Question

我正在编写一个 python selenium 脚本来尝试在谷歌搜索中提取 LinkedIn 个人资料的 URL 链接，但我在缩小 XPath 以仅返回谷歌上的搜索结果链接时遇到问题。

linkedin_urls = driver.find_elements_by_xpath('//div[@class="yuRUbf"]//a[@href]')
for linkedin_url in linkedin_urls:
    url = linkedin_url.get_attribute("href")
    print(url)

    driver.get(url)
    sleep(5)

linkedin_urls 给我的结果

https://uk.linkedin.com/in/roxana-andreea-popescu
https://uk.linkedin.com/in/tunjijabitta
https://www.google.com/search?source=hp&ei=bxjhX4uGC4_ykgXl9pu4Bw&q=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&oq=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&gs_lcp=CgZwc3ktYWIQDFDMZFjhZmCwZ2gAcAB4AIABLogBsAGSAQE0mAEAoAEBqgEHZ3dzLXdpeg&sclient=psy-ab&ved=0ahUKEwjL-dn4huDtAhUPuaQKHWX7BncQ4dUDCA0#
https://www.google.com/search?q=related:https://uk.linkedin.com/in/tunjijabitta&sa=X&ved=2ahUKEwji3qP_huDtAhWAZxUIHTyfAO4QHzABegQIBhAH
https://uk.linkedin.com/in/janomer
https://uk.linkedin.com/in/josephcoker
https://uk.linkedin.com/in/sebemin
https://uk.linkedin.com/in/vicki-marshall-b7433827
https://www.google.com/search?source=hp&ei=bxjhX4uGC4_ykgXl9pu4Bw&q=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&oq=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&gs_lcp=CgZwc3ktYWIQDFDMZFjhZmCwZ2gAcAB4AIABLogBsAGSAQE0mAEAoAEBqgEHZ3dzLXdpeg&sclient=psy-ab&ved=0ahUKEwjL-dn4huDtAhUPuaQKHWX7BncQ4dUDCA0#
https://www.google.com/search?q=related:https://uk.linkedin.com/in/vicki-marshall-b7433827&sa=X&ved=2ahUKEwji3qP_huDtAhWAZxUIHTyfAO4QHzAFegQIARAH
https://uk.linkedin.com/in/andreibodnar
https://www.google.com/search?q=related:https://uk.linkedin.com/in/andreibodnar&sa=X&ved=2ahUKEwji3qP_huDtAhWAZxUIHTyfAO4QHzAGegQIBxAH
https://uk.linkedin.com/in/dmrlawson
https://uk.linkedin.com/in/jack-gilbert-541a251b
https://www.google.com/search?source=hp&ei=bxjhX4uGC4_ykgXl9pu4Bw&q=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&oq=site%3Alinkedin.com%2Fin%2F+AND+%22Software+Developer%22+AND+%22London%22&gs_lcp=CgZwc3ktYWIQDFDMZFjhZmCwZ2gAcAB4AIABLogBsAGSAQE0mAEAoAEBqgEHZ3dzLXdpeg&sclient=psy-ab&ved=0ahUKEwjL-dn4huDtAhUPuaQKHWX7BncQ4dUDCA0#
https://www.google.com/search?q=related:https://uk.linkedin.com/in/jack-gilbert-541a251b&sa=X&ved=2ahUKEwji3qP_huDtAhWAZxUIHTyfAO4QHzAIegQICxAH
https://uk.linkedin.com/in/eren-batu-999068185

我试图找到一种方法将搜索范围缩小到仅 LinkedIn 结果

Answer 1

如果您只想获得 LinkedIn 结果，请在 xpath 下面使用。

使用contains()

linkedin_urls = driver.find_elements_by_xpath('//div[@class="yuRUbf"]//a[contains(@href,"https://uk.linkedin.com")]')

或starts-with()

使用

linkedin_urls = driver.find_elements_by_xpath('//div[@class="yuRUbf"]//a[starts-with(@href,"https://uk.linkedin.com")]')

Answer 2

您想解析 linkedin_url 中的每个字符串，看看它是否提到了 Linkedin。

    if 'linkedin' in linkedin_url:
        print('linkedin')

基本上，把你想在Linkedin地址上执行的驱动代码放在if语句下。

Answer 3

要将搜索限制为仅 LinkedIn 结果，您需要为 visibility_of_all_elements_located() 诱导 WebDriverWait，您可以使用以下任一 Locator Strategies：

使用 CSS_SELECTOR：

print([my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "div.yuRUbf a[href^='https://uk.linkedin.com/in']")))])

使用 XPATH

print([my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//div[@class="yuRUbf"]//a[starts-with(@href, 'https://uk.linkedin.com/in')]")))])

注意：您必须添加以下导入：

from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

检索谷歌搜索链接的 Xpath

3 个答案: