把BeautifulSoup和Selenium混合在一起对我来说效果很好。

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from bs4 import BeautifulSoup as bs

driver = webdriver.Firefox()
driver.get("http://somedomain/url_that_delays_loading")
    try:
        element = WebDriverWait(driver, 10).until(
        EC.presence_of_element_located((By.ID, "myDynamicElement"))) #waits 10 seconds until element is located. Can have other wait conditions  such as visibility_of_element_located or text_to_be_present_in_element

        html = driver.page_source
        soup = bs(html, "lxml")
        dynamic_text = soup.find_all("p", {"class":"class_name"}) #or other attributes, optional
    else:
        print("Couldnt locate element")

附注:你可以在这里找到更多的等待条件

Selenium是抓取JS和Ajax内容的最佳工具。

查看这篇文章，了解如何使用Python从web中提取数据

$ pip install selenium

然后下载Chrome webdriver。

from selenium import webdriver

browser = webdriver.Chrome()

browser.get("https://www.python.org/")

nav = browser.find_element_by_id("mainnav")

print(nav.text)

容易,对吧?

听起来好像你真正要找的数据可以通过主页面上的一些javascript调用的辅助URL访问。

虽然您可以尝试在服务器上运行javascript来处理这个问题，但一种更简单的方法可能是使用Firefox加载页面，并使用Charles或Firebug之类的工具来准确识别辅助URL。然后，您可以直接查询该URL以获得您感兴趣的数据。

使用PyQt5

from PyQt5.QtWidgets import QApplication
from PyQt5.QtCore import QUrl
from PyQt5.QtWebEngineWidgets import QWebEnginePage
import sys
import bs4 as bs
import urllib.request


class Client(QWebEnginePage):
    def __init__(self,url):
        global app
        self.app = QApplication(sys.argv)
        QWebEnginePage.__init__(self)
        self.html = ""
        self.loadFinished.connect(self.on_load_finished)
        self.load(QUrl(url))
        self.app.exec_()

    def on_load_finished(self):
        self.html = self.toHtml(self.Callable)
        print("Load Finished")

    def Callable(self,data):
        self.html = data
        self.app.quit()

# url = ""
# client_response = Client(url)
# print(client_response.html)

我们没有得到正确的结果，因为任何javascript生成的内容都需要在DOM上呈现。当我们获取一个HTML页面时，我们获取初始的，未经javascript修改的DOM。

因此，我们需要在抓取页面之前呈现javascript内容。

由于selenium已经在本线程中多次提到(有时也提到了它的速度有多慢)，我将列出其他两种可能的解决方案。

解决方案1:这是一个关于如何使用Scrapy抓取javascript生成内容的非常好的教程，我们将遵循这一点。

我们需要:

Docker installed in our machine. This is a plus over other solutions until this point, as it utilizes an OS-independent platform. Install Splash following the instruction listed for our corresponding OS.Quoting from splash documentation: Splash is a javascript rendering service. It’s a lightweight web browser with an HTTP API, implemented in Python 3 using Twisted and QT5. Essentially we are going to use Splash to render Javascript generated content. Run the splash server: sudo docker run -p 8050:8050 scrapinghub/splash. Install the scrapy-splash plugin: pip install scrapy-splash Assuming that we already have a Scrapy project created (if not, let's make one), we will follow the guide and update the settings.py: Then go to your scrapy project’s settings.py and set these middlewares: DOWNLOADER_MIDDLEWARES = { 'scrapy_splash.SplashCookiesMiddleware': 723, 'scrapy_splash.SplashMiddleware': 725, 'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810, } The URL of the Splash server(if you’re using Win or OSX this should be the URL of the docker machine: How to get a Docker container's IP address from the host?): SPLASH_URL = 'http://localhost:8050' And finally you need to set these values too: DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter' HTTPCACHE_STORAGE = 'scrapy_splash.SplashAwareFSCacheStorage' Finally, we can use a SplashRequest: In a normal spider you have Request objects which you can use to open URLs. If the page you want to open contains JS generated data you have to use SplashRequest(or SplashFormRequest) to render the page. Here’s a simple example: class MySpider(scrapy.Spider): name = "jsscraper" start_urls = ["http://quotes.toscrape.com/js/"] def start_requests(self): for url in self.start_urls: yield SplashRequest( url=url, callback=self.parse, endpoint='render.html' ) def parse(self, response): for q in response.css("div.quote"): quote = QuoteItem() quote["author"] = q.css(".author::text").extract_first() quote["quote"] = q.css(".text::text").extract_first() yield quote SplashRequest renders the URL as html and returns the response which you can use in the callback(parse) method.

解决方案2:我们暂且称之为实验性的(2018年5月)…… 此解决方案仅适用于Python版本3.6(目前)。

你知道请求模块吗(谁不知道呢)? 现在它有了一个网络爬行的小兄弟:requests-HTML:

这个库旨在使解析HTML(例如抓取网页)尽可能简单和直观。

安装请求-html: pipenv 对页面的url进行请求: 导入HTMLSession 会话= HTMLSession() R = session.get(a_page_url) 渲染响应以获得Javascript生成的比特: r.html.render ()

最后，该模块似乎提供了抓取功能。 或者，我们也可以尝试使用我们刚刚渲染的r.html对象来使用BeautifulSoup。

简单快捷的解决方案:

我也遇到过同样的问题。我想刮一些数据是用JavaScript构建的。如果我只用BeautifulSoup从这个网站抓取文本，那么我就以文本中的标签结束。 我想渲染这个标签，并将从中抓取信息。 另外，我不想使用像Scrapy和selenium这样的笨重框架。

我发现请求模块的get方法接受url，它实际上呈现脚本标签。

例子:

import requests
custom_User_agent = "Mozilla/5.0 (Windows NT 6.1; Win64; x64; rv:47.0) Gecko/20100101 Firefox/47.0"
url = "https://www.abc.xyz/your/url"
response = requests.get(url, headers={"User-Agent": custom_User_agent})
html_text = response.text

这将呈现加载站点和呈现标签。

希望这将有助于作为快速和简单的解决方案，渲染网站加载脚本标签。