将参数传递给URL对象

时间:2019-03-27 22:06:05

标签: python kivy kivy-language

如果要更改on_success函数的值以使用另一个参数,请使用kivy https://kivy.org/doc/stable/api-kivy.network.urlrequest.html中的URL对象,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中具有这样的内容

def url_success(req, result, sensor_id):

1 个答案:

答案 0 :(得分:0)

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))