例如，我有这些文件：
{
"addr": "address1",
"book": "book1"
},
{
"addr": "address2",
"book": "book1"
},
{
"addr": "address1",
"book": "book5"
},
{
"addr": "address3",
"book": "book9"
},
{
"addr": "address2",
"book": "book5"
},
{
"addr": "address2",
"book": "book1"
},
{
"addr": "address1",
"book": "book1"
},
{
"addr": "address15",
"book": "book1"
},
{
"addr": "address9",
"book": "book99"
},
{
"addr": "address90",
"book": "book33"
},
{
"addr": "address4",
"book": "book3"
},
{
"addr": "address5",
"book": "book1"
},
{
"addr": "address77",
"book": "book11"
},
{
"addr": "address1",
"book": "book1"
}
等等。
如何提出一个请求，它将描述每个地址的前N个地址和前M个书籍？
例子预期结果：
地址1 | book_1：5
| book_2：10
| book_3：50
|总计：65
______________________
地址2 | book_1：10
| book_2：10
| | ...
| book_M：10
|总计：M * 10
... ...
______________________
地址N | book_1：20
| book_2：20
| | ...
| book_M：20
|总计：M * 20
答案 0 :(得分：151)
在现代MongoDB版本中，您可以在基本聚合结果之外使用$slice强制执行此操作。对于“大”结果，请为每个分组运行并行查询（演示列表位于答案的末尾），或等待SERVER-9377解析，这将允许对项目数量的“限制” $push到数组。
db.books.aggregate([
{ "$group": {
"_id": {
"addr": "$addr",
"book": "$book"
},
"bookCount": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.addr",
"books": {
"$push": {
"book": "$_id.book",
"count": "$bookCount"
},
},
"count": { "$sum": "$bookCount" }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 },
{ "$project": {
"books": { "$slice": [ "$books", 2 ] },
"count": 1
}}
])
仍未解析SERVER-9377，但在此版本$lookup中允许使用"pipeline"表达式作为参数而不是"localFields"的新“非关联”选项和"foreignFields"选项。然后，这允许与另一个管道表达式“自联接”，我们可以在其中应用$limit以返回“top-n”结果。
db.books.aggregate([
{ "$group": {
"_id": "$addr",
"count": { "$sum": 1 }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 },
{ "$lookup": {
"from": "books",
"let": {
"addr": "$_id"
},
"pipeline": [
{ "$match": {
"$expr": { "$eq": [ "$addr", "$$addr"] }
}},
{ "$group": {
"_id": "$book",
"count": { "$sum": 1 }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 }
],
"as": "books"
}}
])
此处的另一个补充当然是能够通过$expr使用$match插入变量以选择“join”中的匹配项，但一般前提是“管道内的管道” “内部内容可以通过来自父级的匹配进行过滤。由于它们本身都是“管道”，我们可以分别$limit每个结果。
这将是运行并行查询的下一个最佳选择，如果$match被允许并且能够在“子管道”处理中使用索引，实际上会更好。因此，当引用的问题没有使用“限制$push”时，它实际上提供了应该更好的东西。
你似乎偶然发现了顶级的“N”问题。在某种程度上，你的问题很容易解决，但不是你要求的确切限制：
db.books.aggregate([
{ "$group": {
"_id": {
"addr": "$addr",
"book": "$book"
},
"bookCount": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.addr",
"books": {
"$push": {
"book": "$_id.book",
"count": "$bookCount"
},
},
"count": { "$sum": "$bookCount" }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 }
])
现在，这会给你一个这样的结果：
{
"result" : [
{
"_id" : "address1",
"books" : [
{
"book" : "book4",
"count" : 1
},
{
"book" : "book5",
"count" : 1
},
{
"book" : "book1",
"count" : 3
}
],
"count" : 5
},
{
"_id" : "address2",
"books" : [
{
"book" : "book5",
"count" : 1
},
{
"book" : "book1",
"count" : 2
}
],
"count" : 3
}
],
"ok" : 1
}
因此，这与您提出的要求不同，虽然我们确实得到了地址值的最高结果，但基础“书籍”选择并不仅限于所需的结果数量。
事实证明这很难做到，但是可以做到这一点虽然复杂性随着你需要匹配的项目数量的增加而增加。为了简单起见，我们最多可以保持2个匹配：
db.books.aggregate([
{ "$group": {
"_id": {
"addr": "$addr",
"book": "$book"
},
"bookCount": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.addr",
"books": {
"$push": {
