如何使用 python 装饰器修复 "NoneType" 错误？

Question

我正在用装饰器做一个 LRU（最近最少使用）缓存来存储最近使用的本地化，但是当我调用函数从 .json 文件中读取它时，我得到 "NoneType" 错误

def loc_cache(func):
    loc_cache.locale = {} # {lang:(local, count)} local:dict
    loc_cache.MAXLENGTH = 5
    def wrapper(key):
        print(key) #this function wasn't called
        if key not in loc_cache.locale.keys():
            try:
                locale = read_locale(key)
                loc_cache.locale[key] = (locale, 1)
                wrapper.locale = locale
            except KeyError:
                key = "en" # set default locale
                wrapper(key)
        else:
            locale, count = loc_cache.locale[key]
            loc_cache.locale[key] = (locale, count+1)
            wrapper.locale = locale
        return wrapper.locale

@loc_cache
def read_locale(key):
    locale = read_json("./config/locale.json", key)
    return locale

def auth(user:User):
    locale = read_locale(user.locale)
    print(locale["auth"])
    return

u = User(1) # __init__ takes 1 for id
u.locale = "en"
auth(u)

我希望 return 短语 "en" 存储在 .json 文件中，但它说

Traceback (most recent call last):
  File "main.py", line 61, in <module>
    auth(u)
  File "main.py", line 52, in auth
    locale = read_locale(user.locale)
TypeError: 'NoneType' object is not callable

Answer 1

您没有 return 包装函数形成装饰器，因此 Python 像往常一样 returning None 并在您执行 read_locale(user.locale)。您需要：

def loc_cache(func):
    loc_cache.locale = {} # {lang:(local, count)} local:dict
    loc_cache.MAXLENGTH = 5
    def wrapper(key):
        print(key) #this function wasn't called
        if key not in loc_cache.locale.keys():
            try:
                locale = read_locale(key)
                loc_cache.locale[key] = (locale, 1)
                wrapper.locale = locale
            except KeyError:
                key = "en" # set default locale
                wrapper(key)
        else:
            locale, count = loc_cache.locale[key]
            loc_cache.locale[key] = (locale, count+1)
            wrapper.locale = locale
        return wrapper.locale
    return wrapper
    # Here ^^^^

Answer 2

你没有return你的wrapper装饰器的末尾：

def loc_cache(func):
    loc_cache.locale = {} # {lang:(local, count)} local:dict
    loc_cache.MAXLENGTH = 5
    def wrapper(key):
        if key not in loc_cache.locale:
            try:
                locale = read_locale(key)
                loc_cache.locale[key] = (locale, 1)
            except KeyError:
                key = "en" # set default locale
                wrapper(key)
        else:
            locale, count = loc_cache.locale[key]
            loc_cache.locale[key] = (locale, count+1)
        wrapper.locale = locale
        return wrapper.locale
    <b>return wrapper</b>

装饰器通常将一个函数作为输入，returns 是一个函数。在这里你定义了一个内部函数，但是你忘了 return 它。结果，装饰器的输出是 None，你不能调用 None.

如何使用 python 装饰器修复 "NoneType" 错误？

How to fix "NoneType" error with python decorators?

python

python-decorators

python-3.6