Question

我有一个测试用例，我需要检查点击一个元素时，不显示弹出窗口。这段代码工作正常，但是对于NoSuchElementException，它需要太长时间，60秒，打印PASS条件并转到下一个测试用例。在这种情况下，如何减少等待时间？

driver.find_element_by_xpath(
".//*[@id='assetIdDIV']/div/myaots-input/div/div/div[1]/span/i[2]"
).click()

try:
    DUP_popup = driver.find_element_by_xpath(
    ".//*[@id='DuplicateTicketsPopup']/div/div/div/div[1]/span[2]/img"
    )

    if (DUP_popup):
        print ("Duplicate tkts popup is displayed - Fail")

except NoSuchElementException:
        print ("Duplicate popup not displayed - PASS")

Answer 1

您可以使用显式等待。

默认情况下，WebDriverWait每500次调用一次ExpectedCondition 毫秒，直到它在超时内成功返回，否则抛出TimeoutException

from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC

try:
  DUP_popup = WebDriverWait(driver, 10).until(
    EC.presence_of_element_located((By.XPath, 
      "//*[@id='DuplicateTicketsPopup']/div/div/div/div[1]/span[2]/img"))
  )

  # This waits up to 10 seconds before throwing a TimeoutException 
  # unless it finds the element to return within 10 seconds.

  if (DUP_popup):
    print ("Duplicate tkts popup is displayed - Fail")

except TimeoutException:
  print ("Duplicate popup not displayed - PASS")

有关显式等待和隐式等待的更多详细信息，请参见here

Answer 2

不清楚为什么需要60秒才能提升 NoSuchElementException 并打印传递条件，但您可以通过服务员诱导来配置等待时间 WebDriverWait 如下：

driver.find_element_by_xpath(".//*[@id='assetIdDIV']/div/myaots-input/div/div/div[1]/span/i[2]").click()
if(len(WebDriverWait(driver, 20).until(EC.presence_of_all_elements_located((By.XPATH, ".//*[@id='DuplicateTicketsPopup']/div/div/div/div[1]/span[2]/img"))))!= 0):
    print ("Duplicate tkts popup is displayed - Fail")
else:
    print ("Duplicate popup not displayed - PASS")

注意：如果您的程序包含_implicitly_wait（）_，则需要删除_implicitly_wait（）_的实例，因为文档明确提到Do not mix implicit and explicit waits. Doing so can cause unpredictable wait times.

Answer 3

检查您的驱动程序实例的隐式等待值。如果隐式值设置为60secs，那么它会等到60secs抛出任何与selenium相关的异常。

Selenium-Python - 如何减少NoSuchElementException的时间

3 个答案: