python web-crawler猜测解析器警告
python web-crawler猜测解析器警告
提问于 2020-08-23 21:41:41
我正在尝试用python (3.8)做一个网络爬虫，我认为我已经完成了，但是我得到了这个错误，有没有人可以帮助我，提前感谢。
Python代码：
import requests
from bs4 import BeautifulSoup
def aliexpress_spider (max_pages):
page = 1
while page <= max_pages:
url = "https://www.aliexpress.com/af/ps4.html?trafficChannel=af&d=y&CatId=0&SearchText=ps4<ype=affiliate&SortType=default&page=" + str(page)
sourcecode = requests.get(url)
plaintext = sourcecode.text
soup = BeautifulSoup(plaintext)
for link in soup.findAll('a' , {'class' : 'item-title'}):
href = "https://www.aliexpress.com" + link.get("href")
title = link.string
print(href)
print(title)
page += 1
aliexpress_spider(1)错误消息：
GuessedAtParserWarning: No parser was explicitly specified, so I'm using the best available HTML parser for this system ("html.parser"). This usually isn't a problem, but if you run this code on another system, or in a different virtual environment, it may use a different parser and behave differently.
The code that caused this warning is on line 11 of the file C:/Users/moham/PycharmProjects/moh/test.py. To get rid of this warning, pass the additional argument 'features="html.parser"' to the BeautifulSoup constructor.
soup = BeautifulSoup(plaintext)回答 1
Stack Overflow用户
发布于 2021-02-27 00:06:12
import requests
from bs4 import BeautifulSoup
def aliexpress_spider (max_pages):
page = 1
while page <= max_pages:
url = "https://www.aliexpress.com/af/ps4.html?trafficChannel=af&d=y&CatId=0&SearchText=ps4<ype=affiliate&SortType=default&page=" + str(page)
sourcecode = requests.get(url)
soup = BeautifulSoup(sourcecode.text ,"html.parser")
for link in soup.findAll('a' , {'class' : 'item-title'}):
href = "https://www.aliexpress.com" + link.get("href")
title = link.string
print(href)
print(title)
print(soup.title)
page += 1
aliexpress_spider(1)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/63547631
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