我想获得MongoDB集合中所有键的名称。
例如,从这个:
db.things.insert( { type : ['dog', 'cat'] } );
db.things.insert( { egg : ['cat'] } );
db.things.insert( { type : [] } );
db.things.insert( { hello : [] } );
我想获得唯一的键:
type, egg, hello
我想获得MongoDB集合中所有键的名称。
例如,从这个:
db.things.insert( { type : ['dog', 'cat'] } );
db.things.insert( { egg : ['cat'] } );
db.things.insert( { type : [] } );
db.things.insert( { hello : [] } );
我想获得唯一的键:
type, egg, hello
当前回答
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());
其他回答
我知道我来晚了,但如果你想在python中快速找到所有键(甚至嵌套的键),你可以用递归函数来做:
def get_keys(dl, keys=None):
keys = keys or []
if isinstance(dl, dict):
keys += dl.keys()
list(map(lambda x: get_keys(x, keys), dl.values()))
elif isinstance(dl, list):
list(map(lambda x: get_keys(x, keys), dl))
return list(set(keys))
像这样使用它:
dl = db.things.find_one({})
get_keys(dl)
如果你的文件没有相同的密钥,你可以这样做:
dl = db.things.find({})
list(set(list(map(get_keys, dl))[0]))
但是这个解决方案肯定是可以优化的。
一般来说,这个解决方案基本上是解决在嵌套字典中查找键,所以这不是mongodb特定的。
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)
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", ...]
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']
我有一个更简单的工作…
你能做的是插入数据/文档到你的主集合“东西”,你必须在一个单独的集合中插入属性,让我们说“things_attributes”。
所以每次你插入"things"时,你确实会从"things_attributes"中得到该文档的值与你的新文档键的值进行比较,如果有任何新键存在,则将其追加到该文档中并再次重新插入它。
所以things_attributes只有一个唯一键的文档你可以通过findOne()很容易地得到