通过已删除的成员连接到副本集成员

Question

我有一个简单的副本集，有3个成员

rs.status（）的输出

{
    ...,
    "members" : [
        {
            "_id": 1,
            "name": "localhost:27021",
            "stateStr": "PRIMARY",
        },
        {
            "_id": 2,
            "name": "localhost:27022",
            "stateStr": "SECONDARY",
        },
        {
            "_id": 3,
            "name": "localhost:27023",
            "stateStr": "SECONDARY",
        }
    ]
}

我删除了一个成员

rs.remove("localhost:27023")

现在rs.status（）

{
    ...,
    "members" : [
        {
            "_id": 1,
            "name": "localhost:27021",
            "stateStr": "PRIMARY",
        },
        {
            "_id": 2,
            "name": "localhost:27022",
            "stateStr": "SECONDARY",
        },
    ]
}

现在我使用了移除实例的mongodb连接URL。它成功连接到副本集成员!!

var url = 'mongodb://localhost:27023/myproject?replicaSet=rs0';
MongoClient.connect(url, function(err, db) {
  console.log("Connected correctly to server");
  db.close();
});

我可以看到与副本集成员建立连接（localhost：27022和localhost：27023）

2019-04-17T18:59:04.727+0900 I NETWORK  [thread1] connection accepted from 127.0.0.1:53284 #6 (3 connections now open)
2019-04-17T18:59:04.727+0900 I NETWORK  [conn6] received client metadata from 127.0.0.1:53284 conn6: { driver: { name: "nodejs", version: "3.2.3" }, os: { type: "Linux", name: "linux", architecture: "x64", version: "4.15.0-47-generic" }, platform: "Node.js v10.15.0, LE, mongodb-core: 3.2.3" }
2019-04-17T18:59:04.736+0900 I -        [conn6] end connection 127.0.0.1:53284 (3 connections now open)
2019-04-17T18:59:35.334+0900 I ASIO     [NetworkInterfaceASIO-RS-0] Ending idle connection to host localhost:27021 because the pool meets constraints; 1 connections to that host remain open

有人可以解释一下吗？通过已删除的成员连接到副本集成员。在任何文档中都找不到任何解释。

Answer 1

这是一种驾驶行为。默认情况下，连接到副本集的客户端将尝试连接到主节点。在这种情况下，连接到mongodb://localhost:27023/myproject?replicaSet=rs0将首先连接到localhost:27023并使用命令{isMaster: 1}检查连接的节点是否为主/主节点。从结果中，它检测到localhost:27023不是主/主，但它也有其他节点信息可用。接下来的步骤是连接到其他节点，直到找到主节点。

你应该关闭localhost:27023，它甚至不会连接。