javascript - 无法在 'send' : Still in CONNECTING state 上执行 'WebSocket'

Question

我是 React js 开发的新手，尝试将 WebSocket 集成到我的应用程序中。

但在连接期间发送消息时出现错误。

我的代码是

  const url = `${wsApi}/ws/chat/${localStorage.getItem("sID")}/${id}/`;


  const ws = new WebSocket(url);

  ws.onopen = (e) => {
    console.log("connect");
  };

  ws.onmessage = (e) => {
    const msgRes = JSON.parse(e.data);
    setTextMessage(msgRes.type);
    // if (msgRes.success === true) {
    //   setApiMessagesResponse(msgRes);
    // }
    console.log(msgRes);
  };
  // apiMessagesList.push(apiMessagesResponse);

  // console.log("message response", apiMessagesResponse);
  ws.onclose = (e) => {
    console.log("disconnect");
  };

  ws.onerror = (e) => {
    console.log("error");
  };

  const handleSend = () => {
    console.log(message);
    ws.send(message);
  };

得到这个错误

Failed to execute 'send' on 'WebSocket': Still in CONNECTING state

Answer 1

听起来您在套接字完成连接过程之前调用了 ws.send。您需要等待 open 事件/回调，或根据 docs 检查 readyState并在 readyState 更改后排队发送，即在 open 回调触发后。

不建议您这样做，但它可能会有所帮助:

const handleSend = () => {
  if (ws.readyState === WebSocket.OPEN) {
    ws.send()
  } else {
    // Queue a retry
    setTimeout(() => { handleSend() }, 1000)
  }
};

正如 Logan 提到的，我的第一个例子是懒惰的。我只是想让 OP 畅通无阻，我相信读者足够聪明，能够理解如何从那里获取它。因此，请确保适本地处理可用状态，例如，如果 readyState 是 WebSocket.CONNECTING 然后注册一个监听器:

const handleSend = () => {
  if (ws.readyState === WebSocket.OPEN) {
    ws.send()
  } else if (ws.readyState == WebSocket.CONNECTING) {
    // Wait for the open event, maybe do something with promises
    // depending on your use case. I believe in you developer!
    ws.addEventListener('open', () => handleSend())
  } else {
    // etc.
  }
};