Find today's RSS entries with feedparser

[Biks]

I'm trying to create the worlds most basic RSS reader - I just want to know the number (len) of entries that were only created TODAY. I don't care if anything older was modified on today, just entries created.

Should I even be using feedparser do this? It seems I could just findAll <datepub> tags with BeautifulSoup and match it today's date, but everyone insists I should be using feedparser. Note: I'm still new at this, so it's hours of struggling either way.

Here's where I'm at - this shows that I have 50 entries:

1 2 3

import feedparser dgtw = feedparser.parse('https://investorshub.advfn.com/boards/rss.aspx?board_id=22658') print (len(dgtw['entries']))

This just shows the published date of the first entry:

1	print(dgtw.entries[0].published)

I just want to findAll published dates that match today's date and give me a len number/return.

I don't see anything in the docs about this specifically: https://pythonhosted.org/feedparser/date-parsing.html

[snippsat]

It seems I could just findAll <datepub> tags with BeautifulSoup and match it today's date

You could to that.
Now will task work fine from feedparser to.
If just set a range() to number like eg 10 and loop.

1 2 3 4 5 6 7 8 9 10 11 12 13

>>> for n in range(10): ...     dgtw['entries'][n]['published'] ...     'Tue, 11 Jun 2019 16:07:41 GMT' 'Tue, 11 Jun 2019 15:03:52 GMT' 'Tue, 11 Jun 2019 14:43:21 GMT' 'Tue, 11 Jun 2019 13:58:31 GMT' 'Tue, 11 Jun 2019 12:24:25 GMT' 'Tue, 11 Jun 2019 03:13:45 GMT' 'Mon, 10 Jun 2019 21:56:47 GMT' 'Mon, 10 Jun 2019 19:58:24 GMT' 'Mon, 10 Jun 2019 19:38:59 GMT' 'Mon, 10 Jun 2019 14:07:38 GMT'

So see that there is 6 entries today.
Then can just do a quick hack.

1 2 3 4 5 6 7 8

>>> from datetime import datetime >>> >>> today = datetime.today().day >>> today 12 >>> today_match =  f', {today}' >>> today_match ', 12'

when have ', 12' can just use in to look if there is a match in published date string.

1 2 3 4 5 6 7 8 9

>>> today_match =  ', 11' >>> count = 0 >>> for n in range(10): ...     p_date = dgtw['entries'][n]['published']     ...     if today_match in p_date: ...         count += 1       ...         >>> count 6

[Biks]

OK, my grasp of Python code is tenuous. :) I'm having a hard time building the final code from your examples. (sorry)

How do I even see the list of date and times column for:

1 2	for n in range(10):     dgtw['entries'][n]['published']

If I print(dgtw), I see all entries.

I noticed you have today_match listed 3 times. Do I need all three?

1 2 3 4

today_match =  f', {today}'     today_match ', 12'     today_match =  ', 11'

What's the final output supposed to look like?

[snippsat]

Here put together.

1 2 3 4 5 6 7 8 9 10 11 12 13

import feedparser from datetime import datetime   dgtw = feedparser.parse('https://investorshub.advfn.com/boards/rss.aspx?board_id=22658') today = datetime.today().day today_match = f', {today}' count = 0 for n in range(50):     p_date = dgtw['entries'][n]['published']     if today_match in p_date:         count += 1   print(f'published entries {datetime.today().ctime()},til now is <{count}>')

Output:
published entries Wed Jun 12 17:50:04 2019,til now is <2>

Can just set a higher number as here 50,also more entries that will ever be published in day.
Then count will be correct.

[Biks]

Hey this is great! Thanks! The last thing I did was toss the final number into a variable:

total = (f'{count}')
print(total)

I'm tossing that number into a Google sheet. (I managed to figure out how to do that by myself) :P

Thanks again!

[abhishek_k_a]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

#one option might be using of filtering the dates using system dates   from datetime import datetime import feedparser   # Current date time in local system dt = datetime.now() today = datetime.today().day count = 0   dgtw = feedparser.parse('https://investorshub.advfn.com/boards/rss.aspx?board_id=22658')   if dt==today:    for each in range(len(dgtw['entries'])):           count += 1 print(count)