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为什么我的数据库不在我的工作目录中创建.db文件?我在用PyCharm

Youssef Blah • 4 年前 • 956 次点击  

我用的是Pycharm Professional。我有一个名为basic.py的文件,其中包含在数据库中创建模型的模型和代码。代码运行良好,不会遇到任何错误,但是在运行代码后,它不会在我的工作目录中创建.db文件。即使我使用了db.create_all(),也找不到它。我错过了什么?谢谢

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
import os

print(os.getcwd())
basedir = os.path.abspath(os.path.dirname(__file__))

app = Flask(__name__)

app.config["SQLAlCHEMY_DATABASE_URI"] = 'sqlite:////site.db'
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False

db = SQLAlchemy(app)

class Puppy(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.Text)

    to_owner = db.relationship("Owner", backref="puppy", uselist=False)
    to_toy = db.relationship("Toy", backref= "puppy", lazy = "dynamic")

    def __init__(self, name):
        self.name = name

    def __repr__(self):
        return f"This is {self.name}"



class Owner(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.Text)
    pupid = db.Column(db.Integer, db.ForeignKey("puppy.id"))

    def __init__(self, name, pup):
        self.name = name
        self.owner = pup


class Toy(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.Text)
    pupid = db.Column(db.Integer, db.ForeignKey("puppy.id"))

    def __init__(self, name, pup):
        self.name = name
        self.owner = pup

db.create_all()

john = Puppy("john")
jack = Puppy("jack")

print(john.name)
db.session.add_all([john, jack])
db.session.commit()
print(john.name)

print(Puppy.query.all())

john = Puppy.query.filter_by(name="john").first()
print(john)

owner1 = Owner("amr", john.id)


toy1 = Toy("ball", john.id)
toy2 = Toy("frisby", john.id)

db.session.add_all([owner1, toy1, toy2])
db.session.commit()

john = Puppy.query.filter_by(name="john").first()
print(john.name)
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文章 [ 1 ]  |  最新文章 4 年前
Nathan
Reply   •   1 楼
Nathan    5 年前

我觉得你有一个太多了 / 在sqlalchemy uri的sqlite目录规范中。您能否将数据库URI更改为以下内容,并查看是否解决了您的问题:

app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///site.db'