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:硒與Python 3
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。
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:硒與Python 3
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。
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("[email protected]")
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
分享您的代碼,你已經嘗試 – thebadguy
driver.find_element_by_name( 「名字」)。 send_keys( 「瓦薩」) driver.find_element_by_name( 「姓氏」)。send_keys( 「Pupkin」) driver.find_element_by_name( 「電子郵件」)。send_keys( 「[email protected]」) driver.find_element_by_name( 「密碼」 ).send_keys(「VA 。saPupkin1234 「) driver.find_element_by_id(」 註冊提交-BTN「)點擊() #def assert_alert_present(): – Alba
在這個時候,我@ https://www.walmart.com/account/signup – Alba