Question

当我以这种方式定义pymongo集合时，线程数将增加2：

import threading
print threading.activeCount()


def db_bigadevs():
    from pymongo import MongoClient
    con_bigadevs = MongoClient()
    return con_bigadevs['bigadevs']


class collections:
    print 'A1 %s' % threading.activeCount()
    col_webs = db_bigadevs()['webs']
    print 'A2 %s' % threading.activeCount()

输出是：

1
A1 1
A2 3

Answer 1

每个MongoClient都有一个用于定期清理任务的后台线程，以及每个服务器的后台线程，用于监控MongoDB服务器的状态。因此，如果您创建默认的MongoClient，它将连接到localhost上的一个服务器：27017，产生总共两个线程。

A brief explanation is here, where the FAQ says "MongoClient spawns multiple threads to run background tasks such as monitoring connected servers." I wrote a very thorough explanation of PyMongo's use of threads here，主要面向未来的PyMongo维护者。

为什么pymongo集合defenition上升线程数？

1 个答案: