Error binding parameter 4 - probably unsupported type

I should first mention that I'm using SqlAlchemy through Flask-SqlAlchemy. I don't believe this affects the issue but if it does, please let me know. Here is the relevant part of the error message I'm getting when running the create_all function in SqlAlchemy `InterfaceError: (InterfaceError) Error binding parameter 4 - probably unsupported type. u'INSERT INTO podcasts (feed_url, title, url, last_updated, feed_data) VALUES (?, ?, ?, ?, ?)' (u'http://example.com/feed', u'Podcast Show Title', u'http://example.com', '2012-04-17 20:28:49.117000'` Here is my model: ``` class Podcast(db.Model): import datetime __tablename__ = 'podcasts' id = db.Column(db.Integer, primary_key=True) feed_url = db.Column(db.String(150), unique=True) title = db.Column(db.String(200)) url = db.Column(db.String(150)) last_updated = db.Column(db.DateTime, default=datetime.datetime.now) feed_data = db.Column(db.Text) def __init__(self, feed_url): import feedparser self.feed_url = feed_url self.feed_data = feedparser.parse(self.feed_url) self.title = self.feed_data['feed']['title'] self.url = self.feed_data['feed']['link'] ``` Can someone tell me how I can get this to work? I've also tried the following model but that also doesn't work. Same error. ``` class Podcast(db.Model): import datetime __tablename__ = 'podcasts' id = db.Column(db.Integer, primary_key=True) feed_url = db.Column(db.String(150), unique=True) title = db.Column(db.String(200)) url = db.Column(db.String(150)) last_updated = db.Column(db.DateTime) feed_data = db.Column(db.Text) def __init__(self, feed_url): import feedparser self.feed_url = feed_url self.feed_data = feedparser.parse(self.feed_url) self.last_updated = datetime.datetime.now() self.title = self.feed_data['feed']['title'] self.url = self.feed_data['feed']['link'] ```