如何使用Selenium,Python从Google搜索中提取链接

from selenium import webdriver
from selenium.common.exceptions import TimeoutException
from selenium.webdriver.support.ui import WebDriverWait # available since 2.4.0
from selenium.webdriver.support import expected_conditions as EC # available since 2.26.0

query = raw_input("What do you wish to search on Wikipedia?n")
query = " " + query

# Create a new instance of the Firefox driver
driver = webdriver.Firefox()

# go to the google home page
driver.get("https://www.google.com/search?q=site%3Awikipedia.com&ie=utf-8&oe=utf-8")

# the page is ajaxy so the title is originally this:
print driver.title

# find the element that's name attribute is q (the google search box)
inputElement = driver.find_element_by_name("q")

# type in the search
inputElement.send_keys(query)

# submit the form (although google automatically searches now without submitting)
inputElement.submit()

try:
    # we have to wait for the page to refresh,the last thing that seems to be updated is the title

    # You should see "cheese! - Google Search"
    print driver.title

    driver.find_element_by_xpath("//h3[contains(text(),'Wikipedia')]").click()

finally:
    driver.quit()

我正在尝试使用Selenium文档中的示例,因此请原谅这些注释,有时还需要不必要的代码.

我遇到问题的代码行是：

driver.find_element_by_xpath("//h3[contains(text(),'Wikipedia')]").click()

我正在尝试做的是获取相关的维基百科链接,或者更具体地说,获取H3’r’路径所指向的链接.

Here’s a picture of a Google page that I’m describing.

在这种情况下,我希望拉出链接http://en.wikipedia.com/wiki/salary

对不起文字的墙,我想尽可能具体.不管怎样,谢谢您的帮助.

最好的祝福！