如何解决无法从谷歌浏览器中抓取链接
我的代码正在打开标签页,搜索主题并关闭,但它没有向我发送它应该收集的链接。
from selenium import webdriver
pesquisa = input ("o que você quer pesquisar: ")
def get_results(search_term):
url = "https://www.startpage.com"
driver = webdriver.Chrome()
driver.get(url)
search_Box = driver.find_element_by_id("q")
search_Box.send_keys(search_term)
search_Box.submit()
try:
links = driver.find_elements_by_xpath("//ol[@class='web_regular_results']//div//a")
except:
links = driver.find_elements_by_xpath("//div//a")
results = []
for link in links:
href = link.get_attribute("href")
print(href)
results.append(href)
driver.close()
return results
get_results(pesquisa)
解决方法
from selenium import webdriver
pesquisa = input ("o que você quer pesquisar: ")
def get_results(search_term):
url = "https://www.startpage.com"
driver = webdriver.Chrome()
driver.get(url)
search_box = driver.find_element_by_id("q")
search_box.send_keys(search_term)
search_box.submit()
try:
links = driver.find_elements_by_xpath("//ol[@class='web_regular_results']//div//a")
except:
links = driver.find_elements_by_xpath("//div//a")
pass # Here you have to write pass to make following statements run
results = []
for link in links:
href = link.get_attribute("href")
print(href)
results.append(href)
driver.close()
return results
get_results(pesquisa)
在try-except中,需要在异常块之后通过才能运行下面的语句。如果没有,它们将被取消。
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。