[英]Python Flask, SQLAlchemy relationship
我已經嘗試解決這個問題幾個小時了,我無法讓SQLAlchemy工作(它正在工作,直到我把兩個新功能,用戶和注冊)
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)
這是它在運行時給我的回溯:
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' ()
我也試過這個:
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)
我收到了這個錯誤:
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? 謝謝!
我補充說,我修復它的方式很簡單
__tablename__ = "paste_code"
並且一切正常,我認為SQLAlchemy沒有正確檢查表名。
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