假设我有这两个集合
book: {
_id: 'aaa'
name: 'Book 1',
chapters: [
0: {
_id: 'chapter0',
name: 'Chapter 1',
pages: [
0: {
_id: 'page0',
name: 'Page 1',
paragraphs: [
0: {
_id: 'paragraph0',
name: 'Paragraph 1',
bookmarks: [
0: {sentence: 3, reader: 'Foo'},
1: {sentence: 8, reader: 'Bar'},
2: {sentence: 14, reader: 'John'}
]
}
]
}
]
}
]
}
book: {
_id: 'bbb'
name: 'Book 2',
chapters: [
0: {
_id: 'chapter0',
name: 'Chapter 1',
pages: [
0: {
_id: 'page0',
name: 'Page 1',
paragraphs: [
0: {
_id: 'paragraph0',
name: 'Paragraph 1',
bookmarks: []
},
1: {
_id: 'paragraph1',
name: 'Paragraph 2',
bookmarks: [
0: {sentence: 2, reader: 'George'},
1: {sentence: 1, reader: 'Paul'},
2: {sentence: 76, reader: 'John'},
3: {sentence: 54, reader: 'Ringo'}
]
}
]
}
]
}
]
}
我希望能够在获取结果时提取数组 bookmarks 并将它们附加到 book 集合中。像这样的东西会很好:
{
id: 'aaa'
name: 'Book 1'
bookmarks: [{...}, {...}, {...}] //since the first book has 3 bookmarks
},
{
id: 'bbb'
name: 'Book 2'
bookmarks: [{...}, {...}, {...}, {...}] //since the second book has 4 bookmarks
},
如果没有书签，它应该看起来像:
{
id: 'aaa'
name: 'Book 1'
bookmarks: [{...}, {...}, {...}] //since the first book has 3 bookmarks
},
{
id: 'bbb'
name: 'Book 2'
bookmarks: [{...}, {...}, {...}, {...}] //since the second book has 4 bookmarks
},
{
id: 'ccc'
name: 'Book 3'
bookmarks: [] //third book does not have bookmarks for example
},
我尝试过使用此代码进行聚合，但它只是将每本书的每个书签分开并将其插入对象中。
return yield Books.aggregate()
.unwind('chapters')
.unwind('chapters.pages')
.unwind('chapters.pages.paragraphs')
.unwind('chapters.pages.paragraphs.bookmarks')
.group({
_id: '$_id',
books: {
$push: {
_id: '$_id',
name: '$name',
bookmarks: '$chapters.pages.paragraphs.bookmarks'
}
}
}).exec()
有人能指出我正确的方向吗？谢谢!
最佳答案
尝试下面的聚合管道:
Books.aggregate([
{
$unwind: "$book"
},
{
$unwind: "$book.chapters"
},
{
$unwind: "$book.chapters.pages"
},
{
$unwind: "$book.chapters.pages.paragraphs"
},
{
$unwind: {
path: "$book.chapters.pages.paragraphs.bookmarks",
preserveNullAndEmptyArrays: true
}
},
{
$group: {
_id: {
_id: "$_id",
book: "$book.name"
},
bookmarks: {
$push: "$book.chapters.pages.paragraphs.bookmarks"
}
}
}
])
关于mongodb - 从 mongodb 聚合获取深度嵌套数组并包含在结果中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/58076291/