使用pymongo进行清理和可重用的解决方案:

from pymongo import MongoClient
from bson import Code

def get_keys(db, collection):
    client = MongoClient()
    db = client[db]
    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')

用法:

get_keys('dbname', 'collection')
>> ['key1', 'key2', ... ]

使用python。返回集合中所有顶级键的集合:

#Using pymongo and connection named 'db'

reduce(
    lambda all_keys, rec_keys: all_keys | set(rec_keys), 
    map(lambda d: d.keys(), db.things.find()), 
    set()
)

你可以用MapReduce来做:

mr = db.runCommand({
  "mapreduce" : "my_collection",
  "map" : function() {
    for (var key in this) { emit(key, null); }
  },
  "reduce" : function(key, stuff) { return null; }, 
  "out": "my_collection" + "_keys"
})

然后在结果集合上单独运行，以便找到所有的键:

db[mr.result].distinct("_id")
["foo", "bar", "baz", "_id", ...]

这对我来说很有效:

var arrayOfFieldNames = [];

var items = db.NAMECOLLECTION.find();

while(items.hasNext()) {
  var item = items.next();
  for(var index in item) {
    arrayOfFieldNames[index] = index;
   }
}

for (var index in arrayOfFieldNames) {
  print(index);
}

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

如果你的目标集合不是很大，你可以在mongo shell客户端下尝试:

var allKeys = {};

db.YOURCOLLECTION.find().forEach(function(doc){Object.keys(doc).forEach(function(key){allKeys[key]=1})});

allKeys;