PyMongo/Mongoengine相当于mongodump

Question

是否有相同功能PyMongo或mongoengine到MongoDB的mongodump？我似乎无法在文档中找到任何内容.

使用案例:我需要定期备份远程mongo数据库.本地机器是没有安装mongo的生产服务器,我没有管理员权限,所以我不能subprocess用来打电话mongodump.我可以在virtualenv上本地安装mongo客户端,但我更喜欢API调用.

非常感谢 :-).

Answer 1

对于我相对较小的小型数据库,我最终使用了以下解决方案.它不适合大型或复杂的数据库,但它足以满足我的需求.它将所有文档作为json转储到备份目录.它很笨重,但它不依赖于其他东西而不是pymongo.

from os.path import join
import pymongo
from bson.json_utils import dumps

def backup_db(backup_db_dir):
    client = pymongo.MongoClient(host=<host>, port=<port>)
    database = client[<db_name>]
    authenticated = database.authenticate(<uname>,<pwd>)
    assert authenticated, "Could not authenticate to database!"
    collections = database.collection_names()
    for i, collection_name in enumerate(collections):
        col = getattr(database,collections[i])
        collection = col.find()
        jsonpath = collection_name + ".json"
        jsonpath = join(backup_db_dir, jsonpath)
        with open(jsonpath, 'wb') as jsonfile:
            jsonfile.write(dumps(collection))

Answer 2

接受的答案不再有效。这是修改后的代码：

from os.path import join
import pymongo
from bson.json_util import dumps

def backup_db(backup_db_dir):
    client = pymongo.MongoClient(host=..., port=..., username=..., password=...)
    database = client[<db_name>]
    collections = database.collection_names()

    for i, collection_name in enumerate(collections):
        col = getattr(database,collections[i])
        collection = col.find()
        jsonpath = collection_name + ".json"
        jsonpath = join(backup_db_dir, jsonpath)
        with open(jsonpath, 'wb') as jsonfile:
            jsonfile.write(dumps(collection).encode())


backup_db('.')