python - Locate an element using selenium xpath

Question

I am trying to locate an element that has the following line in the chrome inspect code, <href="app/arp/home/profile">.

My line is:

driver.find_element(By.xpath("//a[@href='/app/arp/home/profile']")).click()

But I get the following error:

AttributeError: type object 'By' has no attribute 'xpath'

What is wrong?

Answer 1

Till Selenium v3.141.0 to locate an element using xpath you can use the following syntax:

driver.find_element_by_xpath("//a[@href='/app/arp/home/profile']")

However, in the upcoming releases find_element_by_* commands will be deprecated

def find_element_by_xpath(self, xpath):
    """
    Finds an element by xpath.

    :Args:
     - xpath - The xpath locator of the element to find.

    :Returns:
     - WebElement - the element if it was found

    :Raises:
     - NoSuchElementException - if the element wasn't found

    :Usage:
        ::

            element = driver.find_element_by_xpath('//div/td[1]')
    """
    warnings.warn("find_element_by_* commands are deprecated. Please use find_element() instead")
    return self.find_element(by=By.XPATH, value=xpath)
        

From Selenium v4.x onwards the effective syntax will be:

driver.find_element(By.XPATH, "//a[@href='/app/arp/home/profile']")

An example:

• Code Block:

  from selenium import webdriver
  from selenium.webdriver.common.by import By
  
  driver.get("https://www.google.com/")
  element = driver.find_element(By.NAME, "q")
  print(element)
  driver.quit()
  

• Console Output:

  <selenium.webdriver.remote.webelement.WebElement (session="04a9fac269c3a9cb724cc72769aed4e0", element="1b8ee8d0-b26a-4c67-be10-615286a4d427")>