[英]get google links search results per search in selenium python
Dears, i am trying to get the links of google search results, i was using this profiles = driver.find_elements_by_xpath('//*[@class="r"]/a[1]') this was working correctly, i am using it now, but not giving me results亲爱的,我正在尝试获取谷歌搜索结果的链接,我正在使用这个配置文件 = driver.find_elements_by_xpath('//*[@class="r"]/a[1]')这工作正常,我是现在使用它,但没有给我结果
import csv
from parsel import Selector
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import time
import itertools
import random
listlong = []
driver = webdriver.Chrome('C:/chromedriver.exe')
driver.get('https://www.google.com/search?q=&num=200&start=0&sourceid=chrome')
time.sleep(random.randint(10,11))
search_input = driver.find_element_by_name('q')
search_input.send_keys('site:linkedin.com/in/ AND USA')
time.sleep(3)
search_input.send_keys(Keys.RETURN)
time.sleep(3)
# grab all linkedin profiles from first page at Google
profiles = driver.find_elements_by_xpath('//*[@class="r"]/a[1]')
time.sleep(8)
profiles = [profile.get_attribute('href') for profile in profiles]
listprof = profiles.copy()
listlong.extend(listprof)
print(profiles)
print("f1")
print(listprof)
print("f2")
print(listlong)
print("f3")
driver.quit()
what is the xpath can be used instead of this driver.find_elements_by_xpath('//*[@class="r"]/a[1]')什么是 xpath 可以用来代替这个 driver.find_elements_by_xpath('//*[@class="r"]/a[1]')
based on MBaas answer i tried this.基于 MBaas 的回答,我尝试了这个。 I added it so that will first find the search result, then find based on css selector ("div > div > div.yuRUbf > a").
我添加它是为了首先找到搜索结果,然后根据 css 选择器(“div > div > div.yuRUbf > a”)进行查找。
# grab all linkedin profiles from first page at Google
search_result = driver.find_elements_by_css_selector('div.g')
profiles = driver.find_elements_by_css_selector("div > div > div.yuRUbf > a")
time.sleep(8)
profiles = [profile.get_attribute('href') for profile in profiles]
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