I do urllib2 and I download the htmlSource of the webpage. How do I make this all on 1 line? (Python)

Question

urlReq = urllib2.Request(theurl)
urlReq.add_header('User-Agent',random.choice(agents))
urlResponse = urllib2.urlopen(urlReq)
htmlSource = urlResponse.read()

How do I make htmlSource in 1 line, instead of many lines?

Answer 1

You can't really do that, the only possible thing is put the response and the source on the same line. Or you could use ; between statements, but that's ugly.

But more importantly, why would you do that? Why is it better to have it all in on line?

>>> import this

The Zen of Python, by Tim Peters

...
Readability counts.
...

Answer 2

How to do that in one line? That's what functions are for. Like this:

def getsource(url):
    urlReq = urllib2.Request(url)
    urlReq.add_header('User-Agent',random.choice(agents))
    urlResponse = urllib2.urlopen(urlReq)
    return urlResponse.read()

Now you can do it in one line:

htmlSource = getsource(theurl)

Done!

Update:

Filtering the htmlSource to be one linebreak (as you now claim you want) is done something like this:

htmlSource = htmlSource.replace('\n', '')

And you might need

htmlSource = htmlSource.replace('\r', '')

as well. I sincerely doubt it will speed anything up.