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

nono = driver.current_url
driver.find_element(By.XPATH,"//button[@value='Send']").click()
  while driver.current_url == nono:
      pass
print("page loaded.")

其他回答

正如David Cullen的回答中提到的,我总是看到这样的建议:

element_present = EC.presence_of_element_located((By.ID, 'element_id'))
WebDriverWait(driver, timeout).until(element_present)

对于我来说,很难找到所有可以与By一起使用的定位器,所以我认为在这里提供列表会很有用。 根据Ryan Mitchell的Web Scraping with Python:

ID Used in the example; finds elements by their HTML id attribute CLASS_NAME Used to find elements by their HTML class attribute. Why is this function CLASS_NAME not simply CLASS? Using the form object.CLASS would create problems for Selenium's Java library, where .class is a reserved method. In order to keep the Selenium syntax consistent between different languages, CLASS_NAME was used instead. CSS_SELECTOR Finds elements by their class, id, or tag name, using the #idName, .className, tagName convention. LINK_TEXT Finds HTML tags by the text they contain. For example, a link that says "Next" can be selected using (By.LINK_TEXT, "Next"). PARTIAL_LINK_TEXT Similar to LINK_TEXT, but matches on a partial string. NAME Finds HTML tags by their name attribute. This is handy for HTML forms. TAG_NAME Finds HTML tags by their tag name. XPATH Uses an XPath expression ... to select matching elements.

如果您试图滚动并找到页面上的所有项目。您可以考虑使用以下方法。这是其他人在这里提到的一些方法的组合。它帮我完成了任务:

while True:
    try:
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
        driver.implicitly_wait(30)
        time.sleep(4)
        elem1 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
        len_elem_1 = len(elem1)
        print(f"A list Length {len_elem_1}")
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
        driver.implicitly_wait(30)
        time.sleep(4)
        elem2 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
        len_elem_2 = len(elem2)
        print(f"B list Length {len_elem_2}")
        if len_elem_1 == len_elem_2:
            print(f"final length = {len_elem_1}")
            break
    except TimeoutException:
            print("Loading took too much time!")

这里我使用了一个相当简单的形式:

from selenium import webdriver
browser = webdriver.Firefox()
browser.get("url")
searchTxt=''
while not searchTxt:
    try:    
      searchTxt=browser.find_element_by_name('NAME OF ELEMENT')
      searchTxt.send_keys("USERNAME")
    except:continue

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

试图将find_element_by_id传递给构造函数的presence_of_element_locate(如已接受的答案所示)会引发NoSuchElementException异常。我不得不在fragles的评论中使用语法:

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

driver = webdriver.Firefox()
driver.get('url')
timeout = 5
try:
    element_present = EC.presence_of_element_located((By.ID, 'element_id'))
    WebDriverWait(driver, timeout).until(element_present)
except TimeoutException:
    print "Timed out waiting for page to load"

这与文档中的示例相匹配。这里是By文档的链接。