I'm trying to make an RSS scraper to pull articles from RSS feeds and export
them to text files, to be used either in my BBS bulletins or message areas

I found a web page that supposedly does exactly this, however using the code
and testing I get the Folliwing error:


python scraping.py
File "scraping.py", line 5
def
hackernews_rss('https://news.ycombinator.com/rss'):
^
SyntaxError: invalid
syntax



The code is:

# scraping function
def hackernews_rss('https://news.ycombinator.com/rss'):
try:
r = requests.get(
return print('The scraping job succeeded: ', r.status_code)
except Exception as e:
print('The scraping job failed. See exception: '
print(e)print('Starting scraping'
hackernews_rss(
print('Finished scraping'


Any idea what I flubbed?

Thanks

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Hi Charles,

On 2021-02-05 14:35:27, you wrote to All:

CP> I'm trying to make an RSS scraper to pull articles from RSS feeds and
CP> export them to text files, to be used either in my BBS bulletins or
CP> message areas

CP> I found a web page that supposedly does exactly this, however using the
CP> code and testing I get the Folliwing error:


CP> python scraping.py
CP> File "scraping.py", line 5
CP> def
CP> hackernews_rss('https://news.ycombinator.com/rss'):

CP> ^
CP> SyntaxError: invalid
CP> syntax



CP> The code is:

CP> # scraping function
CP> def hackernews_rss('https://news.ycombinator.com/rss'):

There should be a variable there between the (, not a constant in this
function definition.

CP> try:
CP> r = requests.get(

Is 'requests' defined of set somewhere? I don't see it.

CP> return print('The scraping job succeeded: ', r.status_code)
CP> except Exception as e:
CP> print('The scraping job failed. See exception: '
CP> print(e)print('Starting scraping'

CP> hackernews_rss(

Here where you call it you can give the constant to the function, like:

hackernews_rss('https://news.ycombinator.com/rss'

CP> print('Finished scraping'


CP> Any idea what I flubbed?


Bye, Wilfred.

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