skw*_*eth 5 sqlite node.js express sequelize.js sequelize-cli
我正在尝试使用Sequelize构建一个简单的Node/Express应用程序,但是当我尝试在我的关系数据库中创建一个新记录时,我收到错误Unhandled rejection SequelizeDatabaseError: SQLITE_ERROR: no such table: main.User.基本上,我在Users表中创建一个用户,然后尝试在表中创建一个相关的地址Addresses- 用户已成功创建但在创建地址时失败并出现此错误...其中是main从表名中获取前缀?(下面的完整错误读数)...
首先,这是我的计划的简要介绍......
我的Sequelize版本是Sequelize [Node: 6.8.1, CLI: 2.4.0, ORM: 3.29.0],我使用Sequelize CLI命令sequelize init来设置我项目的这一部分.
我使用SQLite3进行本地开发,并且config/config.json我将开发数据库定义为
"development": {
"storage": "dev.sqlite",
"dialect": "sqlite"
}
Run Code Online (Sandbox Code Playgroud)我的用户迁移:
'use strict';
module.exports = {
up: function(queryInterface, Sequelize) {
return queryInterface.createTable('Users', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
first_name: {
type: Sequelize.STRING
},
last_name: {
type: Sequelize.STRING
},
createdAt: {
allowNull: false,
type: Sequelize.DATE
},
updatedAt: {
allowNull: false,
type: Sequelize.DATE
}
});
},
down: function(queryInterface, Sequelize) {
return queryInterface.dropTable('Users');
}
};
Run Code Online (Sandbox Code Playgroud)和地址迁移(缩写):
module.exports = {
up: function(queryInterface, Sequelize) {
return queryInterface.createTable('Addresses', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
address_line_one: {
type: Sequelize.STRING
},
UserId: {
type: Sequelize.INTEGER,
allowNull: false,
references: {
model: "User",
key: "id"
}
}
})
}
Run Code Online (Sandbox Code Playgroud)用户模型:
'use strict';
module.exports = function(sequelize, DataTypes) {
var User = sequelize.define('User', {
first_name: DataTypes.STRING,
last_name: DataTypes.STRING
}, {
classMethods: {
associate: function(models) {
models.User.hasOne(models.Address);
}
}
});
return User;
};
Run Code Online (Sandbox Code Playgroud)和地址模型:
'use strict';
module.exports = function(sequelize, DataTypes) {
var Address = sequelize.define('Address', {
address_line_one: DataTypes.STRING,
UserId: DataTypes.INTEGER
}, {
classMethods: {
associate: function(models) {
models.Address.hasOne(models.Geometry);
models.Address.belongsTo(models.User, {
onDelete: "CASCADE",
foreignKey: {
allowNull: false
}
});
}
}
});
return Address;
};
Run Code Online (Sandbox Code Playgroud)最后,我的路线index.js:
router.post('/createUser', function(req, res){
var firstName = req.body.first_name;
var lastName = req.body.last_name;
var addressLineOne = req.body.address_line_one;
models.User.create({
'first_name': newUser.firstName,
'last_name': newUser.lastName
}).then(function(user){
return user.createAddress({
'address_line_one': newUser.addressLineOne
})
})
Run Code Online (Sandbox Code Playgroud)因此,当我尝试发布时/createUser,将成功创建用户并且控制台将说明已创建新地址(INSERT INTO 'Addresses'...),但未创建地址并记录以下错误:
Unhandled rejection SequelizeDatabaseError: SQLITE_ERROR: no such table: main.User
at Query.formatError (/Users/darrenklein/Desktop/Darren/NYCDA/WDI/projects/world_table/wt_test_app_1/node_modules/sequelize/lib/dialects/sqlite/query.js:348:14)
at afterExecute (/Users/darrenklein/Desktop/Darren/NYCDA/WDI/projects/world_table/wt_test_app_1/node_modules/sequelize/lib/dialects/sqlite/query.js:112:29)
at Statement.errBack (/Users/darrenklein/Desktop/Darren/NYCDA/WDI/projects/world_table/wt_test_app_1/node_modules/sqlite3/lib/sqlite3.js:16:21)
几个月前我曾经和Sequelize做过一次这样的事情并且成功了,我不能为我的生活找出我在这里缺少的东西.应用程序为什么要查找main.User,如何才能找到正确的表格?谢谢!
啊哈!一个小错误使我的整个操作脱轨.在迁移文件中,references.model必须是多元化的!
references: {
model: "Users",
key: "id"
}
Run Code Online (Sandbox Code Playgroud)
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