我使用一个函数(movies_from_url)从网页上读取总共256个电影。每页包含50部电影。我必须为此阅读前6页(250页电影5页,6电影6页)。
第一个网址:
http://www.imdb.com/search/title?at=0&sort=user_rating&start=1&title_type=feature&year=2005,2014
这是我模糊的想法:
def read_m_by_rating(first_year=2005, last_year=2015, top_number=256):
current_index=1 # current index is start number of a webpage
final_list = []
for _ in xrange(6):
url = http://www.imdb.com/search/title?at=0&sort=user_rating&start=current_index&title_type=feature&year=2005,2014
if top_number==300:
lis = movies_from_url(url, top_number - current_index + 1)
else:
lis = movies_from_url(url, 50)
final_list.append(lis)
current_index=+50
return final_list
答案 0 :(得分:1)
只需在current_index上使用一个简单的循环即可。
while current_index<256:
url = "http://www.imdb.com/search/title?at=0&sort=user_rating&start="\
+str(current_index)+"&title_type=feature&year=2005,2014"
...
...
current_index+=50
return final_list