如何得到“事情要做”的清单?我对网络抓取是新手,我不知道如何循环每一页来得到所有‘事情要做’的href ?告诉我我在哪里做错了什么?任何帮助都会得到高度认可。提前谢谢。
import requests
import re
from bs4 import BeautifulSoup
from urllib.request import urlopen
offset = 0
url = 'https://www.tripadvisor.com/Attractions-g255057-Activities-oa' + str(offset) + '-Canberra_Australian_Capital_Territory-Hotels.html#ATTRACTION_LIST_CONTENTS'
urls = []
r = requests.get(url)
soup = BeautifulSoup(r.text, "html.parser")
for link in soup.find_all('a', {'last'}):
page_number = link.get('data-page-number')
last_offset = int(page_number) * 30
print('last offset:', last_offset)
for offset in range(0, last_offset, 30):
print('--- page offset:', offset, '---')
url = 'https://www.tripadvisor.com/Attractions-g255057-oa' + str(offset) + '-Canberra_Australian_Capital_Territory-Hotels.html#ATTRACTION_LIST_CONTENTS'
r = requests.get(url)
soup = BeautifulSoup(r.text, "html.parser")
for link in soup.find_all('a', {'property_title'}):
iurl='https://www.tripadvisor.com/Attraction_Review-g255057' + link.get('href')
print(iurl)基本上,我想要每个‘事情做’的href。我对“要做的事情”的期望输出是:
https://www.tripadvisor.com/Attraction_Review-g255057-d3377852-Reviews-Weston_Park-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Attraction_Review-g255057-d591972-Reviews-Canberra_Museum_and_Gallery-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Attraction_Review-g255057-d312426-Reviews-Lanyon_Homestead-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Attraction_Review-g255057-d296666-Reviews-Australian_National_University-Canberra_Australian_Capital_Territory.html与下面的示例一样,我使用了下面的代码来获取堪培拉市每一家餐厅的href,我的餐馆代码是:
import requests
import re
from bs4 import BeautifulSoup
from urllib.request import urlopen
with requests.Session() as session:
for offset in range(0, 1050, 30):
url = 'https://www.tripadvisor.com/Restaurants-g255057-oa{0}-Canberra_Australian_Capital_Territory.html#EATERY_LIST_CONTENTS'.format(offset)
soup = BeautifulSoup(session.get(url).content, "html.parser")
for link in soup.select('a.property_title'):
iurl = 'https://www.tripadvisor.com/' + link.get('href')
print(iurl) 餐厅代码的输出是:
https://www.tripadvisor.com/Restaurant_Review-g255057-d1054676-Reviews-Lanterne_Rooms-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Restaurant_Review-g255057-d755055-Reviews-Courgette_Restaurant-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Restaurant_Review-g255057-d6893178-Reviews-Pomegranate-Canberra_Australian_Capital_Territory.html
https://www.tripadvisor.com/Restaurant_Review-g255057-d7262443-Reviews-Les_Bistronomes-Canberra_Australian_Capital_Territory.html
.
.
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.发布于 2017-04-18 11:51:14
好吧,没那么难,你只需要知道要使用哪些标签。
让我用这个例子来解释:
import requests
from bs4 import BeautifulSoup
base_url = 'https://www.tripadvisor.com/' ## we need this to join the links later ##
main_page = 'https://www.tripadvisor.com/Attractions-g255057-Activities-oa{}-Canberra_Australian_Capital_Territory-Hotels.html#ATTRACTION_LIST_CONTENTS'
links = []
## get the initial page to find the number of pages ##
r = requests.get(main_page.format(0))
soup = BeautifulSoup(r.text, "html.parser")
## select the last page from the list of pages ('a', {'class':'pageNum taLnk'}) ##
last_page = max([ int(page.get('data-offset')) for page in soup.find_all('a', {'class':'pageNum taLnk'}) ])
## now iterate over that range (first page, last page, number of links), and extract the links from each page ##
for i in range(0, last_page + 30, 30):
page = main_page.format(i)
soup = BeautifulSoup(requests.get(page).text, "html.parser") ## get the next page and parse it with BeautifulSoup ##
## get the hrefs from ('div', {'class':'listing_title'}), and join them with base_url to make the links ##
links += [ base_url + link.find('a').get('href') for link in soup.find_all('div', {'class':'listing_title'}) ]
for link in links :
print(link)这给了我们8页和212总链接( 30页在每页,2在最后)。
我希望这能把事情弄清楚一点
https://stackoverflow.com/questions/43454459
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