将参数传递给URL对象

Question

在kivy，https://kivy.org/doc/stable/api-kivy.network.urlrequest.html中使用URL对象，如果我想更改on_success函数以接受另一个参数，我将如何将值传递给它？

def generate_images(sensor_id):
    req = UrlRequest(URL, on_success=url_success)

然后在on_success中有这样的东西

def url_success(req, result, sensor_id):

Answer 1

一种解决方案是使用functools.partial()：

from functools import partial

# ...

def generate_images(sensor_id):
    req = UrlRequest(URL, on_success=partial(url_success, sensor_id))

# ...

def url_success(sensor_id, req, result):
    print(sensor_id, req, result)

另一个解决方案是使用lambda函数：

def generate_images(sensor_id):
    req = UrlRequest(URL, on_success= lambda req, result, sensor_id=sensor_id : url_success(req, result, sensor_id))