flask_sqlalchemy create_all无需导入模型 [英] flask_sqlalchemy create_all without having to import models

问题描述

我试图了解如何设置一个调用create_all的独立脚本，而不必将我的所有模型导入其中.以下是相关文件:

I am trying to understand how to set up a standalone script that calls create_all without having to import all my models into it. Below are the relevant files:

db.py

db.py

from flask_sqlalchemy import SQLAlchemy

db = SQLAlchemy()

test_model.py

test_model.py

from db import db

class TestTable(db.model):
    id = db.Column(db.Integer, primary_key=True)
    foo = db.Column(db.String(80))

和 create_all.py

and create_all.py

from db import db

db.create_all()

但是，create_all不会执行任何操作，因为它不会检测到已注册的任何表，除非将它们导入.声明模型时是否有办法将它们添加到注册表"中以绕过显式导入?

However, create_all will not do anything, as it does not detect any tables registered to it, unless they are imported. Is there a way when declaring models to have them added to the "registry" to bypass the explicit import?

推荐答案

总之，没有.您将必须在调用create_all()之前以一种或另一种方式导入这些模型.声明类创建Table实例，并在构造类本身时将其添加到MetaData.如果MetaData不包含表，则create_all将不执行任何操作.

In a word, no. You will have to import those models before the call to create_all(), one way or the other. The declarative classes build the Table instances and add them to the MetaData when the class itself is constructed. If the MetaData contains no tables, create_all will do nothing.

以另一种方式，除非对那些声明进行评估，否则声明模型"将不执行任何操作，因此必须执行该模块.从它导入就可以做到这一点.

Put another way, "declaring models" does nothing unless those declarations are evaluated, so the module must be executed. Importing from it does just that.

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