bs4 python从<span> </span>提取到.csv的值一遍又一遍地打印出相同的结果

时间:2019-01-27 21:57:04

标签: python python-3.x beautifulsoup

我设法建立了一个非常原始的程序,以从活塞头上刮除车辆数据,并通过链接,制造,建模将其打印到.csv文件中,并且正在努力获取价格,这正是我遇到的问题。问题。

我想将价格刮到.csv文件(Price)的第四列,并正确打印网站上每种车辆的价格。

我只能得到它来打印一辆车的价格,并在.csv文件中的每辆车旁边重复一次。

我曾尝试soup.findAllsoup.find_all来查看通过多个元素进行解析是否可行,但这只会造成更大的混乱。

有人可以帮忙吗?

我还试图抓取图像src,并希望将其打印在名为images的另一列(5)上。

import csv ; import requests

from bs4 import BeautifulSoup

outfile = open('pistonheads.csv','w', newline='')
writer = csv.writer(outfile)
writer.writerow(["Link", "Make", "Model", "Price"])

url = 'https://www.pistonheads.com/classifieds?Category=used-cars&Page=1&ResultsPerPage=100'

get_url = requests.get(url)

get_text = get_url.text

soup = BeautifulSoup(get_text, 'html.parser')

car_link = soup.find_all('div', 'listing-headline', 'price')

for div in car_link:
   links = div.findAll('a')
      for a in links:
      link = ("https://www.pistonheads.com" + a['href'])
      make = (a['href'].split('/')[-4])
      model = (a['href'].split('/')[-3])
      price = soup.find('span')
      writer.writerow([link, make, model, price])
      print(link, make, model, price)
outfile.close()

2 个答案:

答案 0 :(得分:3)

您可以尝试以下方法:

import csv, requests, re
from urllib.parse import urlparse
from bs4 import BeautifulSoup as soup
d = soup(requests.get('https://www.pistonheads.com/classifieds?Category=used-cars&ResultsPerPage=100').text, 'html.parser')
def extract_details(_s:soup) -> list:
  _link = _s.find('a', {'href':re.compile('/classifieds/used\-cars/')})['href']
  _, _, make, model, *_ = _link[1:].split('/')
  price, img = _s.find('div', {'class':'price'}).text, [i['src'] for i in _s.find_all('img')]
  return [_link, make, model, price, 'N/A' if not img else img[0]]

with open('filename.csv', 'w') as f:
  _listings = [extract_details(i) for i in d.find_all('div', {'class':'ad-listing'}) if i.find('div', {'class':'price'})]
  write = csv.writer(f)
  write.writerows([['make', 'model', 'price', 'img'], *_listings])

答案 1 :(得分:1)

原因是由于fmt.Printf("%q", Location)

Location = strings.TrimSpace(Location) 将获取其找到的第一个元素。然后,您可以查看price = soup.find('span')对象。但是您想要查看的地方位于.find()内,因为这就是您使用soup遍历的内容

我还添加了a,因为我假设您只是想要文本,而不是整个tag元素。即for a in links:

.text