54

is_visible在 Selenium 文档中找到了该方法,但我不知道如何使用它。我不断收到诸如is_visible needs a selenium instance as the first parameter.

另外,什么是“定位器”?

任何帮助,将不胜感激。

4

2 回答 2

122

您应该is_displayed()改用:

from selenium import webdriver

driver = webdriver.Firefox()
driver.get('http://www.google.com')
element = driver.find_element_by_id('gbqfba') #this element is visible
if element.is_displayed():
  print "Element found"
else:
  print "Element not found"

hidden_element = driver.find_element_by_name('oq') #this one is not
if hidden_element.is_displayed():
  print "Element found"
else:
  print "Element not found"
于 2013-04-11T04:30:35.260 回答
0

从这里:https ://www.selenium.dev/selenium/docs/api/py/webdriver_support/selenium.webdriver.support.expected_conditions.html

只需使用

selenium.webdriver.support.expected_conditions.visibility_of

可见性是指元素不仅显示出来,而且高度和宽度都大于0

于 2020-11-27T09:18:05.273 回答