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MongoDB 用另一个集合中的另一个字段更新字段

[英]MongoDB update field with another field from another collection

I have two collections: books and categories.我有两个 collections:书籍和类别。 Categories collections represent a tree structure which make them nested categories using parents and children.类别 collections 表示树结构,使它们使用父级和子级嵌套类别。

The book can have multiple categories and stores them in an array.这本书可以有多个类别并将它们存储在一个数组中。

Example: Book has category and I want to retire it and set it to the parent category.示例:图书具有类别,我想将其停用并将其设置为父类别。

This is how categories collection is populated.这就是类别集合的填充方式。

db.categories.insertMany([
  {
    _id: "Space Opera",
    ancestors: ["Science Fiction", "Fiction", "Science Fiction & Fantasy"],
    parent: ["Science Fiction"],
  },
  {
    _id: "Dystopian",
    ancestors: ["Science Fiction", "Fiction", "Science Fiction & Fantasy"],
    parent: ["Science Fiction"],
  },
  {
    _id: "Cyberpunk",
    ancestors: ["Science Fiction", "Fiction", "Science Fiction & Fantasy"],
    parent: ["Science Fiction"],
  },
  {
    _id: "Science Fiction",
    ancestors: ["Fiction", "Science Fiction & Fantasy"],
    parent: ["Fiction", "Science Fiction & Fantasy"],
  },
  {
    _id: "Fantasy",
    ancestors: ["Science Fiction & Fantasy"],
    parent: ["Science Fiction & Fantasy"],
  },
  {
    _id: "Science Fiction & Fantasy",
    ancestors: [],
    parent: [],
  },
  {
    _id: "Fiction",
    ancestors: [],
    parent: [],
  },
]);

Also, how do I query this one and get only the value "Science Fiction" (Note that it is stored in an array)?另外,我如何查询这个并只获取值“科幻小说”(注意它存储在一个数组中)?

db.categories.find({_id : "Space Opera"}, {_id : 0, parent : 1})[0].parent // Did not work  

db.categories.find({_id : "Space Opera"}, {_id : 0, parent : 1}) // find parent

// result 

[
  {
    "parent": [
      "Science Fiction"
    ]
  }
]
db.books.update(
    {title : "Book1"}, 
    {$set : {category : [**PARENT CATEGORY**]}}
)

I believe I can use code above inside books.update()我相信我可以在 books.update() 中使用上面的代码

I could store this in a separate variable but in vscode it gives me undefined.我可以将它存储在一个单独的变量中,但在 vscode 中它给了我未定义的。 And the inner query does not give me the right value as stated before, but I think you got the idea.内部查询并没有像前面所说的那样给我正确的值,但我认为你明白了。

db.books.update(
    {title : "Book1"}, 
    {$set : {category : [db.categories.find({_id : "Space Opera"}, {_id : 0, parent : 1})]}}
)

The parent you can get with this aggregation pipeline:您可以使用此聚合管道获得的父级:

db.categories.aggregate([
   { $match: { _id: "Space Opera" } },
   { $project: { _id: 0, parent: { $first: "$parent" } } }
])

or even甚至

db.categories.aggregate([
   { $match: { _id: "Space Opera" } },
   { $project: { _id: 0, parent: { $first: "$parent" } } }
]).toArray().shift().parent

In order to join collections you have to use the $lookup operator.要加入 collections,您必须使用$lookup运算符。 Bear in mind, NoSQL databases like MongoDB are not optimized for join/lookup.请记住,NoSQL 数据库(如 MongoDB)未针对连接/查找进行优化。 In real life you should have a look for a better design.在现实生活中,您应该寻找更好的设计。

db.books.aggregate([
   { $match: { title: "Book1" } },
   {
      $lookup:
         {
            from: "categories",
            pipeline: [
               { $match: { _id: "Space Opera" } },
               { $project: { _id: 0, parent: { $first: "$parent" } } }
            ],
            as: "category"
         }
   },
   { $set: { category: { $first: "$category.parent" } } }
])

If you like to update existing collection, then you have to create a loop for it:如果您想更新现有集合,则必须为其创建一个循环:

db.books.aggregate([...]).forEach(function (doc) {
   db.books.updateOne({ _id: doc._id }, { $set: { category: doc.category } });
})

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