[Tutor] syntax error when attempting simple urllib.request.urlopen item

[Dave Angel]

On 07/09/2013 04:00 PM, Paul Smith wrote:
> Tutor-
>
> Ok newbie to coding here attempting to build controlled web scraper and
> have followed several books-tutorials and am failing at step one.
>
> This is the 3.3.1 code I am trying to run..
> ===
> import urllib.request
>
> htmltext = urllib.request.urlopen("http://google.com").read
>
> print htmltext
> ===
> Other version...
> ===
> #try py url open 7
>
> import urllib.request
> res = urllib.request.urlopen('http://python.org/')
> html = res.read()
> print = html
> close res
> input = ("Press enter to exit")
> ===
>
> so when I run what I think is proper 3.3.1 python code it hangs up with
> syntax error with the idle shell red highlighting the last print reference
> i.e. "htmltext"  or "html".
>
> What is this humble newbie not getting?
>

The first thing you're missing is that print() is a function in Python 
3.3  So you need   print(htmltext)

Once you're past that, you'll find that you need parentheses if you want 
to call the read() method.


-- 
DaveA