5
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I have these 2 functions that are almost the same:

def handle_keywords(self, scan_id):
    keywords = Keyword.objects.all()
    for kwrd in keywords:
        self.process_rss_url(kwrd.rss, SourceType.GOOGLE_ALERTS, kwrd.id, kwrd.last_scan_time, scan_id)
        kwrd.last_scan_time = timezone.now()
        kwrd.save()

def handle_rss(self, scan_id):
    all_rss = RSS.objects.all()
    for rss in all_rss:
        self.process_rss_url(rss.feed_url, SourceType.RSS, rss.id, rss.last_scan_time, scan_id)
        rss.last_scan_time = timezone.now()
        rss.save()

What is a good way to reduce the code duplication?

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migrated from stackoverflow.com Dec 7 '17 at 13:15

This question came from our site for professional and enthusiast programmers.

5
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Try this:

def generate_func(url_attr, type_class, source_type):
    def f(self, scan_id):
        for keyword in type_class.objects.all():
            self.process_rss_url(getattr(keyword, url_attr), source_type, keyword.id, keyword.last_scan_time, scan_id)
            keyword.last_scan_time = timezone.now()
            keyword.save()

    return f

handle_keywords = generate_func('rss', Keyword, SourceType.GOOGLE_ALERTS)
handle_rss = generate_func('feed_url', RSS, SourceType.RSS)
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  • \$\begingroup\$ The "keywords" temp is unnecessary. I think SourceType.RSS and SourceType.GOOGLE_ALERTS are constants, but it's not certain from the code given. \$\endgroup\$ – Snowbody Dec 7 '17 at 14:36
  • \$\begingroup\$ @Snowbody Thanks for pointing out about keywords. Why does it matter if those are constants? \$\endgroup\$ – Solomon Ucko Dec 7 '17 at 16:55
  • \$\begingroup\$ Because then they can be passed in and used directly, saving a call to getattr() \$\endgroup\$ – Snowbody Dec 7 '17 at 18:15
  • \$\begingroup\$ @Snowbody That's what I was wondering. I'll change that. \$\endgroup\$ – Solomon Ucko Dec 7 '17 at 20:16

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