我发现解决这个问题的最简单的方法是选择一个标签，然后发送:

label.sendKeys(Keys.PAGE_DOWN);

希望它有用!

滚动加载页面。例如:medium, quora等

last_height = driver.execute_script("return document.body.scrollHeight")
    while True:
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight-1000);")
        # Wait to load the page.
        driver.implicitly_wait(30) # seconds
        new_height = driver.execute_script("return document.body.scrollHeight")
    
        if new_height == last_height:
            break
        last_height = new_height
        # sleep for 30s
        driver.implicitly_wait(30) # seconds
    driver.quit()

这段代码滚动到底部，但不需要每次都等待。它会不断滚动，然后在底部停止(或超时)

from selenium import webdriver
import time

driver = webdriver.Chrome(executable_path='chromedriver.exe')
driver.get('https://example.com')

pre_scroll_height = driver.execute_script('return document.body.scrollHeight;')
run_time, max_run_time = 0, 1
while True:
    iteration_start = time.time()
    # Scroll webpage, the 100 allows for a more 'aggressive' scroll
    driver.execute_script('window.scrollTo(0, 100*document.body.scrollHeight);')

    post_scroll_height = driver.execute_script('return document.body.scrollHeight;')

    scrolled = post_scroll_height != pre_scroll_height
    timed_out = run_time >= max_run_time

    if scrolled:
        run_time = 0
        pre_scroll_height = post_scroll_height
    elif not scrolled and not timed_out:
        run_time += time.time() - iteration_start
    elif not scrolled and timed_out:
        break

# closing the driver is optional 
driver.close()

这比每次等待0.5-3秒的响应要快得多，因为每次响应可能需要0.1秒

下面是一个示例selenium代码片段，您可以将其用于这种类型的目的。它会转到youtube搜索结果的url“Enumerate python tutorial”，然后向下滚动，直到找到标题为“Enumerate python tutorial(2020)”的视频。

driver.get('https://www.youtube.com/results?search_query=enumerate+python')
target = driver.find_element_by_link_text('Enumerate python tutorial(2020).')
target.location_once_scrolled_into_view

ScrollTo()函数不再工作。这是我用过的，效果很好。

driver.execute_script("document.getElementById('mydiv').scrollIntoView();")

element=find_element_by_xpath("xpath of the li you are trying to access")

element.location_once_scrolled_into_view

当我试图进入一个不可见的“li”时，这很有帮助。