How to dismiss a ui.notification after seconds and do not sleep?

[weinibuliu]

Question

Hi everyone!

I was trying to use ui.notification,it's good.I tried to dismiss it after seconds,but I met a problem.

import asyncio
import time

from nicegui import ui


async def test():
    noti = ui.notification("Test", spinner=True, type="ongoing", timeout=20)
    # do something
    noti.message = "Test Done"

    print(f"Test Done: {time.time()}")
    await asyncio.sleep(3)
    noti.dismiss()


async def on_button_click():
    await test()
    print(f"Now: {time.time()}")


ui.button("Test").on_click(on_button_click)

ui.run()

asyncio.sleep() will block the current task(like time.sleep()).Sorry I just have few knowledge of asyncio,so I try to find another way.I changed my code.

async def test():
    noti = ui.notification("Test", spinner=True, type="ongoing", timeout=20)
    # do something
    noti.message = "Test Done"

    print(f"Test Done: {time.time()}")
    noti._props["options"].update({"timeout": 3000})

I edit the _props of ui.notification.It works well.Now I don't have to wait for sleep.
Can I create a PR to provide a function to set timeout?

[falkoschindler]

Hi @weinibuliu,

What exactly are you trying to do? Why do you create a notification with a 20 seconds timeout and change it to 3 seconds afterwards?

[weinibuliu]

Hi @falkoschindler
I am doing some time consuming calculations,and I can't estimate how long it will take,so I set timeout as 20 (Maybe I should set as None.)
When calculations end up,I will change the notification meaasge,spinner,type.And I want to dismiss the notification.But as I said before,I couldn't use sleep,because it will block the task.So I try to change timeout.

async def test():
    noti = ui.notification("Test", spinner=True, type="ongoing", timeout=20)
    # do some time consuming calculations
    noti.message = "Test Done"

Thanks for your reply.

[falkoschindler]

Ah, now I understand. By setting the timeout to 3 seconds, you're telling the notification to dismiss 3 seconds after that moment. That sounds reasonable.

So this is basically what you're doing:

async def notify():
    n = ui.notification("Running", spinner=True, timeout=None)
    await asyncio.sleep(5.0)
    n.message = "Done"
    n.spinner = False
    n.props["options"]["timeout"] = 3000
    n.update()

ui.button("Test", on_click=notify)

And like message and spinner, it would be great to have a property for timeout so that you don't have to access the props dictionary and call update yourself. A pull request would be very welcome! 👍🏻

[falkoschindler]

By the way, you can also use a timer to dismiss the notification after 3 seconds:

    ui.timer(3.0, n.dismiss, once=True)

[weinibuliu]

Thanks for your quick reply.

I will try to create a PR to let it easier.

Thanks again.