排序聚合addToSet结果 [英] Sorting aggregation addToSet result

问题描述

有没有一种方法可以获取$ addToSet的结果作为排序数组?

Is there a way to get result of $addToSet as sorted array ?

我尝试扩展管道并展开数组，对其进行排序并再次将其分组， 但结果仍未排序.

I tried to expand the pipeline and $unwind the array, sort it and group it again , but still the result isn't sorted.

数组很大，我尽量避免在应用程序中对它们进行排序.

The arrays are pretty big and i try to avoid sort them in the the application.

Document Example :

    {
      "_id" : ObjectId("52a84825cc8391ab188b4567"),
      "id" : 129624
      "message" : "Sample",
      "date" : "12-09-2013,17:34:34",
      "dt" : ISODate("2013-12-09T17:34:34.000Z"),

    }

查询:

    db.uEvents.aggregate(
    [
      {$match : {dt : {$gte : new Date(2014,01,01) , $lt : new Date(2015,01,17)}}}
      ,{$sort : {dt : 1}}
      , {$group : {
        _id : {
                id : "$id"
                , year : {'$year' : "$dt"}
                , month : {'$month' : "$dt"}
                , day : {'$dayOfMonth' : "$dt"}
            }
        ,dt : {$addToSet : "$dt"}

      }}    
    ]

    );

推荐答案

是可以的，但是使用不同的方法.我只是为此提供自己的数据，但是您会明白的.

Yes it is possible, but approach it differently. I'm just provide my own data for this, but you'll get the concept.

我的样本:

{ "array" : [  2,  4,  3,  5,  2,  6,  8,  1,  2,  1,  3,  5,  9,  5 ] }

我要对此CTO进行半报价"，并声明集合被认为是无序.

I'm going to "semi-quote" the CTO on this and state that Sets are considered to be unordered.

Google组有一个实际的JIRA声明，类似这样.因此，让我们从"Elliot"中获取它，并接受这种情况 .

因此，如果要获得有序的结果，则必须在这样的阶段进行按摩

So if you want an ordered result, you have to massage that way with stages like this

db.collection.aggregate([

    // Initial unwind
    {"$unwind": "$array"},

    // Do your $addToSet part
    {"$group": {"_id": null, "array": {"$addToSet": "$array" }}},

    // Unwind it again
    {"$unwind": "$array"},

    // Sort how you want to
    {"$sort": { "array": 1} },

    // Use $push for a regular array
    {"$group": { "_id": null, "array": {"$push": "$array" }}}

])

然后执行任何操作.但是现在您的数组已排序.

And then do whatever. But now your array is sorted.

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