基于@Wolkenarchitekt的回答:https://stackoverflow.com/a/48117846/8808983，我写了一个脚本，可以在db中找到所有键的模式，我认为它可以帮助其他人阅读这个线程:

"""
Python 3
This script get list of patterns and print the collections that contains fields with this patterns.
"""

import argparse

import pymongo
from bson import Code


# initialize mongo connection:
def get_db():
    client = pymongo.MongoClient("172.17.0.2")
    db = client["Data"]
    return db


def get_commandline_options():
    description = "To run use: python db_fields_pattern_finder.py -p <list_of_patterns>"
    parser = argparse.ArgumentParser(description=description)
    parser.add_argument('-p', '--patterns', nargs="+", help='List of patterns to look for in the db.', required=True)
    return parser.parse_args()


def report_matching_fields(relevant_fields_by_collection):
    print("Matches:")

    for collection_name in relevant_fields_by_collection:
        if relevant_fields_by_collection[collection_name]:
            print(f"{collection_name}: {relevant_fields_by_collection[collection_name]}")

    # pprint(relevant_fields_by_collection)


def get_collections_names(db):
    """
    :param pymongo.database.Database db:
    :return list: collections names
    """
    return db.list_collection_names()


def get_keys(db, collection):
    """
    See: https://stackoverflow.com/a/48117846/8808983
    :param db:
    :param collection:
    :return:
    """
    map = Code("function() { for (var key in this) { emit(key, null); } }")
    reduce = Code("function(key, stuff) { return null; }")
    result = db[collection].map_reduce(map, reduce, "myresults")
    return result.distinct('_id')


def get_fields(db, collection_names):
    fields_by_collections = {}
    for collection_name in collection_names:
        fields_by_collections[collection_name] = get_keys(db, collection_name)
    return fields_by_collections


def get_matches_fields(fields_by_collections, patterns):
    relevant_fields_by_collection = {}
    for collection_name in fields_by_collections:
        relevant_fields = [field for field in fields_by_collections[collection_name] if
                           [pattern for pattern in patterns if
                            pattern in field]]
        relevant_fields_by_collection[collection_name] = relevant_fields

    return relevant_fields_by_collection


def main(patterns):
    """
    :param list patterns: List of strings to look for in the db.
    """
    db = get_db()

    collection_names = get_collections_names(db)
    fields_by_collections = get_fields(db, collection_names)
    relevant_fields_by_collection = get_matches_fields(fields_by_collections, patterns)

    report_matching_fields(relevant_fields_by_collection)


if __name__ == '__main__':
    args = get_commandline_options()
    main(args.patterns)

我知道这个问题已经10年了，但是c#没有解决方案，我花了好几个小时才弄清楚。我正在使用。net驱动程序和系统。Linq返回键的列表。

var map = new BsonJavaScript("function() { for (var key in this) { emit(key, null); } }");
var reduce = new BsonJavaScript("function(key, stuff) { return null; }");
var options = new MapReduceOptions<BsonDocument, BsonDocument>();
var result = await collection.MapReduceAsync(map, reduce, options);
var list = result.ToEnumerable().Select(item => item["_id"].ToString());

下面是用Python编写的示例: 这个示例内联返回结果。

from pymongo import MongoClient
from bson.code import Code

mapper = Code("""
    function() {
                  for (var key in this) { emit(key, null); }
               }
""")
reducer = Code("""
    function(key, stuff) { return null; }
""")

distinctThingFields = db.things.map_reduce(mapper, reducer
    , out = {'inline' : 1}
    , full_response = True)
## do something with distinctThingFields['results']

根据mongoldb文档，一个不同的组合

在单个集合或视图中查找指定字段的不同值，并以数组形式返回结果。

索引集合操作将返回给定键或索引的所有可能值:

