我想刮取无限滚动实现的页面的所有数据。下面的python代码可以工作。

for i in range(100):
    driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
    time.sleep(5)

这意味着每当我向下滚动到底部时,我都需要等待5秒,这通常足以让页面完成加载新生成的内容。但是,这可能并不省时。页面可能在5秒内完成新内容的加载。如何在每次向下滚动时检测页面是否完成了新内容的加载?如果我能检测到这一点,一旦我知道页面完成加载,我就可以再次向下滚动以查看更多内容。这样更节省时间。


当前回答

回答得很好。等待XPATH的快速示例。

# wait for sizes to load - 2s timeout
try:
    WebDriverWait(driver, 2).until(expected_conditions.presence_of_element_located(
        (By.XPATH, "//div[@id='stockSizes']//a")))
except TimeoutException:
    pass

其他回答

回答得很好。等待XPATH的快速示例。

# wait for sizes to load - 2s timeout
try:
    WebDriverWait(driver, 2).until(expected_conditions.presence_of_element_located(
        (By.XPATH, "//div[@id='stockSizes']//a")))
except TimeoutException:
    pass

把WebDriverWait放在While循环中并捕获异常如何?

from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.common.exceptions import TimeoutException

browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
while True:
    try:
        WebDriverWait(browser, delay).until(EC.presence_of_element_located(browser.find_element_by_id('IdOfMyElement')))
        print "Page is ready!"
        break # it will break from the loop once the specific element will be present. 
    except TimeoutException:
        print "Loading took too much time!-Try again"

webdriver将在默认情况下通过.get()方法等待页面加载。

正如你可能正在寻找一些特定的元素@user227215所说的,你应该使用WebDriverWait来等待位于你页面中的元素:

from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from selenium.common.exceptions import TimeoutException

browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
try:
    myElem = WebDriverWait(browser, delay).until(EC.presence_of_element_located((By.ID, 'IdOfMyElement')))
    print "Page is ready!"
except TimeoutException:
    print "Loading took too much time!"

我用它来检查提醒。您可以使用任何其他类型方法来查找定位器。

编辑1:

I should mention that the webdriver will wait for a page to load by default. It does not wait for loading inside frames or for ajax requests. It means when you use .get('url'), your browser will wait until the page is completely loaded and then go to the next command in the code. But when you are posting an ajax request, webdriver does not wait and it's your responsibility to wait an appropriate amount of time for the page or a part of page to load; so there is a module named expected_conditions.

ajax页面连续加载数据的解决方案。所述的预览方法无效。我们可以做的是抓取页面dom并对其进行哈希,并在一段时间内比较新旧哈希值。

import time
from selenium import webdriver

def page_has_loaded(driver, sleep_time = 2):
    '''
    Waits for page to completely load by comparing current page hash values.
    '''

    def get_page_hash(driver):
        '''
        Returns html dom hash
        '''
        # can find element by either 'html' tag or by the html 'root' id
        dom = driver.find_element_by_tag_name('html').get_attribute('innerHTML')
        # dom = driver.find_element_by_id('root').get_attribute('innerHTML')
        dom_hash = hash(dom.encode('utf-8'))
        return dom_hash

    page_hash = 'empty'
    page_hash_new = ''
    
    # comparing old and new page DOM hash together to verify the page is fully loaded
    while page_hash != page_hash_new: 
        page_hash = get_page_hash(driver)
        time.sleep(sleep_time)
        page_hash_new = get_page_hash(driver)
        print('<page_has_loaded> - page not loaded')

    print('<page_has_loaded> - page loaded: {}'.format(driver.current_url))

另外,您可以检查DOM是否没有更多的修改,而不是向下滚动100次(在页面底部是AJAX惰性加载的情况下)

def scrollDown(driver, value):
    driver.execute_script("window.scrollBy(0,"+str(value)+")")

# Scroll down the page
def scrollDownAllTheWay(driver):
    old_page = driver.page_source
    while True:
        logging.debug("Scrolling loop")
        for i in range(2):
            scrollDown(driver, 500)
            time.sleep(2)
        new_page = driver.page_source
        if new_page != old_page:
            old_page = new_page
        else:
            break
    return True