将参数传递给 URL 对象

Question

使用kivy中的URL对象，https://kivy.org/doc/stable/api-kivy.network.urlrequest.html，如果我想改变on_success函数接受另一个参数，我该如何将值传递给它？

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

然后在 on_success 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)

另一种解决方案是使用 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))

将参数传递给 URL 对象

Passing arguments to URL object

python

kivy

kivy-language