如何在Selenium python中获取类内的href链接

Question

我正在获取该页面的所有链接 我只想要绿色框中带有绿色下划线的帮助类的href链接。黄色下划线是href链接

from selenium import webdriver;
webpage = "https://download.cms.gov/nppes/NPI_Files.html"
driver = webdriver.Chrome("xx\\xx\\xx\\chromedriver.exe")
driver.get(webpage)
elements = driver.find_elements_by_css_selector("li a")
for element in elements:
    print(element.get_attribute("href"))

[Output] :- https://download.cms.gov/nppes/NPPES_Data_Dissemination_September_2020.zip

Answer 1

我将获得pathname属性：

for element in elements:
    print(element.get_attribute("pathname"))

Answer 2

您可以引用标题文本，然后查找下一个元素

element = driver.find_element_by_xpath("//b[text()='Full Replacement Monthly NPI File']/following::a[1]")
print(element.get_attribute("href"))

或

element = driver.find_element_by_xpath("//tr[./td[./b[text()='Full Replacement Monthly NPI File']]]/following-sibling::tr[1]//a")
print(element.get_attribute("href"))

Answer 3

您可以使用xpath：

links = browser.find_elements_by_xpath("//a[@class='className']")

for link in links:

    print(link.get_attribute('href'))

或者，您可以在div>；下选择一个标记；ul>；li标签（或其他内容）：

links = driver.find_elements_by_xpath("//div[@class='className']/ul/li/a")
for link in links:

    print(link.get_attribute('href'))

这项工作：

browser.get("https://download.cms.gov/nppes/NPI_Files.html")
links = browser.find_elements_by_xpath("//div[@class='bulletlistleft']/ul/li/a")
for link in links:
    print(link.get_attribute('href'))