Python Flask,SQLAlchemy关系

Question

我已经尝试解决这个问题几个小时了,我无法让SQLAlchemy工作(它正在工作,直到我把两个新功能,用户和注册)

from flask.ext.sqlalchemy import SQLAlchemy
from . import app
from datetime import datetime

db = SQLAlchemy(app)

class PasteCode(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    codetitle = db.Column(db.String(60), unique = True)
    codebody = db.Column(db.Text)
    pub_date = db.Column(db.DateTime)
    user_id = db.Column(db.Integer, db.ForeignKey('user.id'))
    parent_id = db.Column(db.Integer, db.ForeignKey('paste_code.id'))
    parent = db.relationship('PasteCode', lazy = True, backref = 'children', uselist = False, remote_side = [id])

    def __init__(self, codetitle, codebody, parent = None):
        self.codetitle = codetitle
        self.codebody = codebody
        self.pub_date = datetime.utcnow()
        self.parent = parent

class User(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    display_name = db.Column(db.String(30))
    #pastes = db.Column(db.Integer, db.ForeignKey('paste_code.id'))
    pastes = db.relationship(PasteCode, lazy = "dynamic", backref = "user")

class Registration(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    username = db.Column(db.String(30), unique = True)
    password = db.Column(db.String(100), unique = False)

这是它在运行时给我的回溯:

OperationalError: (OperationalError) no such table: paste_code u'SELECT paste_code.id AS paste_code_id, paste_code.codetitle AS paste_code_codetitle, paste_code.codebody AS paste_code_codebody, paste_code.pub_date AS paste_code_pub_date, paste_code.user_id AS paste_code_user_id, paste_code.parent_id AS paste_code_parent_id \nFROM paste_code' ()

我也试过这个:

from flask.ext.sqlalchemy import SQLAlchemy
from . import app
from datetime import datetime

db = SQLAlchemy(app)

class PasteCode(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    codetitle = db.Column(db.String(60), unique = True)
    codebody = db.Column(db.Text)
    pub_date = db.Column(db.DateTime)
    user_id = db.Column(db.Integer, db.ForeignKey('user.id'))
    parent_id = db.Column(db.Integer, db.ForeignKey('pastecode.id'))
    parent = db.relationship('PasteCode', lazy = True, backref = 'children', uselist = False, remote_side = [id])

    def __init__(self, codetitle, codebody, parent = None):
        self.codetitle = codetitle
        self.codebody = codebody
        self.pub_date = datetime.utcnow()
        self.parent = parent

class User(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    display_name = db.Column(db.String(30))
    pastes =  db.relationship(PasteCode, lazy = 'dynamic', backref = 'user')
    #pastes = db.relationship(PasteCode, lazy = "dynamic", backref = "user")

class Registration(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    username = db.Column(db.String(30), unique = True)
    password = db.Column(db.String(100), unique = False)

我收到了这个错误:

ArgumentError: Could not determine join condition between parent/child tables on relationship PasteCode.parent. Specify a 'primaryjoin' expression. If 'secondary' is present, 'secondaryjoin' is needed as well.

任何的想法？谢谢!

Answer 1

我补充说,我修复它的方式很简单

__tablename__ = "paste_code"

并且一切正常,我认为SQLAlchemy没有正确检查表名.