why cannot I get the url in one cell ?

[12rambau]

In the example the workflow is the following :

w = Url()
w

w.url

I tested it on my end and it works as expected.

Now if i try to run:

w = Url()
w.url

in 1 cell, I get ''as a result. Why isn't it working in one cell (and by extension in a python script) ?

[davidbrochart]

Because you need to instantiate the widget. That is what this does:

w

This cannot work in a Python script because it is a Jupyter notebook feature.
But there is another way:

url = None
async def foo():
    global url
    w = Url()
    display(w)
    url = await w.get_url()
asyncio.ensure_future(foo())

url will be set after some time it your environment is async.

[12rambau]

so using display is not instanciating anything correct ?

[davidbrochart]

It's displaying the widget w, then you await the URL.

[12rambau]

ok, then If i want to get the url in a notebook let's say like this one:

w = Url()
w

# do stuff with url 
w.url

How can I launch it in voila ?

[davidbrochart]

Voilà is another beast, it will launch all the cells one after the other, so the URL might not be retrieved from the front-end when you do w.url. Something like that could work:

url = None
async def foo():
    global url
    w = Url()
    display(w)
    url = await w.get_url()
await asyncio.ensure_future(foo())

[12rambau]

Voilà is another beast

Unfortunately for me it's the beast I need to focus on ;-)

Something like that could work

arf ok so I tried your last piece of code and even in a jupyter notebook it awaits forever. It seems that the widget cannot be instanciated if I don't leave the cell

[davidbrochart]

Ah yes, you're right, it's deadlocking. See also:

• Async widget methods deadlock due to on_msg not being called ipython/ipykernel#646
• Issue #12786: Create hook for dispatching GUI messages asynchronously  ipython/ipykernel#589

One solution could be to use akernel.

[12rambau]

thank you for the links, I'll explore the solutions and share what work for me before closing the issue

[artttt]

Take a look at https://github.com/Kirill888/jupyter-ui-poll
I believe this will solve this issue

edited to add example using this package now i have my install issues sorted out.

from ipyurl import Url
from IPython.display import display
from jupyter_ui_poll import run_ui_poll_loop

def window_url():
    """returns the url for the window that jupyter is running in"""
    w = Url()
    display(w)
    func = lambda: None if w.url == "" else w.url
    url = run_ui_poll_loop(func)
    return url

out of interest the same can also be done with ipyleaflet providing the url

import ipyleaflet
from IPython.display import display
from jupyter_ui_poll import run_ui_poll_loop


def window_url():
    """returns the url for the window that jupyter is running in"""

    m = ipyleaflet.Map(zoom=0)
    m.layout.display = "none"
    display(m)
    func = lambda: None if m.window_url == "" else m.window_url
    url = run_ui_poll_loop(func)
    m.close()
    return url

[12rambau]

thanks @artttt for the suggestions. I think with this + @davidbrochart insight, I have everything I need to make it work from my side.