硒与Python 3

Question

尝试利用'def assert_alert_present（）：函数来在警报存在时自动断言。想要在Walmart.com的'注册'按钮上使用它，默认情况下是这样的：

您的密码必须包含6到12个字符，并且没有空格。请再试一次。

我故意使用少于6个或超过12个字符，并出现警报。不确定如何为该警报编写函数，以便在警报存在时通过，如果警报不存在则失败。

Answer 1

希望这是你在找什么：

from selenium import webdriver 
from selenium.webdriver.common.by import By 


def assert_alert_present(): 
    driver = webdriver.Chrome() 
    driver.maximize_window() 
    baseurl = "https://www.walmart.com/account/signup" 
    driver.get(baseurl) 

    driver.find_element_by_name("firstName").send_keys("Vasa") 
    driver.find_element_by_name("lastName").send_keys("Pupkin") 
    driver.find_element_by_name("email").send_keys("abc@123.com") 
    driver.find_element_by_name("password").send_keys("123") 

    #this will check and verify the alertpopup 
    try: 
     assert driver.find_element(By.ID, "password-help") 
     print "Alert is present" 
    except: 
     print "Alert is not present" 

    driver.find_element_by_css_selector("button.l-margin-top").click() 

    errormessage = driver.find_element_by_css_selector('.error-label').text 

    if errormessage.strip() == "Your password must contain between 6 and 12 characters, with no spaces. Please try again.": 
     print "Error lable is present there " 
    else: 
     print "Error lable is not present on website, Please check the website " 

assert_alert_present() 

会打印出消息：

Alert is present 
Error lable is present there