关闭 websocket 连接错误

Question

您好，我在我的 node.js 服务器上使用 websockets 来建立 WebRTC 连接。但是当我结束连接时，节点服务器控制台给了我以下错误：

        conn.otherName = null;
                       ^

TypeError: Cannot set property 'otherName' of undefined

Other name 是另一个 Peer 的名称，我在连接中设置如下： 案例“报价”： //例如。用户A想呼叫用户B console.log("发送报价到：", data.name);

        //if UserB exists then send him offer details 
        var conn = users[data.name]; 

        if(conn != null) { 
           //setting that UserA connected with UserB 
           connection.otherName = data.name; 

           sendTo(conn, { 
              type: "offer", 
              offer: data.offer, 
              name: connection.name 
           }); 
        } 

        break;

     case "answer": 
        console.log("Sending answer to: ", data.name); 
        //for ex. UserB answers UserA 
        var conn = users[data.name]; 

        if(conn != null) { 
           connection.otherName = data.name; 
           sendTo(conn, { 
              type: "answer", 
              answer: data.answer 
           });
        } 

然后我像这样关闭连接：

connection.on("close", function() { 

  if(connection.name) { 
     delete users[connection.name]; 

     if(connection.otherName) { 
        console.log("Disconnecting from ", connection.otherName); 
        var conn = users[connection.otherName]; 
        conn.otherName = null;  

        if(conn != null) { 
           sendTo(conn, { 
              type: "leave" 
          }); 
        }  
     } 
  } 
});

如何更改它以关闭连接而不会导致我的节点服务器崩溃？

Answer 1

通过一些防御性代码来阻止它破坏非常简单，改变这个

var conn = users[connection.otherName]; 
conn.otherName = null;  

到这里

if (connection.otherName) {
    var conn = users[connection.otherName]; 
    if (conn) 
         conn.otherName = null;  

    if(conn != null) { 
// The connection is closed, trying to send at this point isn't a good idea
           sendTo(conn, { 
              type: "leave" 
          }); 
        }  
    }

它没有解决为什么 users[connection.otherName] 返回未定义（这是你的练习），但它会阻止它崩溃