我目前使用硒webdriver解析通过facebook用户的朋友页面,并从AJAX脚本提取所有id。但我需要向下滚动来找到所有的朋友。如何向下滚动硒。我正在使用python。


当前回答

你可以使用send_keys来模拟一个END(或PAGE_DOWN)键按下(通常滚动页面):

from selenium.webdriver.common.keys import Keys
html = driver.find_element_by_tag_name('html')
html.send_keys(Keys.END)

其他回答

你可以使用send_keys来模拟一个END(或PAGE_DOWN)键按下(通常滚动页面):

from selenium.webdriver.common.keys import Keys
html = driver.find_element_by_tag_name('html')
html.send_keys(Keys.END)

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

方法如下图所示:

在python中,你可以使用

driver.execute_script("window.scrollTo(0, Y)")

(Y为要滚动到的垂直位置)

我正在寻找一种滚动浏览动态网页的方法,并在到达页面末尾时自动停止,并找到了这个线程。

@Cuong Tran的这篇文章,有一个主要的修改,是我一直在寻找的答案。我认为其他人可能会发现这个修改很有帮助(它对代码的工作方式有明显的影响),因此写了这篇文章。

修改是将捕获最后一页高度的语句移动到循环内部(以便每个检查都与前一页高度进行比较)。

所以,下面的代码:

连续向下滚动一个动态网页(. scrollto()),只有在一次迭代中,页面高度保持相同时才停止。

(还有另一个修改,其中break语句在另一个条件(以防页面“卡住”)中,可以删除)。

    SCROLL_PAUSE_TIME = 0.5


    while True:

        # Get scroll height
        ### This is the difference. Moving this *inside* the loop
        ### means that it checks if scrollTo is still scrolling 
        last_height = driver.execute_script("return document.body.scrollHeight")

        # Scroll down to bottom
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")

        # Wait to load page
        time.sleep(SCROLL_PAUSE_TIME)

        # Calculate new scroll height and compare with last scroll height
        new_height = driver.execute_script("return document.body.scrollHeight")
        if new_height == last_height:

            # try again (can be removed)
            driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")

            # Wait to load page
            time.sleep(SCROLL_PAUSE_TIME)

            # Calculate new scroll height and compare with last scroll height
            new_height = driver.execute_script("return document.body.scrollHeight")

            # check if the page height has remained the same
            if new_height == last_height:
                # if so, you are done
                break
            # if not, move on to the next loop
            else:
                last_height = new_height
                continue

下面是我编写的一个缓慢向下滚动到targets元素的方法

你可以将CSS选择器中元素的y号位置传递给它

它就像我们通过鼠标滚轮一样滚动

一旦这个方法被调用,你用相同的驱动对象再次调用它,但是使用新的目标元素,它将在元素存在的任何地方向上/向下滚动

def slow_scroll_to_element(self, driver, element_selector=None, target_yth_location=None):
    current_scroll_position = int(driver.execute_script("return window.scrollY"))
    
    if element_selector:
        target_yth_location = int(driver.execute_script("return document.querySelector('{}').getBoundingClientRect()['top'] + window.scrollY".format(element_selector)))
    
    scrollSpeed = 100 if target_yth_location-current_scroll_position > 0 else -100

    def chunks(a, n):
        k, m = divmod(len(a), n)
        return (a[i*k+min(i, m):(i+1)*k+min(i+1, m)] for i in range(n))
    
    for l in list(chunks(list(range(current_scroll_position, target_yth_location, scrollSpeed)) + list([target_yth_location+(-scrollSpeed if scrollSpeed > 0 else scrollSpeed)]), 3)):
        for pos in l:
            driver.execute_script("window.scrollTo(0, "+str(pos)+");")
            time.sleep(0.1)
        time.sleep(random.randint(1,3))