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How to use feedparpser to gain all RSS contents?


On Wed, May 17, 2017 at 4:44 PM,  <seraphcj at gmail.com> wrote:
> When using feedparpser object as below, it only contains 25 items, how can i have all contents?
> for example,
> d= feedparser.parse('http://newyork.craigslist.org/stp/index.rss')
> print(len(d['entries'][i]['summary']))
>
> result is 25....

Have you looked at the actual file delivered by craigslist? It seems
to have only 25 items in it. It's not feedparser's limit.

ChrisA