users: {$addToSet: $user_id},
       total_sessions: {$sum: 1}
  }}
)

MongoDB组和不同

1 个答案: