Python console rejects an object reference, having made an object with that reference as its name in previous line

[Steven D'Aprano]

Simon Evans wrote:

> Dear Jussi, and Billy
> I have changed the input in accordance with your advice, re:
> ------------------------------------------------------------------------------
> Python 2.7.6 (default, Nov 10 2013, 19:24:18) [MSC v.1500 32 bit (Intel)]
> on win 32
> Type "help", "copyright", "credits" or "license" for more information.
>>>> from bs4 import BeautifulSoup
>>>> with open("ecologicalpyramid.html","r") as ecological_pyramid:
> ...      soup = next(ecological_pyramid,"lxml")
> ...      producer_entries = soup.find("ul")
> ...      print(producer_entries.li.div.string)
> ... print(producer_entries.li.div.string)


Simon, you are not leaving a blank space between the indented block and the
unindented line.

The interactive interpreter is quite simple, and you MUST leave a blank line
after indented blocks. You need to press the ENTER key *twice* at the end
of the block:

with open("ecologicalpyramid.html","r") as ecological_pyramid: ENTER
TAB soup = next(ecological_pyramid,"lxml") ENTER
TAB producer_entries = soup.find("ul") ENTER
TAB print(producer_entries.li.div.string) ENTER
ENTER
print(producer_entries.li.div.string) ENTER


Don't type the letters E N T E R, press the ENTER key. The same goes for
TAB. Note carefully that after the indented block, there is a blank line.
It should look like this:


>>> with open("ecologicalpyramid.html","r") as ecological_pyramid:
...      soup = next(ecological_pyramid,"lxml")
...      producer_entries = soup.find("ul")
...      print(producer_entries.li.div.string)
... 
>>> print(producer_entries.li.div.string)

If you don't leave the blank line, then *nothing* in that "with" block will
run and consequently the last print line will fail.

Until you fix that, we cannot even begin to help with any other issues you
might be having.



-- 
Steven