JS 代码如何捕获websocket 异常

Question

这个错误异常直接抛到了控制台，如何再JS代码中捕获这个异常，尤其是获取到最后那个code

Answer 1

onerror 捕获不到嘛？

  const websocket = new WebSocket("ws://127.0.0.1:9999")
  websocket.onopen = (evnt) => {
    console.log('onopen', evnt)
  }
  websocket.onerror = (evnt) => {
      console.log('onerror', evnt)
  }

Answer 2

onerror

Answer 3

webscoket 里面好像有内置的方法处理异常

Answer 4

try catch

Answer 5

同问，有解决了吗