如何在 python 中将 JavaScript HTML 渲染为 HTML？
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How can I render JavaScript HTML to HTML in python?
提问by user3928006
I have looked around and only found solutions that render a URL to HTML. However I need a way to be able to render a webpage (That I already have, and that has JavaScript) to proper HTML.
我环顾四周，只找到了将 URL 呈现为 HTML 的解决方案。但是，我需要一种方法来将网页（我已经拥有，并且具有 JavaScript）呈现为正确的 HTML。
Want: Webpage (with JavaScript) ---> HTML
想要：网页（使用 JavaScript）---> HTML
Not: URL --> Webpage (with JavaScript) ---> HTML
不是： URL --> 网页（使用 JavaScript） ---> HTML
I couldn't figure out how to make the other code work the way I wanted.
我不知道如何让其他代码按照我想要的方式工作。
This is the code I was using that renders URLs: http://webscraping.com/blog/Scraping-JavaScript-webpages-with-webkit/
这是我使用的呈现 URL 的代码：http: //webscraping.com/blog/Scraping-JavaScript-webpages-with-webkit/
For clarity, the code above takes a URL of a webpage that has some parts of the page rendered by JavaScript, so if I scrape the page normally using say urllib2 then I won't get all the links etc that are rendered as after the JavaScript.
为清楚起见，上面的代码采用网页的 URL，该网页的某些部分由 JavaScript 呈现，因此如果我通常使用 urllib2 抓取页面，那么我将不会获得在 JavaScript 之后呈现的所有链接等.
However I want to be able to scrape a page, say again with urllib2, and then render that page and get the outcome HTML. (Different to the above code since it takes a URL as it's argument.
但是我希望能够抓取一个页面，用 urllib2 再说一遍，然后呈现该页面并获得结果 HTML。（与上面的代码不同，因为它需要一个 URL 作为它的参数。
Any help is appreciated, thanks guys :)
任何帮助表示赞赏，谢谢大家:)
回答by barak manos
You can pip install seleniumfrom a command line, and then run something like:
您可以pip install selenium从命令行，然后运行类似：
from selenium import webdriver
from urllib2 import urlopen
url = 'http://www.google.com'
file_name = 'C:/Users/Desktop/test.txt'
conn = urlopen(url)
data = conn.read()
conn.close()
file = open(file_name,'wt')
file.write(data)
file.close()
browser = webdriver.Firefox()
browser.get('file:///'+file_name)
html = browser.page_source
browser.quit()
回答by peter wambua
try webdriver.Firefox().get('url')
试试 webdriver.Firefox().get('url')
