Reputation: 1748
I believe that this question is almost the same with this.
But I use websocket in Echo framework instead of Gorilla. So I think the approach will be different.
Echo does provide the example. But it only shows how to connect with single client. When there are more than one client, the other clients don't receive message from server.
How do I make the server broadcasts message to all connected clients?
The accepted answer from referred link says that I have to use connection pool to broadcast messages to all connections. How can I do this in Echo framework?
Upvotes: 4
Views: 4782
Reputation: 6324
You indeed need to follow the same principle of connection pool.
Here is the Echo example, with a very basic implementation of a pool:
package main
import (
"fmt"
"sync"
"github.com/labstack/echo"
mw "github.com/labstack/echo/middleware"
"golang.org/x/net/websocket"
)
var connectionPool = struct {
sync.RWMutex
connections map[*websocket.Conn]struct{}
}{
connections: make(map[*websocket.Conn]struct{}),
}
func main() {
e := echo.New()
e.Use(mw.Logger())
e.Use(mw.Recover())
e.Static("/", "public")
e.WebSocket("/ws", func(c *echo.Context) (err error) {
ws := c.Socket()
connectionPool.Lock()
connectionPool.connections[ws] = struct{}{}
defer func(connection *websocket.Conn){
connectionPool.Lock()
delete(connectionPool.connections, connection)
connectionPool.Unlock()
}(ws)
connectionPool.Unlock()
msg := ""
for {
if err = websocket.Message.Receive(ws, &msg); err != nil {
return err
}
err = sendMessageToAllPool(msg)
if err != nil {
return err
}
fmt.Println(msg)
}
return err
})
e.Run(":1323")
}
func sendMessageToAllPool(message string) error {
connectionPool.RLock()
defer connectionPool.RUnlock()
for connection := range connectionPool.connections {
if err := websocket.Message.Send(connection, message); err != nil {
return err
}
}
return nil
}
Upvotes: 5