我试图插入多行2相关表使用sequelize与mysql。
基于sequelize transaction cannot insert because of foreign key?,我尝试创建一个事务,如果我只执行一次承诺,它似乎可以工作,但是如果我在显示外键问题的循环上执行代码,它就失败了。
下面是代码:
const config = {
"username": "root",
"password": "",
"database": "sequelize_test",
"host": "127.0.0.1",
"dialect": "mysql",
"pool": {
"max": 1,
"min": 0,
"idle": 20000,
"acquire": 20000
}
};
const Sequelize = require('sequelize')
const sequelize = new Sequelize(config.database, config.username, config.password, config)
const Project = sequelize.define('projects', {
id: {
type: Sequelize.INTEGER,
allowNull: false,
autoIncrement: true,
primaryKey: true
},
authorId: {
type: Sequelize.INTEGER
},
name: {
type: Sequelize.STRING
}
})
const Author = sequelize.define('authors', {
id: {
type: Sequelize.INTEGER,
allowNull: false,
autoIncrement: true,
primaryKey: true
},
name: {
type: Sequelize.STRING,
}
})
Author.hasMany(Project, {
foreignKey: 'authorId',
sourceKey: 'id'
})
Project.belongsTo(Author, {
foreignKey: 'id'
})
sequelize
.sync({
force: true
})
.then(() => {
var projectAuthors = [{
projectName: "Proj 1",
authorName: "Author 1"
}, {
projectName: "Proj 2",
authorName: "Author 1"
}, {
projectName: "Proj 3",
authorName: "Author 1"
}, {
projectName: "Proj 4",
authorName: "Author 2"
}, {
projectName: "Proj 5",
authorName: "Author 3"
}];
//Insert all the records on the array
projectAuthors.forEach(function(project) {
sequelize
.transaction(function(t) {
//First insert the author
return Author.findOrCreate({
where: {
name: project.authorName
},
transaction: t
}).spread(function(author) {
//With the id obtained on the previous step, insert the project
return Project.findOrCreate({
where: {
name: project.projectName,
authorId: author.id
},
transaction: t
});
})
});
});
});
- 编辑**
按照Ellebkey的建议,下面是我使用include的控制器。
sequelize
.sync({
force: true
})
.then(() => {
var projectAuthors = [{
projectName: "Proj 1",
authorName: "Author 1"
}, {
projectName: "Proj 2",
authorName: "Author 1"
}, {
projectName: "Proj 3",
authorName: "Author 1"
}, {
projectName: "Proj 4",
authorName: "Author 2"
}, {
projectName: "Proj 5",
authorName: "Author 3"
}];
//Insert all the records on the array
projectAuthors.forEach(function(project) {
sequelize
.transaction(function(t) {
//First insert the author
return Project.findOrCreate({
where: {
name: project.projectName,
author: {
name: project.authorName
}
},
include: [{
association: Project.Author,
}],
transaction: t
});
});
});
});
其失败原因如下:
无效值{名称:"作者3 "}
- 编辑2(表格创建工作)**
多亏了Ellebkey和一些玩来玩去的人的建议,我的模型看起来像这样:
const Project = sequelize.define('projects', {
name: {
type: Sequelize.STRING
}
});
const Author = sequelize.define('authors', {
name: {
type: Sequelize.STRING,
}
});
Project.belongsTo(Author, { as: 'Author', foreignKey : 'authorId'});
Author.hasMany(Project, { as: 'Author', foreignKey : 'authorId'});
但是使用答案中的create()代码,我仍然得到以下结果:
未处理的拒绝错误:无效值{名称:未定义}
顺便说一句,我正在使用续集4. 37. 6
先谢了
1条答案
按热度按时间cbjzeqam1#
首先,根据我的经验,不推荐使用
foreignKey
作为关联,因为会混淆。使用as
代替,而且你也不需要创建每个模型的id
,续集可以为你做。现在,回到关联,这将在
Project
表中创建一个AuthorId
列最后,对于你的
create
和include
,你必须根据你之前声明的方式创建你的对象,就像你用Find
那样,你用Projects
得到你的Author
,所以每个对象应该是这样的:您得到的错误是因为您从未在模型上声明
authorName
。