Selenium with python中的问题（instagram crawler）

Question

我刚刚写了这个instagram爬虫，这是一个大学的小项目。我会告诉你代码并上传一张图片，告诉你我的问题是什么。

from time import sleep
from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains

class App:
    def __init__(self,username="Enter your username here",password="Enter your password here",target_username="shriar.ha"):
        self.username = username
        self.password = password
        self.target_username = target_username
        self.driver = webdriver.Chrome("/Users/Shahriar/Desktop/Selenium and BS projects/chromedriver.exe") #This is the path to webdriver in my PC ,you should change it and give the path of where your webdriver is located.
        self.main_url = "https://www.instagram.com"
        self.driver.get(self.main_url)
        sleep(5)
        self.log_in()
        self.close_notification()
        self.go_to_target_profile()
        sleep(3)
        self.click_on_following()
        self.move_mouse()
        self.scroll_down()
        self.driver.close()

    def move_mouse(self):
        actions = ActionChains(self.driver)
        following_list = self.driver.find_element_by_xpath("//div[@class='isgrP']//div[@role = 'button']")
        actions.move_to_element(following_list).perform()
        sleep(3)

    def scroll_down(self):
        number_of_following = self.driver.find_element_by_xpath("//a[@href='/shriar.ha/following/']/span").get_attribute("innerHTML")
        print(number_of_following)
        number_of_following = int(number_of_following)
        if number_of_following > 7:
            number_of_scrolls = (number_of_following / 7)+3
            for i in range(int(number_of_scrolls)):
                self.driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
                sleep(2)

    def click_on_following(self):
        following_button = self.driver.find_element_by_xpath("//a[@href='/shriar.ha/following/']")
        following_button.click()
        sleep(5)

    def close_notification(self):
        try: 
            sleep(3)
            close_noti_btn = self.driver.find_element_by_xpath("//button[contains(text(),'Not Now')]")
            close_noti_btn.click()
            sleep(2)
        except:
            pass

    def go_to_target_profile(self):
        target_profile_url = self.main_url + "/" + self.target_username + "/"
        self.driver.get(target_profile_url)

    def log_in(self):
        login_button = self.driver.find_element_by_xpath("//a[@href='/accounts/login/?source=auth_switcher']")
        login_button.click()
        sleep(5)
        username_input = self.driver.find_element_by_xpath("//input[@name='username']")
        username_input.send_keys(self.username)
        password_input = self.driver.find_element_by_xpath("//input[@name='password']")
        password_input.send_keys(self.password)
        password_input.submit()

if __name__ == "__main__":
    app = App()

正如您所看到的，它在instagram中登录，然后转到您为程序提供的目标用户名，然后点击以下内容，因此它显示以下列表。这还没有完成，它应该做其他的事情，但现在我陷入了这一步。

我的问题是，当我点击以下。它打开一个小窗口。在那里你可以看到以下列表，我想向下滚动这个列表。见下图：

see this picture

我想向下滚动下面的列表，但我的代码向下滚动主页面，我的意思是背面的页面。我意识到，当我将鼠标光标放在下面的列表上时，我可以用鼠标滚动它，所以我决定编写一个函数将我的鼠标光标放在列表上然后滚动它，但它没有成功。

谁知道我应该怎么做？

谢谢

Answer 1

以下代码适用于我：

def scroll_down(self):
    number_of_following = self.driver.find_element_by_xpath("//a[@href='/shriar.ha/following/']/span").get_attribute("innerHTML")
    print(number_of_following)
    number_of_following = int(number_of_following)
    if number_of_following > 7:
        number_of_scrolls = (number_of_following / 7)+3
        for i in range(int(number_of_scrolls)):
            #scroll by element
            self.driver.execute_script("arguments[0].scrollIntoView(true)",self.driver.find_element_by_xpath("(//div[@role='dialog']//button[text()='Follow'])["+number_of_following+"]"))
            time.sleep(2)

Answer 2

你为什么要打扰移动鼠标并点击？您应该能够使用像requests这样的库来抓取帐户。

或者，有些程序已经在做，你可以从中受到启发。

这里有一些：

• Photo and video crawler
• This one似乎也在追随者

除非你被特别要求，否则我认为这不是一个可行的解决方案，可以移动光标并单击每个链接。