使用Selenium缓慢向下滚动页面

Question

我正试图从航班搜索页面中获取一些数据。

此页面的工作方式如下：

您填写表格，然后单击“搜索”按钮-这是确定的。当你点击按钮时，你会被重定向到有结果的页面，问题就在这里。这个页面不断添加结果，例如一分钟，这不是什么大问题-问题是要获得所有这些结果。当您在real browser中时，您必须向下滚动页面并显示这些结果。所以我尝试使用Selenium向下滚动。它在页面底部向下滚动的速度可能非常快，或者它是一个跳转而不是滚动，因此页面不会加载任何新结果。

当你慢慢向下滚动时，它会重新加载结果，但是如果你非常快地滚动，它会停止加载。

我不确定我的代码是否有助于理解这一点，因此我附加了它。

SEARCH_STRING = """URL"""

class spider():

    def __init__(self):
        self.driver = webdriver.Firefox()

    @staticmethod
    def prepare_get(dep_airport,arr_airport,dep_date,arr_date):
        string = SEARCH_STRING%(dep_airport,arr_airport,arr_airport,dep_airport,dep_date,arr_date)
        return string


    def find_flights_html(self,dep_airport, arr_airport, dep_date, arr_date):
        if isinstance(dep_airport, list):
            airports_string = str(r'%20').join(dep_airport)
            dep_airport = airports_string

        wait = WebDriverWait(self.driver, 60) # wait for results
        self.driver.get(spider.prepare_get(dep_airport, arr_airport, dep_date, arr_date))
        wait.until(EC.invisibility_of_element_located((By.XPATH, '//img[contains(@src, "loading")]')))
        wait.until(EC.invisibility_of_element_located((By.XPATH, u'//div[. = "Poprosíme o trpezlivosť, hľadáme pre Vás ešte viac letov"]/preceding-sibling::img')))
        self.driver.execute_script("window.scrollTo(0,document.body.scrollHeight);")

        self.driver.find_element_by_xpath('//body').send_keys(Keys.CONTROL+Keys.END)
        return self.driver.page_source

    @staticmethod 
    def get_info_from_borderbox(div):
        arrival = div.find('div',class_='departure').text
        price = div.find('div',class_='pricebox').find('div',class_=re.compile('price'))
        departure = div.find_all('div',class_='departure')[1].contents
        date_departure = departure[1].text 
        airport_departure = departure[5].text
        arrival = div.find_all('div', class_= 'arrival')[0].contents
        date_arrival = arrival[1].text
        airport_arrival = arrival[3].text[1:]
        print 'DEPARTURE: ' 
        print date_departure,airport_departure
        print 'ARRIVAL: '
        print date_arrival,airport_arrival

    @staticmethod
    def get_flights_from_result_page(html):

        def match_tag(tag, classes):
            return (tag.name == 'div'
                    and 'class' in tag.attrs
                    and all([c in tag['class'] for c in classes]))

        soup = mLib.getSoup_html(html)
        divs = soup.find_all(lambda t: match_tag(t, ['borderbox', 'flightbox', 'p2']))

        for div in divs:
            spider.get_info_from_borderbox(div)

        print len(divs)


spider_inst = spider() 

print spider.get_flights_from_result_page(spider_inst.find_flights_html(['BTS','BRU','PAR'], 'MAD', '2015-07-15', '2015-08-15'))

所以在我看来，主要的问题是它的滚动速度太快，无法触发新的结果加载。

你知道如何让它工作吗？

Answer 1

这里有一种对我有用的不同方法，包括滚动到最后一个搜索结果的视图中，并在再次滚动之前等待加载其他元素：

# -*- coding: utf-8 -*-
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.common.exceptions import StaleElementReferenceException
from selenium.webdriver.support import expected_conditions as EC


class wait_for_more_than_n_elements(object):
    def __init__(self, locator, count):
        self.locator = locator
        self.count = count

    def __call__(self, driver):
        try:
            count = len(EC._find_elements(driver, self.locator))
            return count >= self.count
        except StaleElementReferenceException:
            return False


driver = webdriver.Firefox()

dep_airport = ['BTS', 'BRU', 'PAR']
arr_airport = 'MAD'
dep_date = '2015-07-15'
arr_date = '2015-08-15'

airports_string = str(r'%20').join(dep_airport)
dep_airport = airports_string

url = "https://www.pelikan.sk/sk/flights/list?dfc=C%s&dtc=C%s&rfc=C%s&rtc=C%s&dd=%s&rd=%s&px=1000&ns=0&prc=&rng=1&rbd=0&ct=0" % (dep_airport, arr_airport, arr_airport, dep_airport, dep_date, arr_date)
driver.maximize_window()
driver.get(url)

wait = WebDriverWait(driver, 60)
wait.until(EC.invisibility_of_element_located((By.XPATH, '//img[contains(@src, "loading")]')))
wait.until(EC.invisibility_of_element_located((By.XPATH,
                                               u'//div[. = "Poprosíme o trpezlivosť, hľadáme pre Vás ešte viac letov"]/preceding-sibling::img')))

while True:  # TODO: make the endless loop end
    results = driver.find_elements_by_css_selector("div.flightbox")
    print "Results count: %d" % len(results)

    # scroll to the last element
    driver.execute_script("arguments[0].scrollIntoView();", results[-1])

    # wait for more results to load
    wait.until(wait_for_more_than_n_elements((By.CSS_SELECTOR, 'div.flightbox'), len(results)))

注意事项：

• 您需要确定何时停止循环-例如，在特定的len（results）value
• wait\u for\u more\u\n\u元素是一种自定义的预期条件，有助于确定何时加载下一部分，我们可以再次滚动

Answer 2

经过一些实验，我终于找到了一个很好的解决方案：

    def __scroll_down_page(self, speed=8):
    current_scroll_position, new_height= 0, 1
    while current_scroll_position <= new_height:
        current_scroll_position += speed
        self.__driver.execute_script("window.scrollTo(0, {});".format(current_scroll_position))
        new_height = self.__driver.execute_script("return document.body.scrollHeight")

Answer 3

同样的问题我需要它我需要刮一个社交媒体网站

y = 1000
    for timer in range(0,50):
         driver.execute_script("window.scrollTo(0, "+str(y)+")")
         y += 1000  
         time.sleep(1)

每1000次睡眠允许加载