审查tripadvisor的刮表

Question

我是新的网络刮刮在python3。我想刮一下迪拜所有酒店的评论，但问题是我只能刮掉我在网址中描述的酒店评论。有人能告诉我，我怎样才能得到所有的酒店评论，而不含蓄地提供每个酒店的网址？

import requests
from bs4 import BeautifulSoup


importurl = 'https://www.tripadvisor.com/Hotel_Review-g295424-d302778-Reviews-Roda_Al_Bustan_Dubai_Airport-Dubai_Emirate_of_Dubai.html'
r = requests.get(importurl)
soup = BeautifulSoup(r.content, "lxml")
 resultsoup = soup.find_all("p", {"class" : "partial_entry"})
#save the reviews to a test text file locally
for review in resultsoup:
review_list = review.get_text()
print(review_list)
with open('testreview.txt', 'w') as fid: 
    for review in resultsoup:
        review_list = review.get_text()
        fid.write(review_list)

Answer 1

您应该找到所有酒店的索引页，将所有链接放入列表中，而不是循环url列表以获得评论。

import bs4, requests
index_pages = ('http://www.tripadvisor.cn/Hotels-g295424-oa{}-Dubai_Emirate_of_Dubai-Hotels.html#ACCOM_OVERVIEW'.format(i) for i in range(0, 540, 30))
urls = []
with requests.session() as s:
    for index in index_pages:
        r = s.get(index)
        soup = bs4.BeautifulSoup(r.text, 'lxml')
        url_list = [i.get('href') for i in soup.select('.property_title')]
        urls.append(url_list)

退出：

len(urls): 540