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如何从属性 url 获取 url 我的意思是 url 本身的风格？style="width: 433px; height: 510px; background-image: url( https://cs7056.vk.me/c635104/v635104607/1c316/ADzy-2WY8pw.jpg )" Selenium3 Python3对你来说很容易！

import requests
from bs4 import BeautifulSoup
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.common.keys import Keys
from selenium.common.exceptions import NoSuchElementException
from selenium.common.exceptions import NoAlertPresentException
import re
import time


url = 'https://vk.com/uporols_you'
driver                          = webdriver.Firefox(executable_path='C:/Users/PANDEMIC/AppData/Local/Mozilla/geckodriver.exe')

def login(driver):
    log_page                    = driver.get('https://login.vk.com/?act=login')
    find_login_input            = driver.find_element_by_id('login_form').find_element_by_id('email').send_keys('+77782303865')
    find_password_input         = driver.find_element_by_id('login_form').find_element_by_id('pass').send_keys('pass')
    find_button                 = driver.find_element_by_xpath('//button[@id="login_button"]').click()
    time.sleep(5)



def get_photo_from_page(driver):
    driver.get(url)
    try:
        driver.find_element_by_class_name('popup_box_container').find_element_by_class_name('box_title_wrap').find_element_by_class_name('box_x_button').click()
    except:
        print('nope nothing')

    for i in range(2):
        scrol_down = driver.find_element_by_id('public_wall').find_element_by_id('wall_more_link').click()
        time.sleep(2)

    tut = []
    #t = (a[@class="page_post_thumb_wrap image_cover  page_post_thumb_last_column page_post_thumb_last_row"])
    for ii in driver.find_elements_by_xpath('//a[@style]'):
        o = ii.get_attribute('style')
        print(o)
    #soup = BeautifulSoup(htlm, 'lxml')
    #im = soup.find_all('a', class_="'page_post_thumb_wrap image_cover  page_post_thumb_last_column page_post_thumb_last_row'")
    #print(htlm)
    #for a in im:
    #   s = a.get('data-src_big').split('|')[0]
    #   tut.append(s)
    #print(tut) 

    #for num, link in enumerate(tut, start=1):
    #   p = requests.get(link)
    #   out = open("img%s.jpg" % (num), 'wb')
    #   out.write(p.content)
    #   out.close()


def main():
    login(driver)
    get_photo_from_page(driver)


if __name__ == '__main__':
    main()
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1 回答 1

2

在这种特殊情况下，您可以只解析您已经能够使用脚本收集的样式字符串。

只需将此函数添加到您的代码中：

def parse_style_attribute(style_string):
    if 'background-image' in style_string:
        style_string = style_string.split(' url("')[1].replace('");', '')
        return style_string
    return None

这是一个简单的字符串解析，如果字符串中有“背景图像”，则提取 url，如果没有图像，则返回 None。

然后，您可以在代码中使用它：

links = list()
for ii in driver.find_elements_by_xpath('//a[@style]'):
    o = ii.get_attribute('style')
    links.append(parse_style_attribute(o))
links = [link for link in links if link is not None]
于 2017-02-12T16:19:42.577 回答