无法从亚马逊获得电话评级

Question

对于此代码，我将获得电话评级的多个输出，但我只需要输出作为评论数量及其评级的输出。

import requests
import re
from bs4 import BeautifulSoup
import time
x=[]
def rating_amazon(url):
    #url="http://www.amazon.in/Samsung-Galaxy-Smartphone-Prime-SM-G610F/dp/B01LY4V6IM"
    request =  requests.get(url).text
    soup = BeautifulSoup(request,"html.parser")
    for i in soup.find_all('span', attrs={'class':'a-size-base','id':'acrCustomerReviewText'}):
        print i.text.replace("customer ","")
    for i in soup.find_all('i', attrs={'class':'a-icon a-icon-star a-star-3'}):
        #print i.find('span', attrs={'class':'a-icon-alt'})
        print i.text

url=["http://www.amazon.in/Samsung-Galaxy-Smartphone-Prime-SM-G610F/dp/B01LY4V6IM","http://www.amazon.in/SAMSUNG-Galaxy-J7-2016-White/dp/B01JEH8ND0"]
counter=0
while counter<len(url):
    rating_amazon(url[counter])
    counter+=1

Answer 1

如果您只想要一个结果，那么在rating_amazon中使用第二个for循环是什么。只需将findall存储在列表中并使用第一个实例即可。

for i in soup.find_all('i', attrs={'class':'a-icon a-icon-star a-star-3'}):
    print i.text

就这样改变它，

a=soup.find_all('i', attrs={'class':'a-icon a-icon-star a-star-3'})
print a[0].text

我得到了这样的结果，

47 reviews
3.2 out of 5 stars
28 reviews
3.0 out of 5 stars