"book": "$_id.book",
"count": "$bookCount"
},
},
"count": { "$sum": "$bookCount" }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 },
{ "$unwind": "$books" },
{ "$sort": { "count": 1, "books.count": -1 } },
{ "$group": {
"_id": "$_id",
"books": { "$push": "$books" },
"count": { "$first": "$count" }
}},
{ "$project": {
"_id": {
"_id": "$_id",
"books": "$books",
"count": "$count"
},
"newBooks": "$books"
}},
{ "$unwind": "$newBooks" },
{ "$group": {
"_id": "$_id",
"num1": { "$first": "$newBooks" }
}},
{ "$project": {
"_id": "$_id",
"newBooks": "$_id.books",
"num1": 1
}},
{ "$unwind": "$newBooks" },
{ "$project": {
"_id": "$_id",
"num1": 1,
"newBooks": 1,
"seen": { "$eq": [
"$num1",
"$newBooks"
]}
}},
{ "$match": { "seen": false } },
{ "$group":{
"_id": "$_id._id",
"num1": { "$first": "$num1" },
"num2": { "$first": "$newBooks" },
"count": { "$first": "$_id.count" }
}},
{ "$project": {
"num1": 1,
"num2": 1,
"count": 1,
"type": { "$cond": [ 1, [true,false],0 ] }
}},
{ "$unwind": "$type" },
{ "$project": {
"books": { "$cond": [
"$type",
"$num1",
"$num2"
]},
"count": 1
}},
{ "$group": {
"_id": "$_id",
"count": { "$first": "$count" },
"books": { "$push": "$books" }
}},
{ "$sort": { "count": -1 } }
])
这样，实际上会从前两个“地址”条目中为您提供前两本“书籍”。
但是对于我的钱，请保留第一个表单然后简单地“切片”返回的数组元素以获取第一个“N”元素。
演示代码适用于v8.x和v10.x版本的当前LTS版本的NodeJS。这主要是针对async/await语法，但是在一般流程中没有任何内容存在任何此类限制，并且只需对简单的承诺进行少量修改，甚至回到简单的回调实现。
<强> index.js
const { MongoClient } = require('mongodb');
const fs = require('mz/fs');
const uri = 'mongodb://localhost:27017';
const log = data => console.log(JSON.stringify(data, undefined, 2));
(async function() {
try {
const client = await MongoClient.connect(uri);
const db = client.db('bookDemo');
const books = db.collection('books');
let { version } = await db.command({ buildInfo: 1 });
version = parseFloat(version.match(new RegExp(/(?:(?!-).)*/))[0]);
// Clear and load books
await books.deleteMany({});
await books.insertMany(
(await fs.readFile('books.json'))
.toString()
.replace(/\n$/,"")
.split("\n")
.map(JSON.parse)
);
if ( version >= 3.6 ) {
// Non-correlated pipeline with limits
let result = await books.aggregate([
{ "$group": {
"_id": "$addr",
"count": { "$sum": 1 }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 },
{ "$lookup": {
"from": "books",
"as": "books",
"let": { "addr": "$_id" },
"pipeline": [
{ "$match": {
"$expr": { "$eq": [ "$addr", "$$addr" ] }
}},
{ "$group": {
"_id": "$book",
"count": { "$sum": 1 },
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 }
]
}}
]).toArray();
log({ result });
}
// Serial result procesing with parallel fetch
// First get top addr items
let topaddr = await books.aggregate([
{ "$group": {
"_id": "$addr",
"count": { "$sum": 1 }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 }
]).toArray();
// Run parallel top books for each addr
let topbooks = await Promise.all(
topaddr.map(({ _id: addr }) =>
books.aggregate([
{ "$match": { addr } },
{ "$group": {
"_id": "$book",
"count": { "$sum": 1 }
}},
{ "$sort": { "count": -1 } },
{ "$limit": 2 }
]).toArray()
)
);
// Merge output
