内部服务器错误 Flask - Python

Question

这是我在 Python - Flask 中的代码。在这里，我正在为剧院（剧院 table）输入数据，然后获取相应的 ID，然后将屏幕添加到屏幕 table.

中的相应 theatre_id

问题是 Theatre 被添加到数据库中，我可以获取 ID。 Screen 的代码似乎不起作用（如果我注释掉 session.screen 添加和提交语句，for 循环确实起作用）如果屏幕提交没有发生，甚至回滚也不会发生。

session_theatre = Session_theatre()
session_screen = Session_screen()
id = 1
if request.method == "POST":
    if form.validate_on_submit():
        name = str(form.name.data)
        city = str(form.location.data)
        address = str(form.address.data)
        no_of_screen = int(form.total_no_screen.data)
        if (name !="" and name!=" " and city != "" and city != " " and address != ""and address != " " and no_of_screen != None):
            t = Theatre(name,city,address,1)
            try:
                session_theatre.add(t)
                session_theatre.commit()
                query = session_theatre.query(Theatre).filter_by(name=name,city =city).all()
                for i in query :
                    id = i
                for i in range (0,no_of_screen):
                    flash(id)
                    screen = Screen(str(i+1),1,20,1,20,id)
                    session_screen.add(screen)
                    session_screen.commit()
                flash("Successfully added !!")
            except :
                session_screen.rollback()
                session_theatre.rollback()
                flash("Oops something went wrong !!")
            finally:
                session_screen.close()
                session_theatre.close()
        else :
            flash("Please fill the input")
return render_template('admin/add_theatre.html',form = form)

屏幕模型

class Screen(db.Model):
    __tablename__ = "screens"
    id = db.Column(db.Integer,primary_key=True)
    screen_num = db.Column(db.String(1))
    seat_row_start = db.Column(db.Integer)
    seat_row_end = db.Column(db.Integer)
    seat_col_start = db.Column(db.Integer)
    seat_col_end = db.Column(db.Integer)
    theatre_id =  db.Column(db.Integer, db.ForeignKey('theatres.id'))

    def __init__(self, screen_num, 
    seat_row_start,seat_row_end,seat_col_start,seat_col_end,theatre_id):
        self.screen_num = screen_num
        self.seat_row_start = seat_row_start
        self.seat_row_end = seat_row_end
        self.seat_col_start = seat_col_start
        self.seat_col_end = seat_col_end
        self.theatre_id = theatre_id

Answer 1

要从 session_theatre 的查询中获取列表，您必须在末尾添加 all()：

query = session_theatre.query(Theatre).filter_by(name=name,city =city).all()

查询 returns Theatre 对象的列表，您可以访问每个 Theatre 对象的 id 属性，遍历列表并通过其访问属性姓名：

for obj in query :
    for i in range (0,no_of_screen):
        flash(obj.id)
        screen = Screen(i+1,1,20,1,20,obj.id)
        session_screen.add(screen)
        session_screen.commit()

内部服务器错误 Flask - Python

internal server error Flask - Python

python

sqlalchemy

flask

flask-sqlalchemy