使用查询结果作为射野选择的Flask WTForm

Question

参考以下表格。在

CONTROL_CHOICES = Session.query(schema.OfficeType).order_by(schema.OfficeType.descr).all()
CONTROL_CHOICES = [(office.id, office.descr) for office in CONTROL_CHOICES]

class ControlForm(Form):

    institution = RadioField('Institution', choices=CONTROL_CHOICES)

    date = DateField('Date')
    submit = SubmitField('SUBMIT')

简单HTML

^{pr2}$

无线电字段已成功打印。

但是，在填写表单时，validate_on_submit()不会打印错误，但不会在if中执行代码

validate()打印以下错误。在

{'institution': ['Not a valid choice']}

@app.route('/composition_profile', methods=['GET', 'POST'])
def composition_profiles():
    if request.method == 'GET':
        if request.args.get('institution') and request.args.get('date'):
            form = ControlForm(request.args)
            print(form.institution.data)

            if form.validate():

                print('terms')
            print(form.errors)

你知道这是怎么导致错误的选择吗？我不太清楚发生了什么事。把它改成QuerySelectField可以，但是我想要收音机功能

Answer 1

这个问题似乎是由于WTForm期望Value字段是String而不是{}这一事实造成的

改变

CONTROL_CHOICES = [(office.id, office.descr) for office in CONTROL_CHOICES]

到

CONTROL_CHOICES = [(str(office.id), office.descr) for office in CONTROL_CHOICES]

缓解问题

Answer 2

考虑重新分解：

def my_view():
    class F(MyBaseForm):
        pass

    F.username = TextField('username')
    for name in iterate_some_model_dynamically():
        setattr(F, name, TextField(name.title()))

    form = F(request.POST, ...)
    # do view stuff

来源：http://wtforms.simplecodes.com/docs/1.0.1/specific_problems.html#dynamic-form-composition