topaddr = topaddr.map((d,i) => ({ ...d, books: topbooks[i] }));
log({ topaddr });
client.close();
} catch(e) {
console.error(e)
} finally {
process.exit()
}
})()
<强> books.json
{ "addr": "address1", "book": "book1" }
{ "addr": "address2", "book": "book1" }
{ "addr": "address1", "book": "book5" }
{ "addr": "address3", "book": "book9" }
{ "addr": "address2", "book": "book5" }
{ "addr": "address2", "book": "book1" }
{ "addr": "address1", "book": "book1" }
{ "addr": "address15", "book": "book1" }
{ "addr": "address9", "book": "book99" }
{ "addr": "address90", "book": "book33" }
{ "addr": "address4", "book": "book3" }
{ "addr": "address5", "book": "book1" }
{ "addr": "address77", "book": "book11" }
{ "addr": "address1", "book": "book1" }
答案 1 :(得分：33)
使用如下聚合函数：
[
{$group: {_id : {book : '$book',address:'$addr'}, total:{$sum :1}}},
{$project : {book : '$_id.book', address : '$_id.address', total : '$total', _id : 0}}
]
它会给你如下结果：
{
"total" : 1,
"book" : "book33",
"address" : "address90"
},
{
"total" : 1,
"book" : "book5",
"address" : "address1"
},
{
"total" : 1,
"book" : "book99",
"address" : "address9"
},
{
"total" : 1,
"book" : "book1",
"address" : "address5"
},
{
"total" : 1,
"book" : "book5",
"address" : "address2"
},
{
"total" : 1,
"book" : "book3",
"address" : "address4"
},
{
"total" : 1,
"book" : "book11",
"address" : "address77"
},
{
"total" : 1,
"book" : "book9",
"address" : "address3"
},
{
"total" : 1,
"book" : "book1",
"address" : "address15"
},
{
"total" : 2,
"book" : "book1",
"address" : "address2"
},
{
"total" : 3,
"book" : "book1",
"address" : "address1"
}
我没有完全得到您期望的结果格式，因此请随意将其修改为您需要的格式。
答案 2 :(得分：1)
以下查询将提供与所需响应完全相同的结果：
db.books.aggregate([
{
$group: {
_id: { addresses: "$addr", books: "$book" },
num: { $sum :1 }
}
},
{
$group: {
_id: "$_id.addresses",
bookCounts: { $push: { bookName: "$_id.books",count: "$num" } }
}
},
{
$project: {
_id: 1,
bookCounts:1,
"totalBookAtAddress": {
"$sum": "$bookCounts.count"
}
}
}
])
响应如下所示：
/* 1 */
{
"_id" : "address4",
"bookCounts" : [
{
"bookName" : "book3",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 2 */
{
"_id" : "address90",
"bookCounts" : [
{
"bookName" : "book33",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 3 */
{
"_id" : "address15",
"bookCounts" : [
{
"bookName" : "book1",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 4 */
{
"_id" : "address3",
"bookCounts" : [
{
"bookName" : "book9",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 5 */
{
"_id" : "address5",
"bookCounts" : [
{
"bookName" : "book1",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 6 */
{
"_id" : "address1",
"bookCounts" : [
{
"bookName" : "book1",
"count" : 3
},
{
"bookName" : "book5",
"count" : 1
}
],
"totalBookAtAddress" : 4
},
/* 7 */
{
"_id" : "address2",
"bookCounts" : [
{
"bookName" : "book1",
"count" : 2
},
{
"bookName" : "book5",
"count" : 1
}
],
"totalBookAtAddress" : 3
},
/* 8 */
{
"_id" : "address77",
"bookCounts" : [
{
"bookName" : "book11",
"count" : 1
}
],
"totalBookAtAddress" : 1
},
/* 9 */
{
"_id" : "address9",
"bookCounts" : [
{
"bookName" : "book99",
"count" : 1
}
],
"totalBookAtAddress" : 1
}