How to pass a variable out of a lambda

Question

from playwright.sync_api import sync_playwright

url = "https://hls-js.netlify.app/demo/" 
with sync_playwright() as p: 

    browser = p.chromium.launch(channel = "chrome")
    page = browser.new_page() 
    page.on("response", lambda response: print(response.url) if (".m3u8" in response.url) else False) 
    page.goto(url) 
    page.wait_for_timeout(10*1000) 
    browser.close()

Hi guys, i need to pass the "response.url" inside a variable that I can use further in the code. I don't want response.url printed out, I just want to save that string in a variable. How can I do this?

Possible duplicate of: https://stackoverflow.com/questions/6282042/assignment-inside-lambda-expression-in-python — Ashok Arora, Nov 25 '21 at 12:26
If you're on python 3.8 or above, this might work: https://stackoverflow.com/a/14617232/6389787 — Ashok Arora, Nov 25 '21 at 12:26
function assigned to event can't return value - so you may have to assign result to global variable, or append to global list if you want to get many urls. — furas, Nov 25 '21 at 14:19
maybe better solution is to put rest of code inside this function and then you don't have to return value outside function. — furas, Nov 25 '21 at 14:23
@AshokArora it's not exactly the same, indeed those solutions do not fit with my code — Francesco Nestorini, Nov 25 '21 at 15:55
@furas I tried to assign it to a global variable but it does not work out. The rest of the code needs to be outside unluckily — Francesco Nestorini, Nov 25 '21 at 15:57
write normal function with `global url` and assing `url = response.url`. In `lambda` you can't use `global` and can't assing values to variables. You may only `append` to list. — furas, Nov 25 '21 at 16:02

score 1 · Answer 1 · answered Nov 25 '21 at 16:15

Functions assigned to events (in any module) can't return value and you may need to use global variable for this. But lambda can't use global and can't assign value to variable so it needs normal function

url = ""

def on_response(response):
    global url  # inform functio to use external variable (instead of local variable) when you use `=` to assign value
    
    if ".m3u8" in response.url:
        url = response.url

page.on("response", on_response)

Thanks it worked out – Francesco Nestorini Nov 25 '21 at 16:17 — Francesco Nestorini, Nov 25 '21 at 16:17

How to pass a variable out of a lambda

1 Answers1