Question

我有一个包含这样的对象的集合：

{
  "_id": ObjectId("52ed12c144aecc4bf004d0b6"),
  "title": "myBook",
  "summary": "This is a book summary",
  "chapters": [
    {
      "number": 1
      "created_at": ISODate("2013-12-17T23:00:00Z")
    },
    {
      "number": 2
      "created_at": ISODate("2014-12-17T23:00:00Z")
    }
  ]
  "covers": [
    "http://url.com/cover1.jpg",
    "http://url.com/cover2.jpg",
  ],
  "urls": [
    "http://url.com/myBook",
    "http://url.com/myBook/option2"
  ],
  "genres": [
    "comedy",
    "romantic"
  ],
  "created_at": ISODate("2012-12-17T23:00:00Z"),
  "modify_at": ISODate("2014-02-01T15:41:48.149Z")
}

我只想获得'crated_at'＆gt; ='2012-12-17T23：00：00Z'的书籍，只有'created_at'＆gt;的章节“2013-12-17T23：00：00Z”。（作为过滤器）

我的输出json应该是这样的：

{
  "_id": ObjectId("52ed12c144aecc4bf004d0b6"),
  "title": "myBook",
  "summary": "This is a book summary",
  "chapters": [
    {
      "number": 2
      "created_at": ISODate("2014-12-17T23:00:00Z")
    }
  ]
  "covers": [
    "http://url.com/cover1.jpg",
    "http://url.com/cover2.jpg",
  ],
  "urls": [
    "http://url.com/myBook",
    "http://url.com/myBook/option2"
  ],
  "genres": [
    "comedy",
    "romantic"
  ],
  "created_at": ISODate("2012-12-17T23:00:00Z"),
  "modify_at": ISODate("2014-02-01T15:41:48.149Z")
}

在查询输出中可能是没有任何章节的书籍，其他的则包含所有章节的子集或书籍及其所有章节。

要按'created_at'搜索图书，请按以下方式查询：

db.books.find({ created_at: {$gte: ISODate("2012-12-17T23:00:00Z")} })

但我不知道我要添加什么来过滤此结果输出中的章节。

Answer 1

您可以尝试使用聚合框架。

db.chapters.aggregate(
{ $match: { created_at: {$gte: ISODate("2012-12-17T23:00:00Z")} } },
{ $unwind: "$chapters" },
{ $match: { "chapters.created_at": {$gte: ISODate("2013-12-17T23:00:00Z")} } },
{ $group: 
    {
        _id: 
        {
            id: "$_id",
            title: "$title",
            summary: "$summary",
            covers: "$covers",
            urls: "$urls",
            genres: "$genres",
            created_at: "$created_at",
            modify_at: "$modify_at"
        },
        chapters: { $push: "$chapters" }
    }
},
{ $project:
    {
        _id: "$_id.id",
        title: "$_id.title",
        summary: "$_id.summary",
        chapters: "$chapters",
        covers: "$_id.covers",
        urls: "$_id.urls",
        genres: "$_id.genres",
        created_at: "$_id.created_at",
        modify_at: "$_id.modify_at"
    }
}
)

按日期获取mongo集合中的特定值嵌套数组元素

1 个答案: