我正在使用带有装饰器的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,会出现“ 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)
我希望它会返回存储在.json文件中的“ en”中的词组,但是它说
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
答案 0 :(得分:2)
您没有从装饰器返回包装函数,因此Python照常返回None
并尝试在执行read_locale(user.locale)
时调用它。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 ^^^^
答案 1 :(得分:0)
您没有在装饰器的结尾退回您的wrapper
:
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
return wrapper
装饰器通常将一个函数作为输入,然后返回一个函数。在这里,您定义了一个内部函数,但是您忘了返回它。结果,装饰器的输出为None
,并且您无法调用None
。