Pymongo - 如何获取作为代表集的python字典的状态

Question

我是mongo的新手,我正在使用pymongo.我发现pymongo的文档到处都是.

1)我有一个代表.从mongo shell,如果我在下面运行,我得到了我需要的东西.

sudo mongo 111.111.111.111 --eval "printjson(rs.status())"
MongoDB shell version: 2.2.1
connecting to: 111.111.111.111/test
{
    "set" : "hey",
    "date" : ISODate("2012-11-10T11:47:58Z"),
    "myState" : 1,
    "members" : [
        {
            "_id" : 0,
            "name" : "111.111.111.111:27017",
            "health" : 1,
            "state" : 1,
            "stateStr" : "PRIMARY",
            "uptime" : 69189,
            "optime" : Timestamp(1352478921000, 1),
            "optimeDate" : ISODate("2012-11-09T16:35:21Z"),
            "self" : true
        }
    ],
    "ok" : 1
}

我需要从pymongo获取该信息.

1)我连接到主要.我得到none但是根据上面的节点是主要的:

c = ReplicaSetConnection("111.111.111.111:27017", replicaSet='heythat')
print c.primary
None

2)我想运行c.command("status")但不支持该操作.

那么,如何使用pymongo返回主要,并返回状态作为Python字典？

Answer 1

rs.status() 调用'replSetGetStatus'管理命令,在python中你可以这样做:

conn = ReplicaSetConnection("111.111.111.111:27017", replicaSet='heythat')
conn.admin.command('replSetGetStatus')

Protip:如果你想知道shell帮助器调用什么命令,那么将括号从函数中删除以查看代码,例如:

heythat:PRIMARY> rs.status
function () {
    return db._adminCommand("replSetGetStatus");
}