Python Web Crawlers and “getting” html source code

萝らか妹 提交于 2019-11-30 01:52:31

Use Python 2.7, is has more 3rd party libs at the moment. (Edit: see below).

I recommend you using the stdlib module urllib2, it will allow you to comfortably get web resources. Example:

import urllib2

response = urllib2.urlopen("http://google.de")
page_source = response.read()

For parsing the code, have a look at BeautifulSoup.

BTW: what exactly do you want to do:

Just for background, I need to download a page and replace any img with ones I have

Edit: It's 2014 now, most of the important libraries have been ported, and you should definitely use Python 3 if you can. python-requests is a very nice high-level library which is easier to use than urllib2.

An Example with python3 and the requests library as mentioned by @leoluk:

pip install requests

Script req.py:

import requests

url='http://localhost'

# in case you need a session
cd = { 'sessionid': '123..'}

r = requests.get(url, cookies=cd)
# or without a session: r = requests.get(url)
r.content

Now,execute it and you will get the html source of localhost!

python3 req.py

The first thing you need to do is read the HTTP spec which will explain what you can expect to receive over the wire. The data returned inside the content will be the "rendered" web page, not the source. The source could be a JSP, a servlet, a CGI script, in short, just about anything, and you have no access to that. You only get the HTML that the server sent you. In the case of a static HTML page, then yes, you will be seeing the "source". But for anything else you see the generated HTML, not the source.

When you say modify the page and return the modified page what do you mean?

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