Question

嗨我在mongodb中有以下数据结构：

{
"_id" : "4087322f-1ad0-4595-935f-b41ef6d87306",
"lastModifiedDate" : ISODate("2016-11-22T20:48:05.904Z"),
"notes" : " I do not like him that much",
"peopleId" : "1121",
"status" : "asses"
}

我希望查询数据，使其返回每个peopleId的最新值，该值也与所选状态匹配。我一直在研究$ group，但是做同等地位会导致一些问题。

我尽可能使用peopleId对值进行分组：

.aggregate([      
            {$group : {_id : "$peopleId", "allData" : {$push : "$$ROOT"}}},
            {$sort : { lastModifiedDate : -1}}     
           ]).pretty()

给我这些结果：

{
    "_id" : "1123",
    "allData" : [
        {
            "_id" : "c9d6a6ce-104d-414a-8d89-512e556d7aba",
            "lastModifiedDate" : ISODate("2016-11-22T20:55:35.662Z"),
            "notes" : "He's in!",
            "peopleId" : "1123",
            "status" : "done"
        },
        {
            "_id" : "14e522d8-6cae-42ee-ad52-d5cf3cae0c29",
            "lastModifiedDate" : ISODate("2016-11-22T20:56:46.128Z"),
            "notes" : "He's in!",
            "peopleId" : "1123",
            "status" : "asses"
        }
    ]
}
{
    "_id" : "1121",
    "allData" : [
        {
            "_id" : "d2c3e5da-8696-4dcc-a5cd-1f2657f0192c",
            "lastModifiedDate" : ISODate("2016-11-22T20:46:35.097Z"),
            "notes" : " I do not like him",
            "peopleId" : "1121",
            "status" : "pending"
        },
        {
            "_id" : "4087322f-1ad0-4595-935f-b41ef6d87306",
            "lastModifiedDate" : ISODate("2016-11-22T20:48:05.904Z"),
            "notes" : " I do not like him that much",
            "peopleId" : "1121",
            "status" : "asses"
        },
        {
            "_id" : "be4fa5bd-da8c-4c1a-9010-6a86afd11dbc",
            "lastModifiedDate" : ISODate("2016-11-22T20:49:27.268Z"),
            "notes" : "He's in!\nTue Nov 22 12:49:27 PST 2016 He's TF OUT",
            "peopleId" : "1121",
            "status" : "done"
        }
    ]
}

由此我需要根据lastModifiedDate提取最新的条目，然后检查它是否与给定的状态匹配。

Answer 1

以下是解决方案。

解决方案1： -

如果您想要输出中集合中的所有字段。

db.collection.aggregate([
    {$match : {"status" : "asses"}},
    {$sort : {peopleId : 1, lastModifiedDate : -1}},
    {$group : {_id : "$peopleId", "allData" : {$push : "$$ROOT"}}},
    {$project: {_id :0, allData : {$slice : ["$allData", 0, 1]}}}
]);

<强>输出： -

{
    "allData" : [ 
        {
            "_id" : ObjectId("5834c004ba41f1f22e600c7f"),
            "lastModifiedDate" : ISODate("2016-11-24T20:48:05.904Z"),
            "notes" : " I do not like him that much",
            "peopleId" : "1",
            "status" : "asses"
        }
    ]
}

解决方案2： -

如果您不想要输出中集合中的所有字段。

db.collection.aggregate([
    {$match : {"status" : "asses"}},
    {$sort : {peopleId : 1, lastModifiedDate : -1}},
    {$group : {_id : "$peopleId", lastModDate : {$first : "$lastModifiedDate"}}}
])

<强>输出： -

{
    "_id" : "1",
    "lastModDate" : ISODate("2016-11-24T20:48:05.904Z")
}

Answer 2

我创建了一个包含以下文档的集合

MatchCollection matches = Regex.Matches(sString, @"<b>(.*)<\/b>", RegexOptions.Singleline);

如果我执行以下查询

{ "_id" : "4087322f-1ad0-4595-935f-b41ef6d87306", "lastModifiedDate" : ISODate("2016-11-22T20:48:05.904Z"), "notes" : " I do not like him that much",
"peopleId" : "1121", "status" : "asses" }
{ "_id" : "4087322f-1ad0-4595-935f-b41ef6d87305", "lastModifiedDate" : ISODate("2016-10-22T20:48:05.904Z"), "notes" : " I do not like him that much",
"peopleId" : "1121", "status" : "asses" }
{ "_id" : "4087322f-1ad0-4595-935f-b41ef6d87304", "lastModifiedDate" : ISODate("2016-11-21T20:48:05.904Z"), "notes" : " I do not like him that much",
"peopleId" : "1121", "status" : "asses" }

它给了我

db.mel.find().sort({lastModifiedDate:-1}).limit(1)

根据lastmodifeddate最新。

Mongo查询返回最新条目

2 个答案: