MongoDB自定义排序 [英] MongoDB custom sorting

问题描述

我有如下记录的集合

{
"_id":417,
"ptime":ISODate("2013-11-26T11:18:42.961Z"),
"p":{

    "type":"1",
    "txt":"test message"
},  

"users":[
    {
        "uid":"52872ed59542f",
        "pt":ISODate("2013-11-26T11:18:42.961Z")
    },
    {
        "uid":"524eb460986e4",
        "pt":ISODate("2013-11-26T11:18:42.961Z")
    },
    {
        "uid":"524179060781e",
        "pt":ISODate("2013-11-27T12:48:35Z")
    }
],

},

{
"_id":418,

"ptime":ISODate("2013-11-25T11:18:42.961Z"),
"p":{

    "type":"1",
    "txt":"test message 2"
},  

"users":[
    {
        "uid":"524eb460986e4",
        "pt":ISODate("2013-11-23T11:18:42.961Z")
    },
    {
        "uid":"52872ed59542f",
        "pt":ISODate("2013-11-24T11:18:42.961Z")
    },

    {
        "uid":"524179060781e",
        "pt":ISODate("2013-11-22T12:48:35Z")
    }
],

}

如何按照 ptime 和 pt 的降序对上述记录进行排序，其中 users uid ="52872ed59542f" ?

How to sort the above records with descending order of ptime and pt where users uid ="52872ed59542f" ?

推荐答案

您可以使用聚合框架按如下方式先按 ptime 再按 users.pt 字段排序.

You can use the Aggregation Framework to sort by first ptime and then users.pt field as follows.

db.users.aggregate(
  {$sort : {'ptime' : 1}},
  {$unwind : "$users"},
  {$match: {"users.uid" : "52872ed59542f"}},
  {$sort : {'users.pt' : 1}},
  {$group : {_id : {id : "$_id", "ptime" : "$ptime", "p" : "$p"}, users : {$push : "$users"}}},
  {$group : {_id : "$_id.id", "ptime" : {$first : "$_id.ptime"}, "p" : {$first : "$_id.p"}, users : {$push : "$users"}}}
);

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