我有两个收藏:书籍和类别。类别集合表示一个树结构,使它们使用父级和子级进行嵌套类别。
书籍可以有多个类别并将它们存储在数组中。
示例:书籍有类别,我想取消它并将其设置为父类别。
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: [],
}]);
另外,如何查询这个并仅获取值“Science Fiction”(请注意,它存储在数组中)?
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**]
}
})
我相信我可以在 books.update() 中使用上面的代码
我可以将其存储在一个单独的变量中,但在 vscode 中它给了我未定义的值。 并且内部查询没有给我前面所说的正确值,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
})
]
}
})
您可以通过此聚合管道获得的父级:
db.categories.aggregate([
{ $match: { _id: "Space Opera" } },
{ $project: { _id: 0, parent: { $first: "$parent" } } }
])
甚至
db.categories.aggregate([
{ $match: { _id: "Space Opera" } },
{ $project: { _id: 0, parent: { $first: "$parent" } } }
]).toArray().shift().parent
为了加入集合,您必须使用 $lookup 运算符。请记住,像 MongoDB 这样的 NoSQL 数据库并未针对连接/查找进行优化。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" } } }
])
如果您想更新现有集合,那么您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 } });
})
我必须更新城市表中的州 ID,所以我尝试了这个,它正在工作:
州名称匹配的地方
从州表id更新城市表中的州id
db.states.find({}, { state:1, _id:1}).forEach(function (sts) {
db.cities.updateMany(
{ state: sts.state },
{ $set: { state_id: sts._id}}
)
})
SQL
Update Cities
Set State_Id = States._Id
From Cities
INNER JOIN States ON Cities.State = States.State