【问题标题】:I want to know advanced mongodb aggregation我想知道高级mongodb聚合
【发布时间】:2019-11-10 13:13:49
【问题描述】:

我想知道如何使这个 mysql 查询的 MongoDB 查询相同

"SELECT * FROM `folders` `F` JOIN `files` `I` ON I._id IN F.filesId;"

【问题讨论】:

标签: mysql mongodb aggregation


【解决方案1】:
db.folders.aggregate([
{ 
    $lookup: { 
        from: "files",
        localField: "_id", 
        foreignField: "filesId", //_id is ObjectId() so  filesId fields on 
        as: "docs"               //  files table mustbe ObjectId Save
    } 
}
])

https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/

【讨论】:

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