如何以自定义顺序对Mongo中的数据进行排序

问题描述 投票:0回答:3

我如何使用Mongo功能对这些数据进行排序:

输入

{ "_id" : 1, "domainName" : "test1.com", "hosting" : "hostgator.com" }
{ "_id" : 2, "domainName" : "test2.com", "hosting" : "aws.amazon.com"}
{ "_id" : 3, "domainName" : "test3.com", "hosting" : "aws.amazon.com" }
{ "_id" : 4, "domainName" : "test4.com", "hosting" : "hostgator.com" }
{ "_id" : 5, "domainName" : "test5.com", "hosting" : "aws.amazon.com" }
{ "_id" : 6, "domainName" : "test6.com", "hosting" : "cloud.google.com" }
{ "_id" : 7, "domainName" : "test7.com", "hosting" : "aws.amazon.com" }
{ "_id" : 8, "domainName" : "test8.com", "hosting" : "hostgator.com" }
{ "_id" : 9, "domainName" : "test9.com", "hosting" : "cloud.google.com" }
{ "_id" : 10, "domainName" : "test10.com", "hosting" : "godaddy.com" }

按托管字段顺序获取结果,例如按此顺序:

1)首先 - 所有的教父;

2)第二 - 所有AWS;

3)接下来,其他一切。

产量

{
    "result" : [
            {

                    "_id" : 10, 
                    "domainName" : "test10.com",
                    "hosting" : "godaddy.com"
            },
            {
                    "_id" : 2, 
                    "domainName" : "test2.com",
                    "hosting" : "aws.amazon.com"
            },
            {
                    "_id" : 3, 
                    "domainName" : "test3.com",
                    "hosting" : "aws.amazon.com"
            },
            {
                    "_id" : 5, 
                    "domainName" : "test5.com",
                    "hosting" : "aws.amazon.com"
            },
            {
                    "_id" : 7, 
                    "domainName" : "test7.com",
                    "hosting" : "aws.amazon.com"
            },
            {
                    "_id" : 1, 
                    "domainName" : "test1.com",
                    "hosting" : "hostgator.com"
            },
            {
                    "_id" : 4, 
                    "domainName" : "test4.com",
                    "hosting" : "hostgator.com"
            },
            {
                    "_id" : 6, 
                    "domainName" : "test6.com",
                    "hosting" : "cloud.google.com"
            },
            {
                    "_id" : 8, 
                    "domainName" : "test8.com",
                    "hosting" : "hostgator.com"
            },
            {
                    "_id" : 9, 
                    "domainName" : "test9.com",
                    "hosting" : "cloud.google.com"
            },
    ]
    }

通过这个例子,我想以更相关的方式为用户返回结果。在origanl任务中,我想使用少量其他集合对其进行排序,这些集合提供其他信息。

但是,如果你帮我完成以前的任务,那就足够了吗?

UPD:关于问题的第二部分。

另一个任务是如何从一个集合中依赖另一个集合返回可排序数据。

例:

第一个集合与之前给出的相同:

 { "_id" : 1, "domainName" : "test1.com", "hosting" : "hostgator.com" }
 ...

第二个集合,提供有关托管的一些其他信息:

{ '_id': 123, 'quality':'best', 'hostings': ["hostgator.com",  "aws.amazon.com"]},
{ '_id': 321, 'quality':'good', 'hostings': ["cloud.google.com"]},
{ '_id': 345, 'quality':'bad', 'hostings': ["godaddy.com"]},

结果,我需要按以下顺序从第一个集合数据返回:

1)首先所有好的主机2)所有好的3)第三个都坏

输出:

{
"result" : [
        //Best:
        {
                "_id" : 1, 
                "domainName" : "test1.com",
                "hosting" : "hostgator.com"
        },
        {
                "_id" : 4, 
                "domainName" : "test4.com",
                "hosting" : "hostgator.com"
        },
        {
                "_id" : 8, 
                "domainName" : "test8.com",
                "hosting" : "hostgator.com"
        },
        {
                "_id" : 2, 
                "domainName" : "test2.com",
                "hosting" : "aws.amazon.com"
        },
        {
                "_id" : 3, 
                "domainName" : "test3.com",
                "hosting" : "aws.amazon.com"
        },
        {
                "_id" : 5, 
                "domainName" : "test5.com",
                "hosting" : "aws.amazon.com"
        },
        {
                "_id" : 7, 
                "domainName" : "test7.com",
                "hosting" : "aws.amazon.com"
        },

       // Good:
        {
                "_id" : 9, 
                "domainName" : "test9.com",
                "hosting" : "cloud.google.com"
        },
        {
                "_id" : 6, 
                "domainName" : "test6.com",
                "hosting" : "cloud.google.com"
        },

       //Bad
        {
                "_id" : 10, 
                "domainName" : "test10.com",
                "hosting" : "godaddy.com"
        }
]
}

更新2

我在上一个例子中得到了很好的答案和例子。非常感谢!但我用另一个例子堆叠。

我需要比较3个集合的ID以按顺序排序 - 第一个:朋友,第二个:'requests',以及:其他用户。

输入

db.friends.find({userId: currentUser});
    // {"_id" : "PgC7LrtaZtQsShtzT", "userId" : "tHuxnWxFLHvcpRgHb", "friendId" : "jZagPF7bd4aW8agXb",}
db.requests.find({userId: currentUser});
   // looks like friend but with 'requesterId'

现在我需要聚合'用户'集合,定义与前两个集合(朋友,请求)匹配的分数。

使用提供的答案,我管理结果,但只有一个集合。如何使用3个或多个?

mongodb mongodb-query
3个回答
1
投票

您可以将每个hosting投影为由整数表示的单独类型,最后对这些整数进行排序。在下面的聚合管道中说明

[
{$lookup: {
    from: 'secondCollectionStoringQuality',
    localField: 'hosting',
    foreignField: 'hostings',
    as: 'nw'
    }},
{$unwind: '$nw'},
{$project: {
        domainName: 1,
        hosting: 1,
        type: {
            $cond: [
                {$eq: ['$nw.quality', 'best']},
                0,
                {$cond: [
                    {$eq: ['$nw.quality', 'good']},
                    1,
                    2
                    ]}
            ]
        }
    }},
    {$sort: {type: 1}}
]

0
投票

这个解决方案怎么样?

1)N个条件。

2)必需的排序顺序可以通过参数传递。

[{
        "$lookup": {
            "from": "secondCollectionStoringQuality",
            "localField": "hosting",
            "foreignField": "hostings",
            "as": "nw"
        }
    },
    {
        "$unwind": "$nw"
    },
    {
        "$project": {
            "domainName": 1,
            "hosting": 1,
            "type": {
                "$indexOfArray": [
                    ["best", "good"], "$nw.quality"
                ]
            }
        }
    },
    {
      "$project": {
          "domainName": 1,
          "hosting": 1,
          "type": {
              "$cond": [ {"$eq":["$type", -1]},  10000, "$type"]
          }
      }
    },
    {
        "$sort": {
            "type": 1
        }
    }
]

0
投票

如果数据存在于数组中,则上述答案有效。你可以在没有$lookup$project的情况下实现这一目标。

.aggregate([
   {$addFields: {
      sortId: {
         $cond: [{$eq: ['$hosting', 'godaddy.com']},0,
            {$cond: [{$eq: ['$hosting', 'aws.amazon.com']},1,2]
      }
   }},
  {$sort: {sortId: 1}}
])
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