我有一个函数,可以从API调用返回字典列表。 API每个请求最多可返回100条记录。记录存储在页面中,每页包含100条记录。
def call_api(page):
return list_of_records
我需要的是一个装饰器,该装饰器每次在返回带有页参数= + 1的100条记录时都可以重新运行此函数。
类似:
if len(list_of_records) == 100:
call_api(page +=1)
最终输出应为包含所有记录的单个列表
答案 0 :(得分:2)
定义一个可以递归调用一个函数的装饰器当然是可能的。
在您的情况下,这样的装饰器可能是这样:
import functools
def with_pages(func):
@functools.wraps(func)
def wrapper(page):
list_of_records = []
partial_list = func(page)
list_of_records.extend(partial_list)
while len(partial_list) == 100: # there is probably a better condition to check if you are at the last page (e.g. next_page URI in api reply)
page += 1
partial_list = func(page)
list_of_records.extend(partial_list)
return list_of_records
return wrapper
@with_pages
def call_api(page):
# api interaction creating list_of_records_on_a_page
return list_of_records_on_a_page
或者,您可以调整call_api
以便可以递归调用。
类似的东西:
def call_api(page, list_of_records=[]):
partial_list = ... # the records you get from the api
list_of_records.extend(partial_list)
# often api provide the url for the next page or some more reliable info
# about whether or not you are on the last page, i.e. testing for the length
# of the reply is probably not the best strategy
if len(partial_list) == 100:
# time.sleep(0.2) # just to not end up with to many requests in a short period of time
call_api(page + 1, list_of_records)
return list_of_records
只需调用call_api(1)
就可以获取所有页面。