Python Selenium單擊可見的元素

Question

我正在使用Python Selenium嘗試檢查某個元素是否可見，然后單擊（如果存在）...

# Check to see if element is visible
    myelement = driver.find_element_by_xpath("//a[@id='form1']")

    if myelement.is_displayed():
        print (" ")
    else:
        driver.find_element_by_xpath("//a[@id='form1']").click

這不起作用，我要去哪里錯了？

Answer 1

假設您的xpath正確，則應使用click()而不是click 。 這是一種方法，而不是屬性。

Answer 2

你有兩個問題

• click是一種方法，應該是click()

• 當前，您正在嘗試單擊是否未顯示按鈕。 它應該是

   if myelement.is_displayed(): driver.find_element_by_xpath("//a[@id='form1']").click() else: print (" ") 

您也不必重新定位元素即可單擊它

myelement = driver.find_element_by_xpath("//a[@id='form1']")

if myelement.is_displayed():
    myelement.click()
else:
    print (" ")

Answer 3

最好的方法是創建一個基類，然后重新定義click and find方法，並改用以下方法：

from selenium                                 import webdriver
from selenium.webdriver.support.ui            import WebDriverWait
from selenium.webdriver.support.select        import Select
from selenium.webdriver.support               import expected_conditions as EC
from selenium.webdriver.common.by             import By
from abc                                      import abstractmethod




class LocatorMode:

 XPATH = "xpath"
 CSS_SELECTOR = "cssSelector"
 NAME = "name"
 ID = "id"
 TAG_NAME = "tagName"


class BasePage(object):

 def __init__(self, driver):
     self.driver = driver


 def wait_for_element_visibility(self, waitTime, locatorMode, Locator):
     element = None
     if   locatorMode == LocatorMode.ID:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.visibility_of_element_located((By.ID, Locator)))
     elif locatorMode == LocatorMode.NAME:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.visibility_of_element_located((By.NAME, Locator)))
     elif locatorMode == LocatorMode.XPATH:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.visibility_of_element_located((By.XPATH, Locator)))
     elif locatorMode == LocatorMode.CSS_SELECTOR:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.visibility_of_element_located((By.CSS_SELECTOR, Locator)))
     else:
         raise Exception("Unsupported locator strategy.")
     return element


 def wait_until_element_clickable(self, waitTime, locatorMode, Locator):
     element = None
     if   locatorMode == LocatorMode.ID:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.element_to_be_clickable((By.ID, Locator)))
     elif locatorMode == LocatorMode.NAME:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.element_to_be_clickable((By.NAME, Locator)))
     elif locatorMode == LocatorMode.XPATH:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.element_to_be_clickable((By.XPATH, Locator)))
     elif locatorMode == LocatorMode.CSS_SELECTOR:
          element = WebDriverWait(self.driver, waitTime).\
                  until(EC.element_to_be_clickable((By.CSS_SELECTOR, Locator)))
     else:
         raise Exception("Unsupported locator strategy.")
     return element



 def find_element(self, locatorMode, Locator):
     element = None
     if locatorMode == LocatorMode.ID:
        element = self.driver.find_element_by_id(Locator)
     elif locatorMode == LocatorMode.NAME:
        element = self.driver.find_element_by_name(Locator)
     elif locatorMode == LocatorMode.XPATH:
        element = self.driver.find_element_by_xpath(Locator)
     elif locatorMode == LocatorMode.CSS_SELECTOR:
        element = self.driver.find_element_by_css_selector(Locator)
     else:
        raise Exception("Unsupported locator strategy.")
     return element


 def fill_out_field(self, locatorMode, Locator, text):
     self.find_element(locatorMode, Locator).clear()
     self.find_element(locatorMode, Locator).send_keys(text)

 def click(self, waitTime, locatorMode, Locator):
     self.wait_until_element_clickable(waitTime, locatorMode, Locator).click()

Answer 4

您也可以嘗試/除外：

try:
    driver.find_element_by_xpath("//a[@id='form1']").click()  # will click element if visible

except:
    print "Element not visible."

Answer 5

is_displayed（）！=可見

因此，如果元素不在屏幕上，則.click（）無效，但仍“顯示”

正確的步驟是您必須將元素滾動到屏幕中並且可見，然后單擊（）