无法从亚马逊获得 phone 的评分
not able to get rating of phone from amazon
对于此代码,我得到了 phone 评级的多个输出,但我只需要输出评论数量及其评级
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
如果您只想要一个结果,那么 rating_amazon
中的第二个 for 循环有什么用?
只需将 findall
存储在列表中并使用第一个实例。
for i in soup.find_all('i', attrs={'class':'a-icon a-icon-star a-star-3'}):
print i.text
改成这样,
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
对于此代码,我得到了 phone 评级的多个输出,但我只需要输出评论数量及其评级
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
如果您只想要一个结果,那么 rating_amazon
中的第二个 for 循环有什么用?
只需将 findall
存储在列表中并使用第一个实例。
for i in soup.find_all('i', attrs={'class':'a-icon a-icon-star a-star-3'}):
print i.text
改成这样,
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