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名录的HREF我最近发布的要求,以刮去名录和@alecxe数据帮助了吨在我面前展现了一些新的方法来提取数据,但我坚持再次,想凑数据为每个链接在黄页,所以我可以得到有更多数据的黄页页面。我想添加一个名为“url”的变量,并获取业务的href,而不是实际的业务网站,而是业务的黄页页面。我尝试了各种各样的东西,但似乎没有任何工作。 href在“class = business-name”之下。抓取与蟒蛇
import csv
import requests
from bs4 import BeautifulSoup
with open('cities_louisiana.csv','r') as cities:
lines = cities.read().splitlines()
cities.close()
for city in lines:
print(city)
url = "http://www.yellowpages.com/search?search_terms=businesses&geo_location_terms="baton%rouge+LA&page="+str(count)
for city in lines:
for x in range (0, 50):
print("http://www.yellowpages.com/search?search_terms=businesses&geo_location_terms=baton%rouge+LA&page="+str(x))
page = requests.get("http://www.yellowpages.com/search?search_terms=businesses&geo_location_terms=baton%rouge+LA&page="+str(x))
soup = BeautifulSoup(page.text, "html.parser")
for result in soup.select(".search-results .result"):
try:
name = result.select_one(".business-name").get_text(strip=True, separator=" ")
except:
pass
try:
streetAddress = result.select_one(".street-address").get_text(strip=True, separator=" ")
except:
pass
try:
city = result.select_one(".locality").get_text(strip=True, separator=" ")
city = city.replace(",", "")
state = "LA"
zip = result.select_one('span[itemprop$="postalCode"]').get_text(strip=True, separator=" ")
except:
pass
try:
telephone = result.select_one(".phones").get_text(strip=True, separator=" ")
except:
telephone = "No Telephone"
try:
categories = result.select_one(".categories").get_text(strip=True, separator=" ")
except:
categories = "No Categories"
completeData = name, streetAddress, city, state, zip, telephone, categories
print(completeData)
with open("yellowpages_businesses_louisiana.csv", "a", newline="") as write:
wrt = csv.writer(write)
wrt.writerow(completeData)
write.close()
很不错的!我仍然对Python和编程一般都很陌生。尽管我只是通过添加'business_name_element = result.select_one(“。business-name”)'和link = urljoin(page.url,business_name_element [“href”])''做了一些小改动,您的解决方案仍然很棒。当我阅读你的代码时,我会对其进行逆向工程,这样才有意义。感谢您的支持! –