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allow getting crawl from API as rss feed
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parent
eae7ed8447
commit
c374d7695e
3 changed files with 23 additions and 2 deletions
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@ -97,8 +97,8 @@ class CrawlSchema(Schema):
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def get_crawls(request):
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return Crawl.objects.all().distinct()
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@router.get("/crawl/{crawl_id}", response=CrawlSchema, url_name="get_crawl")
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def get_crawl(request, crawl_id: str, with_snapshots: bool=False, with_archiveresults: bool=False):
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@router.get("/crawl/{crawl_id}", response=CrawlSchema | str, url_name="get_crawl")
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def get_crawl(request, crawl_id: str, as_rss: bool=False, with_snapshots: bool=False, with_archiveresults: bool=False):
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"""Get a specific Crawl by id or abid."""
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crawl = None
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@ -114,5 +114,18 @@ def get_crawl(request, crawl_id: str, with_snapshots: bool=False, with_archivere
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crawl = crawl or Crawl.objects.get(id__icontains=crawl_id)
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except Exception:
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pass
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if crawl and as_rss:
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# return snapshots as XML rss feed
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urls = [
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{'url': snapshot.url, 'title': snapshot.title, 'bookmarked_at': snapshot.bookmarked_at, 'tags': snapshot.tags_str}
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for snapshot in crawl.snapshot_set.all()
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]
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xml = '<rss version="2.0"><channel>'
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for url in urls:
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xml += f'<item><url>{url["url"]}</url><title>{url["title"]}</title><bookmarked_at>{url["bookmarked_at"]}</bookmarked_at><tags>{url["tags"]}</tags></item>'
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xml += '</channel></rss>'
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return xml
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return crawl
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