MongoDB按文档数组中的最大值查找 [英] MongoDB find by max value in array of documents

问题描述

提供包含以下文件的集合:

Given a collection with documents such as:

{
    "host" : "example.com",
    "ips" : [
        {
            "ip" : NumberLong("1111111111"),
            "timestamp" : NumberLong(1373970044)
        },
        {
            "ip" : NumberLong("2222222222"),
            "timestamp" : NumberLong(1234978746)
        }
    ]
}

我需要返回所有ip值为X的文档，但前提是X的关联时间戳是ips数组中最高时间戳(因此，上述示例文档不应 匹配搜索"2222222222"，因为这不是不是具有最新时间戳的IP.

I need to return all documents with an ip value of X, but only if the associated timestamp for X is the highest timestamp in the ips array (so the above example document should not match a search for "2222222222" because that is not the IP with the most recent timestamp).

这是我第一次在MongoDB中做一些超出基本内容的事情，所以我能得到的最接近的是:

This is my first time doing anything much beyond fairly basic stuff in MongoDB so the closest I've been able to get is:

coll.aggregate({$ match:{"ips.ip":X}}，{$ group:{"_ id":"$ host"， "max":{$ max:"$ ips.timestamp"}}}，{$ sort:{"ips.timestamp":-1}}).

coll.aggregate({$match:{"ips.ip":X}},{$group:{"_id":"$host", "max":{$max:"$ips.timestamp"}}},{$sort:{"ips.timestamp":-1}}).result

这显然不能满足我的要求，它会返回ips.ip值为X的任何内容.仅当X的关联时间戳最高时，我才能仅返回ip.ip为X的文档. ips数组?

Which obviously doesn't give me what I'm looking for, it returns anything with an ips.ip value of X. How do I return only documents where ip.ip is X only if X's associated timestamp is the highest for that ips array?

推荐答案

如果host是唯一的，则以下代码可以完成此工作.否则，您可以在分组操作中将host替换为_id:

If host is unique, the following code should do the job. Otherwise, you can simply replace host by _id in the grouping operation:

coll.aggregate([
  {$unwind: "$ips"},
  {$project:{host:"$host",ip:"$ips.ip", ts:"$ips.timestamp"} },
  {$sort:{ts:1} },
  {$group: {_id: "$host", IPOfMaxTS:{$last: "$ip"}, ts:{$last: "$ts"} } }
])

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