我正在尝试创建登录测试,因此第一步是,如果用户成功登录脚本,则应在登录后的主页中查找元素。
我的问题是,如果无法登录python的用户抛出NoSuchElementException
异常,而不会转到其他异常。
from selenium import webdriver
from selenium.common.exceptions import NoSuchElementException
def login_test(self):
driver_location = 'D:\\chromedriver.exe'
os.environ["webdriver.chrome.driver"] = driver_location
driver = webdriver.Chrome(driver_location)
driver.maximize_window()
driver.implicitly_wait(3)
driver.get("http://www.example.com/")
prof_icon= driver.find_element_by_xpath("//button[contains(@class,'button')]")
if prof_icon.is_displayed():
print("Success: logged in!")
else:
print("Failure: Unable to login!")
我也尝试过:
prof_icon= driver.find_element_by_xpath("//button[contains(@class,'button')]")
try:
if prof_icon.is_displayed():
print("Success: logged in!")
except NoSuchElementException :
print("Failure: Unable to login")
但是脚本总是崩溃并引发异常。我只需要它来打印消息,以防万一该元素未显示。
答案 0 :(得分:0)
应该是:
except NoSuchElementException: #correct name of exception
print("Failure: Unable to login")
您可以查看元素是否存在,如果不存在,则显示“失败:无法登录”。
注意.find_elements_*
中的复数“ s”。
prof_icon = driver.find_elements_by_xpath("//button[contains(@class,'button')]")
if len(prof_icon) > 0
print("Success: logged in!")
else:
print("Failure: Unable to login")
希望有帮助!
答案 1 :(得分:0)
你很近。要找到要为visibility_of_element_located()
生成 WebDriverWait 的元素,可以使用以下Locator Strategies之一:
使用CSS_SELECTOR
:
try:
WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "button.button")))
print("Success: logged in!")
except TimeoutException:
print("Failure: Unable to login")
使用XPATH
:
try:
WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//button[contains(@class,'button')]")))
print("Success: logged in!")
except TimeoutException:
print("Failure: Unable to login")
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
from selenium.common.exceptions import TimeoutException