Chrome 和 Web 串行 API。 如何在不询问的情况下获取端口

Question

I need your help.

Is there a way to take control of a serial port, previously authorized, without using the popup window? (without port = await navigator.serial.requestPort(); )

Thank you and sorry for my english, italian speaking here.

Answer 1

I believe JS code below should work fine for you:

// Get all serial ports the user has previously granted the website access to.
const ports = await navigator.serial.getPorts();

Source: https://web.dev/serial/#open-port