Provide link back to source in fetch result

intake/app.py

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