我正在使用requests
包与BeautifulSoup
一起搜索Google新闻,了解查询的搜索结果数量。我有两种类型的IndexError
,我想区分:
#resultStats
返回空字符串'[]'
。似乎正在进行的是,当查询字符串太长时,谷歌甚至不会说" 0搜索结果&#34 ;;它只是没有说什么。IndexError
是谷歌给我一个验证码。我需要区分这些情况,因为当谷歌发给我一个验证码时,我希望我的刮刀等待五分钟,但不是当它只是一个空的结果字符串时。
我目前有一个陪审团操纵的方法,我发送另一个查询,其中包含已知的非零数量的搜索结果,这使我可以区分这两个IndexErrors
。我想知道使用BeautifulSoup
是否有更优雅和直接的做法。
这是我的代码:
import requests, bs4, lxml, re, time, random
import pandas as pd
import numpy as np
URL = 'https://www.google.com/search?tbm=nws&q={query}&tbs=cdr%3A1%2Ccd_min%3A{year}%2Ccd_max%3A{year}&authuser=0'
headers = {
"User-Agent":
"Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/44.0.2403.157 Safari/537.36"
}
def tester(): # test for captcha
test = requests.get('https://www.google.ca/search?q=donald+trump&safe=off&client=ubuntu&espv=2&biw=1910&bih=969&source=lnt&tbs=cdr%3A1%2Ccd_min%3A2016%2Ccd_max%3A&tbm=nws', headers=headers)
dump = bs4.BeautifulSoup(test.text,"lxml")
result = dump.select('#resultStats')
num = result[0].getText()
num = re.search(r"\b\d[\d,.]*\b",num).group() # regex
num = int(num.replace(',',''))
num = (num > 0)
return num
def search(**params):
response = requests.get(URL.format(**params),headers=headers)
print(response.content, response.status_code) # check this for google requiring Captcha
soup = bs4.BeautifulSoup(response.text,"lxml")
elems = soup.select('#resultStats')
try: # want code to flag if I get a Captcha
hits = elems[0].getText()
hits = re.search(r"\b\d[\d,.]*\b",hits).group() # regex
hits = int(hits.replace(',',''))
print(hits)
return hits
except IndexError:
try:
tester() > 0 # if captcha, this will throw up another IndexError
print("Empty results!")
hits = 0
return hits
except IndexError:
print("Captcha'd!")
time.sleep(120) # should make it rotate IP when captcha'd
hits = 0
return hits
for qry in list:
hits = search(query= qry, year=2016)
答案 0 :(得分:2)
我只是搜索"验证码"例如,如果这是Google Recaptcha,则可以搜索包含令牌的隐藏输入:
is_captcha_on_page = soup.find("input", id="recaptcha-token") is not None