From 9a9b5c19d8c07026b104ed81838145454bfa9fc9 Mon Sep 17 00:00:00 2001 From: Cédric Bonhomme Date: Sun, 12 Apr 2015 17:03:20 +0200 Subject: It is now possible to add a new feed from any page via a dropdown menu. --- pyaggr3g470r/views/feed.py | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) (limited to 'pyaggr3g470r/views/feed.py') diff --git a/pyaggr3g470r/views/feed.py b/pyaggr3g470r/views/feed.py index e4c0dc9a..af43d6f0 100644 --- a/pyaggr3g470r/views/feed.py +++ b/pyaggr3g470r/views/feed.py @@ -134,6 +134,8 @@ def form(feed_id=None): if request.method == 'POST': if not form.validate(): + print(dir(form)) + print("oups") return render_template('edit_feed.html', form=form) existing_feeds = list(feed_contr.read(link=form.link.data)) if existing_feeds and feed_id is None: @@ -141,8 +143,8 @@ def form(feed_id=None): "warning") return redirect(url_for('feed.form', feed_id=existing_feeds[0].id)) - # Edit an existing feed + print("new...") if feed_id is not None: feed_contr.update({'id': feed_id}, {'title': form.title.data, @@ -154,6 +156,7 @@ def form(feed_id=None): return redirect(url_for('feed.form', feed_id=feed_id)) # Create a new feed + print("new feed") new_feed = FeedController(g.user.id).create( title=form.title.data, description="", -- cgit