flask-admin

问题 I'm new to the flask-admin library so please forgive me if this is trivial. When I click 'Save' to create a new row for a model, I also want to do some custom things. In my case, I'll create a table dynamically whose name is the string entered in the form. This will be in addition to what flask-admin does for me i.e. add a new row to the model table. So where will I put the custom logic to do what I want to do? I saw this post on so: Customize (override) Flask-Admin's Submit method from edit

问题 I'm new to the flask-admin library so please forgive me if this is trivial. When I click 'Save' to create a new row for a model, I also want to do some custom things. In my case, I'll create a table dynamically whose name is the string entered in the form. This will be in addition to what flask-admin does for me i.e. add a new row to the model table. So where will I put the custom logic to do what I want to do? I saw this post on so: Customize (override) Flask-Admin's Submit method from edit

问题 I am trying to create a one-to-one relationship between a Department & Ticket table. This way when looking inside of Flask-Admin a user would be able to see the Department name instead of the ID. I have tried to setup the relationship as follows: # Ticket Table class Tickets(db.Model): __tablename__ = 'tickets' ticketID = db.Column(db.Integer, nullable=False, primary_key=True, autoincrement=True, unique=True) cust_name = db.Column(db.String(50), nullable=False) cust_email = db.Column(db

问题 I'm currently using Flask-Admin to create an admin screen but I'm running into a roadblock. Suppose I have two classes that look like this: class Part(db.Model): id = db.Column(db.Integer, primary_key=True, unique=True) part_data = db.relationship("PartHistory", backref="part") name = db.Column(db.String(255)) class PartHistory(db.Model): id = db.Column(db.Integer, primary_key=True, unique=True) part_id = db.Column(db.Integer, db.ForeignKey('part.id')) date_checked = db.Column(db.Date) Part

问题 I'm currently using Flask-Admin to create an admin screen but I'm running into a roadblock. Suppose I have two classes that look like this: class Part(db.Model): id = db.Column(db.Integer, primary_key=True, unique=True) part_data = db.relationship("PartHistory", backref="part") name = db.Column(db.String(255)) class PartHistory(db.Model): id = db.Column(db.Integer, primary_key=True, unique=True) part_id = db.Column(db.Integer, db.ForeignKey('part.id')) date_checked = db.Column(db.Date) Part

问题 i have an aplication homestay reservation built with flask.. every homestay have an user login, this login built with flask-security and every owner of the homestay have role User . and every user can input their homestay data with flask-admin. but unfortunately if a user input their data, the others user which have role User can seing the data have input too.. so.. my question how to separate the data if a user have the same role..? user A can just see his data, and user B so too.. this is

问题 I am facing to performance problem in Flask-Admin, although performance of Flask application is good. my model is: class Ck(Base): __tablename__ = "ck" id = Column(Integer, primary_key=True, autoincrement=True) nazev = Column(String(100)) kontakt = Column(Text) terms = relationship(Term, backref=backref('ck', lazy='noload'), lazy='dynamic') class Term(Base): __tablename__ = "term" id = Column(Integer, primary_key=True, autoincrement=True) hotel_id = Column(Integer, ForeignKey('hotel.id')) ck

问题 Hi there fellow Flask developers! In Flask-admin, I currently try to implement inline model editing into my model view. On the model side, I have a simple tree structure that represents a set of content pages. Each node has several child nodes and also several content data models associated with it. The models are named ContentNode and ContentData . If I use the inline_models property on the node view class as described in the Docs here, it seems to work fine at first. # AuthModelView is

问题 I've checked the docs and It's pretty vague how the is_accessible method should be implemented. Here is what the docs of flask admin showed class MicroBlogModelView(sqla.ModelView): def is_accessible(self): return login.current_user.is_authenticated() def inaccessible_callback(self, name, **kwargs): # redirect to login page if user doesn't have access return redirect(url_for('login', next=request.url)) what I don't get though is how do you call it is it automatically called or do you have to

问题 As title says, I have small web app, without using database and models. I'd like interface to change some of Flask own config parameters, and thought that flask-admin may bring me there quickly. Is this easily possible? 回答1: You can't generally change configuration after starting the application without restarting the server. The application (at least in production) will be served with multiple processes, possibly even on multiple servers. Changes to the configuration will only effect the