使用filter和limit连接objectid的数组字段 - MongoDB

Question

有谁知道如何在某些要求下获得所需的结果？我想不出有效的方法，希望专家可以帮助我。 ：）

非常感谢你。

播放列表文件

"_id" : "playlist1", "title" : "Sample Playlist", 
"items" : ["item3", "item2", "item1", "item5", "item4"] // These are ObjectId type

项目集合

{ "_id" : "item1", "type" : "video", "playlist": "playlist1" }
{ "_id" : "item2", "type" : "music", "playlist": "playlist1" }
{ "_id" : "item3", "type" : "video", "playlist": "playlist1" }
{ "_id" : "item4", "type" : "music", "playlist": "playlist1" }
{ "_id" : "item5", "type" : "video", "playlist": "playlist1" }
{ "_id" : "item6", "type" : "video", "playlist": "playlist2" } // Different playlist

要求

• 使用跳过
限制为 2
• 仅限视频
的类型
• 按播放列表文档中的项目数组字段维护订单
• 高效的方式，因为物品可以是巨大的阵列（如1000 +）

期望结果

第1页：

{ "_id" : "item3", "type" : "video", "playlist": "playlist1" } // Maintain order
{ "_id" : "item1", "type" : "video", "playlist": "playlist1" }

第2页：

{ "_id" : "item5", "type" : "video", "playlist": "playlist1" }

Answer 1

这似乎对我有用：

db.test.aggregate([
    {$match: {_id: "playlist1"}},
    {$unwind: {path: "$items", includeArrayIndex: 'item_index'}},
    {$limit: 2},
    {
        $lookup: {
            from: "items",
            localField: "items",
            foreignField: "_id",
            as: "items"
        }
    },
    {$sort: {item_index: 1}},
    {$replaceRoot: {newRoot: {$arrayElemAt: ["$items", 0]}}}
])

Answer 2

您可以使用$ sort $ skip $ limit of mongodb来获取所需的对象

db.Items.aggregate({
            $match: {"type":"video"}
        }, {
            $sort: {
                KEY_TO_SORT: -1
            }
        }, {
            $skip: parseInt(skip)
        }, {
            $limit: parseInt(limit)
        },
        function (err, success) {
       // Error and success comes here
        });