返回一个数组，其中包含标识和描述集合上现有索引的文档列表

所以在一个给定的方法中，你可以使用下面的方法，为了查询一个集合中所有注册的索引，并返回一个对象，比如一个键的索引(这个例子使用async/await用于NodeJS，但显然你可以使用任何其他异步方法):

async function GetFor(collection, index) {

    let currentIndexes;
    let indexNames = [];
    let final = {};
    let vals = [];

    try {
        currentIndexes = await collection.indexes();
        await ParseIndexes();
        //Check if a specific index was queried, otherwise, iterate for all existing indexes
        if (index && typeof index === "string") return await ParseFor(index, indexNames);
        await ParseDoc(indexNames);
        await Promise.all(vals);
        return final;
    } catch (e) {
        throw e;
    }

    function ParseIndexes() {
        return new Promise(function (result) {
            let err;
            for (let ind in currentIndexes) {
                let index = currentIndexes[ind];
                if (!index) {
                    err = "No Key For Index "+index; break;
                }
                let Name = Object.keys(index.key);
                if (Name.length === 0) {
                    err = "No Name For Index"; break;
                }
                indexNames.push(Name[0]);
            }
            return result(err ? Promise.reject(err) : Promise.resolve());
        })
    }

    async function ParseFor(index, inDoc) {
        if (inDoc.indexOf(index) === -1) throw "No Such Index In Collection";
        try {
            await DistinctFor(index);
            return final;
        } catch (e) {
            throw e
        }
    }
    function ParseDoc(doc) {
        return new Promise(function (result) {
            let err;
            for (let index in doc) {
                let key = doc[index];
                if (!key) {
                    err = "No Key For Index "+index; break;
                }
                vals.push(new Promise(function (pushed) {
                    DistinctFor(key)
                        .then(pushed)
                        .catch(function (err) {
                            return pushed(Promise.resolve());
                        })
                }))
            }
            return result(err ? Promise.reject(err) : Promise.resolve());
        })
    }

    async function DistinctFor(key) {
        if (!key) throw "Key Is Undefined";
        try {
            final[key] = await collection.distinct(key);
        } catch (e) {
            final[key] = 'failed';
            throw e;
        }
    }
}

因此，使用基本_id索引查询一个集合，将返回以下内容(测试集合在测试时只有一个文档):

Mongo.MongoClient.connect(url, function (err, client) {
    assert.equal(null, err);

    let collection = client.db('my db').collection('the targeted collection');

    GetFor(collection, '_id')
        .then(function () {
            //returns
            // { _id: [ 5ae901e77e322342de1fb701 ] }
        })
        .catch(function (err) {
            //manage your error..
        })
});

注意，这使用了NodeJS Driver原生的方法。正如其他一些答案所建议的那样，还有其他方法，例如聚合框架。我个人认为这种方法更灵活，因为您可以轻松地创建和微调如何返回结果。显然，这只处理顶级属性，而不是嵌套属性。 此外，为了保证所有文档都被表示，如果有二级索引(除了主_id索引)，这些索引应该根据需要设置。

要获得所有键减去_id的列表，可以考虑运行以下聚合管道:

var keys = db.collection.aggregate([
    { "$project": {
       "hashmaps": { "$objectToArray": "$$ROOT" } 
    } }, 
    { "$group": {
        "_id": null,
        "fields": { "$addToSet": "$hashmaps.k" }
    } },
    { "$project": {
            "keys": {
                "$setDifference": [
                    {
                        "$reduce": {
                            "input": "$fields",
                            "initialValue": [],
                            "in": { "$setUnion" : ["$$value", "$$this"] }
                        }
                    },
                    ["_id"]
                ]
            }
        }
    }
]).toArray()[0]["keys"];

这一行将集合中的所有键提取到一个逗号分隔的排序字符串中:

db.<collection>.find().map((x) => Object.keys(x)).reduce((a, e) => {for (el of e) { if(!a.includes(el)) { a.push(el) }  }; return a}, []).sort((a, b) => a.toLowerCase() > b.toLowerCase()).join(", ")

这个查询的结果通常是这样的:

_class, _id, address, city, companyName, country, emailId, firstName, isAssigned, isLoggedIn, lastLoggedIn, lastName, location, mobile, printName, roleName, route, state, status, token