Question

首先让我说我是mongo和node.js的新手。对于我工作的每个客户，我正在查看过去一周的流量，以查找六个变量的最大值和最小值以及它们的日期。现在mongo聚合负责一些工作，但我相信它可以做更多。现在我的管道如下：

$ match - 仅返回过去一周内的文件
$ group - 按客户和按日期分组
$ project - 计算一些值

文件以下列庄园退回：

{ _id: 
   { client: "aRandomClientNameWouldBeHere",
     date: Thu Jun 04 46370 00:00:00 GMT-0400 (EDT) },
  impSales: 0,
  pixSales: 0,
  apiSales: 0,
  s2sSales: 0,
  clicks: 3,
  impressions: 0 }

我无法在管道中添加另一个$ group阶段，只对客户进行分组并获取每个字段的最大值和最小值的日期。

我需要这样的东西：

$group: { 
  _id: "$client" ,
  maxImpSales: { $max: impSales },
  maxImpSalesDate : { ??? },
  minImpSales: { $min: impSales },
  minImpSalesDate : { ??? },
  ...
}

然而，由于两个原因，我无法让它工作。

我不知道如何按客户端进行分组，因为客户端现在与日期一起位于_id字段中。（即我使用_id：“$ client”或_id：_id.client还是别的什么？）
如何嵌入min和max发生的日期？

我如何做1＆amp; 2？谢谢。下面是我的3阶段管道

{
  $match:
  {
    date:
    {
      $gte: _.last(dutil.lastXDates(6))
    }
  }
},
{
  $group:
  {
    _id:
    {
      client: "$client",
      date: "$date"
    },
    z1is: { $sum: "$actions.sales.import.z1.count" },
    z2is: { $sum: "$actions.sales.import.z2.count" },
    z3is: { $sum: "$actions.sales.import.z3.count" },
    z1ps: { $sum: "$actions.sales.pixel.z1.count" },
    z2ps: { $sum: "$actions.sales.pixel.z2.count" },
    z3ps: { $sum: "$actions.sales.pixel.z3.count" },
    z1as: { $sum: "$actions.sales.apiOnly.z1.count" },
    z2as: { $sum: "$actions.sales.apiOnly.z2.count" },
    z3as: { $sum: "$actions.sales.apiOnly.z3.count" },
    z1ss: { $sum: "$actions.sales.s2s.z1.count" },
    z2ss: { $sum: "$actions.sales.s2s.z2.count" },
    z3ss: { $sum: "$actions.sales.s2s.z3.count" },
    z1ic: { $sum: "$actions.click.import.z1.count" },
    z2ic: { $sum: "$actions.click.import.z2.count" },
    z3ic: { $sum: "$actions.click.import.z3.count" },
    z1pc: { $sum: "$actions.click.pixel.z1.count" },
    z2pc: { $sum: "$actions.click.pixel.z2.count" },
    z3pc: { $sum: "$actions.click.pixel.z3.count" },
    z1ac: { $sum: "$actions.click.apiOnly.z1.count" },
    z2ac: { $sum: "$actions.click.apiOnly.z2.count" },
    z3ac: { $sum: "$actions.click.apiOnly.z3.count" },
    z1sc: { $sum: "$actions.click.s2s.z1.count" },
    z2sc: { $sum: "$actions.click.s2s.z2.count" },
    z3sc: { $sum: "$actions.click.s2s.z3.count" },
    z1ii: { $sum: "$actions.display.import.z1.count" },
    z2ii: { $sum: "$actions.display.import.z2.count" },
    z3ii: { $sum: "$actions.display.import.z3.count" },
    z1pi: { $sum: "$actions.display.pixel.z1.count" },
    z2pi: { $sum: "$actions.display.pixel.z2.count" },
    z3pi: { $sum: "$actions.display.pixel.z3.count" },
    z1ai: { $sum: "$actions.display.apiOnly.z1.count" },
    z2ai: { $sum: "$actions.display.apiOnly.z2.count" },
    z3ai: { $sum: "$actions.display.apiOnly.z3.count" },
    z1si: { $sum: "$actions.display.s2s.z1.count" },
    z2si: { $sum: "$actions.display.s2s.z2.count" },
    z3si: { $sum: "$actions.display.s2s.z3.count" }
  }
},
{
  $project:
  {
    totalCountImpSales: { $add: ["$z1is", "$z2is", "$z3is"] },
    totalCountImpClicks: { $add: ["$z1ic", "$z2ic", "$z3ic"] },
    totalCountImpImpressions: { $add: ["$z1ii", "$z2ii", "$z3ii"] },
    totalCountPixSales: { $add: ["$z1ps", "$z2ps", "$z3ps"] },
    totalCountPixClicks: { $add: ["$z1pc", "$z2pc", "$z3pc"] },
    totalCountPixImpressions: { $add: ["$z1pi", "$z2pi", "$z3pi"] },
    totalCountApiSales: { $add: ["$z1as", "$z2as", "$z3as"] },
    totalCountApiClicks: { $add: ["$z1ac", "$z2ac", "$z3ac"] },
    totalCountApiImpressions: { $add: ["$z1ai", "$z2ai", "$z3ai"] },
    totalCountS2sSales: { $add: ["$z1ss", "$z2ss", "$z3ss"] },
    totalCountS2sClicks: { $add: ["$z1sc", "$z2sc", "$z3sc"] },
    totalCountS2sImpressions: { $add: ["$z1si", "$z2si", "$z3si"] }
  }
}

Answer 1

有一种解决方法。在$sort之后添加$project阶段，然后使用$group阶段中的$first和$last运算符来获取最小值和最大值。那就是：

...
// Sort by impSales ascending
{
    $sort: { impSales: 1}
},
{
$group: { 
  _id: "$_id.client" ,
  maxImpSales: { $last: "$impSales" },
  maxImpSalesDate : { $last: "$_id.date" },
  minImpSales: { $first: "$impSales" },
  minImpSalesDate : { $first: "$_id.date" },
  ...
}
}
...

麻烦在mongodb聚合查询中使用$ group两次

1 个答案: