Question

我想获取Mongoose Model.aggregate命令的结果并对其执行其他聚合方法。我的想法是，我可以只查询一次数据库，聚合到某个点，然后对这些结果执行其他聚合方法，而无需再次重新查询数据库。（以下代码为Coffeescript）

firstAggregation = [
    {
      $geoNear: {
        near:
          type: "Point"
          coordinates: geo
        includeLocs: "geo"
        distanceField: "distance"
        maxDistance: distance
        spherical: true
        uniqueDocs: true
      }
    },
    {
      $match: { "active" : true }
    }
  ]

secondAggregation = firstAggregation.concat [
        {
          $unwind: "$offers"
        },
        {
          $group: {
            _id: "$offers"
          }
        }
      ]

app.MyModel.aggregate firstAggregation, (err, results) ->

  # ... do something with the first results ...

  # **** apply secondAggregation here! ****
  # the below doesn't work, but its what I want to achieve

  results.aggregate secondAggregation, (err, secondResults) ->
    # ... do something with the second results without having requeried the DB ...

这可能吗？

Answer 1

不，您无法将结果从一个聚合传递到第二个聚合。虽然您可以传递文档_id的列表以使用$in进行过滤，但除此之外，您可以做的其他事情并不多。

{ $match : { _id : { $in: [ /* list of _ids */] } } }

$match reference
$in reference

虽然给定了聚合框架结果的当前v2.4限制（上限为16MB），但您可以执行类似于客户端内存数据库（在本例中为NodeJS）的逻辑。。

在您问题中的上述示例中，看起来只是为了尝试查找字段offers的值数组的所有不同值？

offerNames = {}
for result in results
    for offer in result.offers
        offerNames[i] = (offerNames[i] or 0) + 1

for name, value of offerNames
    console.log name + " = " + value

对现有的Mongoose聚合结果执行其他聚合

1 个答案: