客户端如何向 graphql 订阅服务器发送关闭信号？

问：

我正在 graphql 服务器上工作，并且有一个订阅 API。这是我在 gqlgen 文档上找到的起始代码：

// CurrentTime is the resolver for the currentTime field.
func (r *subscriptionResolver) CurrentTime(ctx context.Context) (<-chan *model.Time, error) {
    // First you'll need to `make()` your channel. Use your type here!
    ch := make(chan *model.Time)

    // You can (and probably should) handle your channels in a central place outside of `schema.resolvers.go`.
    // For this example we'll simply use a Goroutine with a simple loop.
    go func() {
        // Handle deregistration of the channel here. Note the `defer`
    defer close(ch)

        for {
            // In our example we'll send the current time every second.
            time.Sleep(1 * time.Second)
            fmt.Println("Tick")

            // Prepare your object.
            currentTime := time.Now()
            t := &model.Time{
                UnixTime:  int(currentTime.Unix()),
                TimeStamp: currentTime.Format(time.RFC3339),
            }

            // The subscription may have got closed due to the client disconnecting.
            // Hence we do send in a select block with a check for context cancellation.
            // This avoids goroutine getting blocked forever or panicking,
            select {
            case <-ctx.Done(): // This runs when context gets cancelled. Subscription closes.
                fmt.Println("Subscription Closed")
                // Handle deregistration of the channel here. `close(ch)`
                return // Remember to return to end the routine.
            
            case ch <- t: // This is the actual send.
                // Our message went through, do nothing 
            }
        }
    }()

    // We return the channel and no error.
    return ch, nil
}

我想知道收到 ctx 后会发生什么。Done（） 信号。客户端是否通过取消订阅或关闭订阅来发送此信号？或者它可以在一段时间后自动发生？（我的意思是设置一些空闲的超时参数。 另外，我想知道我这边（服务器端）的超时可以触发 Done（） 信号吗？