如何在没有猫鼬的情况下填充 MongoDB？答案

【问题标题】：How to populate in MongoDB without mongoose?如何在没有猫鼬的情况下填充 MongoDB？
【发布时间】：2022-11-22 03:29:10
【问题描述】：

我正在做一个大学项目，我需要用 ObjectId 填充一个对象数组，但我不能在我的项目中使用 mongoose。我有两个系列 -主题和学习计划.

例子学习计划文档：

{
  _id: ObjectId('111'),
  name: "Study program 1"
  description: "Lorem ipsum dolor sit amet",
  language: "en",
  subjects: [
    {
      id: ObjectId('222'),
      optionality: "selective",
      credits: 8,
    },
    {
      id: ObjectId('333'),
      optionality: "selective",
      credits: 5
    },
  ],
}

例子主题文件：

{
  _id: ObjectId('222'),
  name: "Subject A",
  description: "Subject A description.",
},
{
  _id: ObjectId('333'),
  name: "Subject B",
  description: "Subject B description.",
}

我需要使用来自的适当文档填充 subjects 数组中的对象主题收藏基于id。基本上我正在寻找的是这个结果：

{
  _id: ObjectId('111'),
  name: "Study program 1"
  description: "Lorem ipsum dolor sit amet",
  language: "en",
  subjects: [
    {
      
      _id: ObjectId('222'),
      name: "Subject A",
      description: "Subject A description.",
      optionality: "selective",
      credits: 8,
    },
    {
      _id: ObjectId('333'),
      name: "Subject B",
      description: "Subject B description.",
      optionality: "selective",
      credits: 5
    },
  ],
}

到目前为止，我已尝试使用以下 $lookup：

{
  $lookup: {
    from: "subject",
    localField: "subjects.id",
    foreignField: "_id",
    as: "subjects",
  }
}

但这会删除 optionality 和 credits 属性。有没有办法在不使用猫鼬的情况下实现这一目标？谢谢你。

【问题讨论】：

标签： javascript mongodb mongoose lookup populate

【解决方案1】：

这是一种方法。

db.studyProgram.aggregate([
  {
    "$lookup": {
      "from": "subject",
      "localField": "subjects.id",
      "foreignField": "_id",
      "as": "subjectDocs"
    }
  },
  {
    "$set": {
      "subjects": {
        "$map": {
          "input": "$subjects",
          "as": "subject",
          "in": {
            "$mergeObjects": [
              {
                "$first": {
                  "$filter": {
                    "input": "$subjectDocs",
                    "as": "doc",
                    "cond": {"$eq": ["$$doc._id", "$$subject.id"]}
                  }
                }
              },
              {
                "optionality": "$$subject.optionality",
                "credits": "$$subject.credits"
              }
            ]
          }
        }
      },
      "subjectDocs": "$$REMOVE"
    }
  }
])

试试mongoplayground.net。

【讨论】：

【解决方案2】：

我也很好奇这个问题的答案。我很期待解决方案

【讨论】：