Jaculabilis
/
intake
1
0
Fork 0
Code Issues Pull Requests Projects Releases Wiki Activity

Provide link back to source in fetch result

This commit is contained in:
Tim Van Baak 2024-09-21 15:07:05 -07:00
parent 5cf7e8ef48
commit 256a52e902
1 changed files with 3 additions and 5 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

8
intake/app.py
View File

@ -16,7 +16,6 @@ from flask import (
redirect, redirect,
url_for, url_for,
current_app, current_app,
make_response,
) )
from intake.core import intake_data_dir from intake.core import intake_data_dir
@ -397,11 +396,10 @@ def fetch(source_name: str):
try: try:
items = fetch_items(source) items = fetch_items(source)
titles = "\n".join(item.display_title for item in items)
update_items(source, items) update_items(source, items)
response = make_response(f"Update returned {len(items)} items:\n{titles}", 200) titles = "\n".join(f"<li>{item.display_title}</li>" for item in items)
response.mimetype = "text/plain" source_url = url_for("source_feed", name=source_name)
return response return f"Update returned {len(items)} items:<ul>{titles}</ul><p><a href=\"{source_url}\">{source_name}</a></p>"
except InvalidConfigException as ex: except InvalidConfigException as ex:
abort(500, f"Could not fetch {source_name}:\n{ex}") abort(500, f"Could not fetch {source_name}:\n{ex}")
except SourceUpdateException as ex: except SourceUpdateException as ex: