Question

我有一个这样的集合，但有更多的数据。

{
  _id: ObjectId("db759d014f70743495ef1000"),
  tracked_item_origin: "winword",
  tracked_item_type: "Software",
  machine_user: "mmm.mmm",
  organization_id: ObjectId("a91864df4f7074b33b020000"),
  group_id: ObjectId("20ea74df4f7074b33b520000"),
  tracked_item_id: ObjectId("1a050df94f70748419140000"),
  tracked_item_name: "Word",
  duration: 9540,
}

{
  _id: ObjectId("2b769d014f70743495fa1000"),
  tracked_item_origin: "http://www.facebook.com",
  tracked_item_type: "Site",
  machine_user: "gabriel.mello",
  organization_id: ObjectId("a91864df4f7074b33b020000"),
  group_id: ObjectId("3f6a64df4f7074b33b040000"),
  tracked_item_id: ObjectId("6f3466df4f7074b33b080000"),
  tracked_item_name: "Facebook",
  duration: 7920,
}

我做了一个聚合，请返回这样的分组数据：

{"_id"=>{"tracked_item_type"=>"Site", "tracked_item_name"=>"Twitter"}, "duration"=>288540},
{"_id"=>{"tracked_item_type"=>"Site", "tracked_item_name"=>"ANoticia"}, "duration"=>237300},
{"_id"=>{"tracked_item_type"=>"Site", "tracked_item_name"=>"Facebook"}, "duration"=>203460},
{"_id"=>{"tracked_item_type"=>"Software", "tracked_item_name"=>"Word"}, "duration"=>269760},
{"_id"=>{"tracked_item_type"=>"Software", "tracked_item_name"=>"Excel"}, "duration"=>204240}

简单聚合代码：

AgentCollector.collection.aggregate(
  {'$match' => {group_id: '20ea74df4f7074b33b520000'}},
  {'$group' => {
    _id: {tracked_item_type: '$tracked_item_type', tracked_item_name: '$tracked_item_name'},
    duration: {'$sum' => '$duration'}
  }},
  {'$sort' => {
    '_id.tracked_item_type' => 1,
    duration: -1
  }}
)

有一种方法可以通过tracked_item_type键限制只有2个项目吗？防爆。 2个站点和2个软件。

Answer 1

由于你的问题目前尚不清楚，我真的希望你的意思是你要指定两个Site键和2个Software键，因为这是一个很好而简单的答案，你可以添加到你的$匹配阶段如：

{$match: {
    group_id: "20ea74df4f7074b33b520000",
    tracked_item_name: {$in: ['Twitter', 'Facebook', 'Word', 'Excel' ] }
}},

我们都欢呼和快乐;）

但是，如果您的问题更加恶魔般，例如，根据持续时间从结果中获得前2个Sites和Software条目，那么我们非常感谢您产生憎恶< /强>

警告：

您的里程数可能因您实际想做的事情而有所不同，或者您的结果是否会因为大小而爆炸。但这是作为你所处的目的的一个例子：

db.collection.aggregate([ // Match items first to reduce the set {$match: {group_id: "20ea74df4f7074b33b520000" }}, // Group on the types and "sum" of duration {$group: { _id: { tracked_item_type: "$tracked_item_type", tracked_item_name: "$tracked_item_name" }, duration: {$sum: "$duration"} }}, // Sort by type and duration descending {$sort: { "_id.tracked_item_type": 1, duration: -1 }}, /* The fun part */ // Re-shape results to "sites" and "software" arrays {$group: { _id: null, sites: {$push: {$cond: [ {$eq: ["$_id.tracked_item_type", "Site" ]}, { _id: "$_id", duration: "$duration" }, null ]} }, software: {$push: {$cond: [ {$eq: ["$_id.tracked_item_type", "Software" ]}, { _id: "$_id", duration: "$duration" }, null ]} } }}, // Remove the null values for "software" {$unwind: "$software"}, {$match: { software: {$ne: null} }}, {$group: { _id: "$_id", software: {$push: "$software"}, sites: {$first: "$sites"} }}, // Remove the null values for "sites" {$unwind: "$sites"}, {$match: { sites: {$ne: null} }}, {$group: { _id: "$_id", software: {$first: "$software"}, sites: {$push: "$sites"} }}, // Project out software and limit to the *top* 2 results {$unwind: "$software"}, {$project: { _id: 0, _id: { _id: "$software._id", duration: "$software.duration" }, sites: "$sites" }}, {$limit : 2}, // Project sites, grouping multiple software per key, requires a sort // then limit the *top* 2 results {$unwind: "$sites"}, {$group: { _id: { _id: "$sites._id", duration: "$sites.duration" }, software: {$push: "$_id" } }}, {$sort: { "_id.duration": -1 }}, {$limit: 2} ])

现在导致的结果是* 并不完全那些理想的结果，但是它可以以编程方式使用，并且比在循环中过滤先前的结果更好。（我的测试数据）

{ "result" : [ { "_id" : { "_id" : { "tracked_item_type" : "Site", "tracked_item_name" : "Digital Blasphemy" }, "duration" : 8000 }, "software" : [ { "_id" : { "tracked_item_type" : "Software", "tracked_item_name" : "Word" }, "duration" : 9540 }, { "_id" : { "tracked_item_type" : "Software", "tracked_item_name" : "Notepad" }, "duration" : 4000 } ] }, { "_id" : { "_id" : { "tracked_item_type" : "Site", "tracked_item_name" : "Facebook" }, "duration" : 7920 }, "software" : [ { "_id" : { "tracked_item_type" : "Software", "tracked_item_name" : "Word" }, "duration" : 9540 }, { "_id" : { "tracked_item_type" : "Software", "tracked_item_name" : "Notepad" }, "duration" : 4000 } ] } ], "ok" : 1 }

所以你看到你得到了数组中的前2 Sites，其中每个都嵌入了前2个Software项。聚合本身，无法进一步明确这一点，因为我们需要重新合并我们拆分的项目才能做到这一点，而且还没有我们可以用来执行此操作的运算符动作。

但这很有趣。它不是全部完成的方式，但大多数的方式，并将其变成4文档响应将是相对简单的代码。但是我的头已经疼了。

限制聚合聚合中的聚合

1 个答案:

警告：