硒與Python 3

Question

嘗試利用'def assert_alert_present（）：函數來在警報存在時自動斷言。想要在Walmart.com的'註冊'按鈕上使用它，默認情況下是這樣的：

您的密碼必須包含6到12個字符，並且沒有空格。請再試一次。

我故意使用少於6個或超過12個字符，並出現警報。不確定如何爲該警報編寫函數，以便在警報存在時通過，如果警報不存在則失敗。

Answer 1

希望這是你在找什麼：

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() 

會打印出消息：

Alert is present 
Error lable is present there