如何使用if-elif-else检查元素是否存在

Question

我有一个下拉菜单，我需要检查它的条目。如果没有条目，我可以添加新条目；如果只有一个条目，我将其删除并添加新条目；当我有多个条目（>；=2）时，我无法继续添加条目。我可以通过person_rem_btn查看。如果我只有一个按钮doc_person_table:0:person_rem_btn我可以继续，如果我有第二个按钮doc_person_table:1:person_rem_btn我不能继续

我得到一个例外：

selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"xpath","selector":"//*[@id='frmMain:doc_person_table:1:person_rem_btn']/span[1]"}

然而，这就是重点，这个元素可能不可用，我只是检查它的存在。我将感谢任何帮助。这是我的密码：

if driver.find_element_by_xpath("//*[@id='frmMain:doc_person_table:1:person_rem_btn']/span[1]") == True:
   print ("there are already many entries")
   driver.close()
elif (driver.find_element_by_xpath("//*[@id='frmMain:doc_person_table:1:person_rem_btn']/span[1]") == False and driver.find_element_by_xpath("//*[@id='frmMain:doc_person_table:0:person_rem_btn']/span[1]") == True):
   print ("there is only one entry, it will be removed to proceed")
   WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, "//*[@id='frmMain:doc_person_table:0:person_rem_btn']/ span[1]"))).click()
else:
    print ("there is no entry, you can proceed")

Answer 1

您可以采用几种方法，其中一种方法是为visibility_of_all_elements_located()创建一个下拉项的列表，并探测列表大小：

• 使用CSS_SELECTOR：

  list = WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "[id*='doc_person_table'][id$='person_rem_btn']>span")))
  if not list:
      print ("there is no entry, you can proceed")
      # other steps
  elif len(list) == 1
      print ("there is only one entry, it will be removed to proceed")
      # other steps
  else:
      print ("there are already many entries")
      break
  

• 使用XPATH：

  list = WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//*[contains(@id, 'doc_person_table') and contains(@id, 'person_rem_btn')]/span")))
  if not list:
      print ("there is no entry, you can proceed")
      # other steps
  elif len(list) == 1
      print ("there is only one entry, it will be removed to proceed")
      # other steps
  else:
      print ("there are already many entries")
      break