如何使用列表项的名称创建函数，

如何解决如何使用列表项的名称创建函数，

在kivy应用程序启动时，它会接收json值，并将其分配给变量，并将其附加到列表中。在每次启动应用程序时，它都会收到不同的值，因此列表始终是不同的。

Ch1_name = channel.json()['items']['channels'][0]['channelName']
Ch2_name = channel.json()['items']['channels'][1]['channelName']
Ch3_name = channel.json()['items']['channels'][2]['channelName']
Ch4_name = channel.json()['items']['channels'][3]['channelName']

channels = []

channels.append(Ch1_name)
channels.append(Ch2_name)
channels.append(Ch3_name)
channels.append(Ch4_name)

print(channels)
['cnn','fox','sky sports','fsc']

问题： 我需要为每个频道创建函数，该函数将保留列表项的名称，即频道名称。像这样：

def cnn()
    print('this function was named cnn just like an item list')

也许上课有另外一种方法吗？谢谢您的任何想法。

解决方法

for channel_name in channels:
  exec(f"""
def {channel_name}():
  print(f'this function was named {channel_name} just like an item list')
"""
  )

,

您可以做一些元编程工作来动态添加功能。这是一些示例代码，它使用setattr根据通道名称在类上设置函数。

class ChannelPrinter:
    pass

channel_printer = ChannelPrinter()

for channel in ['cnn','fox','sky sports','fsc']:
    def make_func(name):
        def print_channel_info():
            print('this function was named {} just like an item list'.format(name))
        return print_channel_info

    setattr(channel_printer,channel,make_func(name=channel))

> channel_printer.cnn()
this function was named cnn just like an item list
> channel_printer.fox()
this function was named fox just like an item list