Question

使用Python Selenium我试图检查一个元素是否可见，然后点击它，如果它... ...

# 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

这不起作用，我哪里错了？

Answer 1

假设您的xpath是正确的，您应该使用click()，而不是click。这是一种方法，而不是属性。

Answer 2

你有两个问题

click是一种方法，它应该是click()

目前，您正在尝试点击按钮不。它应该是

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以点击它

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

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

Answer 3

最好的方法是创建一个基类并重新定义click and find方法并改为使用它：

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

您也可以尝试/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（）！=可见

因此，如果元素不在屏幕上，但仍然显示＆＃34;

，则.click（）不起作用

正确的步骤是您必须将元素滚动到屏幕并可见，然后单击（）

如果可见，Python Selenium会单击一个元素

5 个答案: