问题
I am using QuerySelectField in my forms.py and when submitting it I get the following error:
InterfaceError: (InterfaceError) Error binding parameter 8 - probably unsupported
type. u'INSERT INTO menu (title, title_eng, alias, menu_type, ordering,
check_out_time, access, published, parent_id, image, content, content_eng,
metades, metakey) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)'
(u'\u0423\u0441\u043b\u043e\u0432\u0438\u044f \u043f\u043e\u0434\u043a\u043b\u044e\u0447\u0435\u043d\u0438\u044f',
u'terms of connecting', u'terms', u'simple', 4, '2014-01-23 00:00:00.000000',
u'public', u'1', <app.models.Menu object at 0x7fe158171990>, u'url/folder/image.jpg',
u'asd', u'asd', u'asd', u'asd')
I googled it and found out that the issue is that the returned value of QuerySelectField is object and I have to convert it to string, but I couldnt. Can you please help me with that issue?
here is my forms.py:
def menu_list():
return Menu.query
class Add_menu_form(Form):
"""Add_menu_form is used to add/edit menu"""
title = TextField(u'Название меню', [validators.Length(min=1, max=250), validators.Required()])
title_eng = TextField(u'Название меню на английском', [validators.Length(min=1, max=250), validators.Required()])
alias = TextField(u'Короткое название')
menu_type = SelectField(u'Тип меню',
choices=[('simple', u'обычное'),
('blog', u'блог'),
('products', u'продукция'),
('gallery', u'галерея')])
ordering = IntegerField(u'Позиция')
check_out_time = DateField(u'Дата публикации')
access = SelectField(u'Доступ',
choices=[('public', u'открытый'),
('registered', u'для зарегистрированных'),
('admin', u'для администратора')])
published = SelectField(u'Опубликовать',
choices=[('1', u'да'),
('0', u'нет')])
parent_id = QuerySelectField(u'Родительская группа',
query_factory = menu_list,
get_pk = lambda a: a.id,
get_label = lambda a: a.title,
allow_blank=True)
image = TextField(u'Заглавная картинка')
content = TextAreaField(u'Содержание', [validators.Required()])
content_eng = TextAreaField(u'Содержание на английском', [validators.Required()] )
metades = TextAreaField(u'HTML описание')
metakey = TextAreaField(u'HTML ключевые слова')
this is my models.py:
class Menu(db.Model):
"""Menu is used for websites navigation titles.
eg. Home/About Us/Blog/Contacts/and etc"""
id = db.Column(db.Integer, primary_key = True)
title = db.Column(db.String(255))
title_eng = db.Column(db.String(255))
alias = db.Column(db.String(255))
menu_type = db.Column(db.String(10))
#menu type: simple, blog, gallery, contacts, products
ordering = db.Column(db.SmallInteger, default = '1')
check_out_time = db.Column(db.DateTime)
access = db.Column(db.String(30))
#access: user, reductor, manager, administrator
published = db.Column(db.SmallInteger, default = '1')
parent_id = db.Column(db.Integer)
image = db.Column(db.String(350))
content = db.Column(db.String)
content_eng = db.Column(db.String)
metades = db.Column(db.String(350))
metakey = db.Column(db.String(350))
def __init__(self, title, title_eng, alias,
menu_type, ordering, check_out_time, access,
published, parent_id, image, content, content_eng,
metades, metakey):
self.title = title
self.title_eng = title_eng
self.alias = alias
self.menu_type = menu_type
self.ordering = ordering
self.check_out_time = check_out_time
self.access = access
self.published = published
self.parent_id = parent_id
self.image = image
self.content = content
self.content_eng = content_eng
self.metades = metades
self.metakey = metakey
# __str__ is a special method, like __init__, that is
# supposed to return a string representation of an object.
def __str__(self):
return '%.d' % (self.id)
and the views.py:
@admin.route('/manage/add_menu', methods = ['GET', 'POST'])
@login_required
def add_menu(parent = ''):
form = Add_menu_form()
if form.validate_on_submit():
new_menu = Menu(
form.title.data,
form.title_eng.data,
form.alias.data,
form.menu_type.data,
form.ordering.data,
form.check_out_time.data,
form.access.data,
form.published.data,
form.parent_id.data,
form.image.data,
form.content.data,
form.content_eng.data,
form.metades.data,
form.metakey.data)
form.populate_obj(new_menu)
db.session.add(new_menu)
db.session.commit()
flash('New menu was added successfully.')
return redirect(url_for('cabinet.manage', current = 'menu_settings'))
return render_template('admin/manage/site_figuration/add_menu.html',
title = 'Internet market',
parent = parent,
form = form)
回答1:
Finally the issue is solved after hours of googling. the issue was about QuerySelectField. The problems was that when retrieving form.parent_id.data it actually returned a query object, whie I need a string vaue. So I converted the value to string and added submitted it to database:
a = str(form.parent_id.data)
if form.validate_on_submit():
new_menu = Menu(
form.title.data,
form.title_eng.data,
form.alias.data,
form.menu_type.data,
form.ordering.data,
form.check_out_time.data,
form.access.data,
form.published.data,
a, #form.parent_id.data,
form.image.data,
form.content.data,
form.content_eng.data,
form.metades.data,
form.metakey.data)
form.populate_obj(new_menu)
db.session.add(new_menu)
db.session.commit()
来源:https://stackoverflow.com/questions/24732871/flask-sqlalchemy-exc-interfaceerror-app-models-menu-object-at-0x7f287026dd10