如何使用列表项的名称创建函数，答案

【问题标题】：How to create functions with a name of the list item,如何使用列表项的名称创建函数，
【发布时间】：2021-01-18 14:32:22
【问题描述】：

在 kivy 应用程序中，它在启动时接收一个 json 值，将它们分配给一个变量并将它们附加到一个列表中。在每次应用启动时，它都会收到不同的值，因此列表总是不同的。

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，这将保留作为频道名称的列表项的名称。比如：

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

也许有一个不同的方式来做一个类？感谢您的任何想法。

【问题讨论】：

标签： python function variables kivy kivymd

【解决方案1】：

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

【讨论】：

非常感谢，非常简单的解决方案！

【解决方案2】：

你可以做一些元编程的东西来动态添加函数。下面是一些示例代码，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

【讨论】：