我找不到要单击的按钮元素

Question

我试图点击“查看所有细节”按钮，从OpenTable中展开一家餐厅的细节，但我一直得到一个无元素的例外。

from selenium import webdriver

driver = webdriver.Chrome(
    '/Library/Python/2.7/site-packages/chromedriver')
url = "https://www.opentable.com/chicago-illinois-restaurant-listings"
driver.get(url)

element = driver.find_element_by_xpath(
    '//*[@id="search_results"]/div[2]/div[1]/div/div[2]/div[1]/a')
element.click()
driver.find_element_by_css_selector(
    '#overview-section > div:nth-child(4) > div.f9f46391 > button').click()

driver.quit()

Answer 1

每个结果链接都有target='_blank'属性。这意味着，如果要单击链接详细信息页面将在新选项卡中打开。要处理“新建”选项卡上的元素，应切换到该选项卡：

driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])

请注意，您还应等待按钮变为可单击：

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

wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()