向下滚动列表 instagram selenium 和 python

Question

我为 instagram 写了一个脚本。我需要一个方法来返回我的关注者列表。我的追随者不可见（只有 10 个），我必须向下滚动页面。我正在使用 selenium webdriver 和 python 来自动化这个过程。但不幸的是它没有向下滚动。这是我的代码

    def get_followers(self):
    try:
        driver.find_elements_by_css_selector('a._t98z6')[0].click()
    except Exception as e:
        print("Sorry, i don't have access to your followers: {0}".format(e))
    else:
    followers = []
                while True:
            driver.execute_script("window.scrollTo(0, document.body.scrollHeight)")
            try:
                WebDriverWait(driver, 20).until(lambda x: x.find_element_by_css_selector("li._6e4x5"))
            except:
                break
        followers = driver.find_elements_by_css_selector("a._2g7d5.notranslate._o5iw8")

     return followers

任何解决方案将不胜感激。谢谢。

Answer 1

followers_panel = browser.find_element_by_xpath(XPATH)
i = 1
while i < number_of_followers:
    try:
        follower = browser.find_element_by...
        i += 1
    except NoSuchElementException:
        self.browser.execute_script(
            "arguments[0].scrollTop = arguments[0].scrollHeight",followers_panel
        )

这样你就可以抓取所有的关注者了。

Answer 2

我设法使用 ActionChain 滚动滚动。但是随着列表的增长和你电脑+互联网的速度，它变得很慢。起初只有 20-24 个名字，每个卷轴你可以得到 n 10 个名字。然后我习惯在单击最后一个元素时单击最后一个元素，再次单击最后一个元素时会出现 10 个新用户。就这样吧。

from selenium.webdriver.common.action_chains import ActionChains

def get_list():
    element=[]
    ran_num=int(random.randint(0,len(subject)-1))
    search(subject[ran_num])
    br.find_element_by_class_name("_e3il2").click() #open first image
    time.sleep(2)
    br.find_element_by_partial_link_text('likes').click()
    time.sleep(2)

    while len(element)<150:
            element=br.find_elements_by_xpath("//*[@class='_9mmn5']")
            i=len(element)-1
            element[i].click()
            time.sleep(1.50)

    likers=br.find_elements_by_xpath("//*[@class='_2g7d5 notranslate _o5iw8']") #get the username
    for i in range(len(likers)):
            insta_id=likers[i].text
            if (insta_id not in main_list):
                main_list.append(insta_id)
                with open ('to_like.txt','a') as f:
                    f.write('%s\n'%insta_id)

    return()

Answer 3

这是有效的。 不要更改睡眠时间，因为它们允许滚动条重新加载新的关注者，而无需再次将其滚动回顶部。

    FList = driver.find_element_by_css_selector('div[role=\'dialog\'] ul')
    numberOfFollowersInList = len(FList.find_elements_by_css_selector('li'))

    FList.click()
    actionChain = webdriver.ActionChains(driver)
    time.sleep(random.randint(2,4))

    while (numberOfFollowersInList < max):
        actionChain.key_down(Keys.SPACE).key_up(Keys.SPACE).perform()        
        numberOfFollowersInList = len(FList.find_elements_by_css_selector('li'))
        time.sleep(0.4)
        print(numberOfFollowersInList)
        actionChain.key_down(Keys.SPACE).key_up(Keys.SPACE).perform()            
        time.sleep(1)

Answer 4

我也一直在尝试找到一种方法来滚动关注者弹出窗口或对话框，但还没有找到一种方法来关注关注者框以使用我通常在 WebDriver 中使用的滚动功能。我通过在单击关注者链接后简单地发送 ARROW_DOWN 键直到它一直向下滚动来解决这个问题，计数列表和完整列表在最后一个 ARROW_DOWN 键之后保持不变。我相信关于对话框的部分是无关紧要的，但是嗯。这是我的代码：

    def listfollowers (instaURL):
actions = ActionChains(driver)
assert isinstance(instaURL, object)
driver.get(instaURL)
time.sleep(3)  # Let the user actually see something!
followersbutton = wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, "a[href*='followers']")))
followersbutton.click()
time.sleep(2)
dialoguebox = wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, "body > div:nth-child(14) > div > div.zZYga > div > div.j6cq2 > ul > div")))
actions.move_to_element(dialoguebox)
actions.click()
actions.perform()
actions.reset_actions()

###Below 是您列出关注者并滚动关注者对话框所需的内容

followerlist = []
scrollfollowercount = driver.find_elements_by_class_name("UYK0S")
while len(followerlist) < len(scrollfollowercount):
profiles = driver.find_elements_by_class_name("UYK0S") #the followers
    for profile in profiles:
        profileurl = profile.get_attribute('href')
        followerlist.append(profileurl)
        actions.send_keys(Keys.ARROW_DOWN)
        actions.send_keys(Keys.ARROW_DOWN)
        actions.send_keys(Keys.ARROW_DOWN)
        actions.send_keys(Keys.ARROW_DOWN) #included a few times for good measure
        actions.perform()
        actions.reset_actions()
        scrollfollowercount = driver.find_elements_by_class_name("UYK0S")
        if len(scrollfollowercount) == len(followerlist):
            break
print(followerlist)

所以对于我的 MAIN 部分，我有登录位，然后调用我的函数 listfollowers()

    actions = ActionChains(driver)
driver.get("https://www.instagram.com/kapow.fitness/followers")
time.sleep(2)
username = wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, "#react-root > section > main > div > article > div > div:nth-child(1) > div > form > div:nth-child(1) > div")))
username.click()
actions.send_keys("your-username")
actions.perform()
password = wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, "#react-root > section > main > div > article > div > div:nth-child(1) > div > form > div:nth-child(2) > div > div.f0n8F")))
password.click()
actions.reset_actions()
actions.send_keys("your-password")
actions.send_keys(Keys.RETURN)
actions.perform()
time.sleep(2)

listfollowers("https://www.instagram.com/kapow.fitness/")