我正在尝试与Sequelize建立一对多关联:
这是我的项目:
'use strict';
import { models } from '../sequelize/index';
var ProjectImages = require('./projectImages');
module.exports = (sequelize, DataTypes) => {
const Projects = sequelize.define(
'Projects',
{
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: DataTypes.INTEGER,
},
title: {
type: DataTypes.STRING,
},
},
{}
);
Projects.associate = function(models) {
// associations can be defined here
Projects.belongsToMany(models.Tags, { through: models.TagsProjects });
Projects.hasMany(models.ProjectImages);
};
return Projects;
};
这里是项目的图像:
'use strict';
import { models } from '../sequelize/index';
module.exports = (sequelize, DataTypes) => {
const ProjectImages = sequelize.define(
'ProjectImages',
{
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: DataTypes.INTEGER,
},
img: {
type: DataTypes.STRING,
},
createdAt: {
allowNull: false,
type: DataTypes.DATE,
defaultValue: DataTypes.literal('CURRENT_TIMESTAMP'),
},
updatedAt: {
allowNull: false,
type: DataTypes.DATE,
defaultValue: DataTypes.literal('CURRENT_TIMESTAMP'),
},
},
{}
);
return ProjectImages;
};
我正在使用类似于this snippet的方式同步数据库。如果我不尝试在ProjectImages中创建任何行,则不会出现任何错误;但是,如果我尝试以此创建行(ID为1的项目存在):
"ProjectImages": [
{
"id": 1,
"img": "https://fakeimg.pl/500x500/",
"ProjectsId": 1
}
],
我收到此错误:
Executing (default): INSERT INTO `ProjectImages`
(`id`,`img`,`createdAt`,`updatedAt`,`ProjectId`)
VALUES
(1,'https://fakeimg.pl/500x500/',CURRENT_TIMESTAMP,CURRENT_TIMESTAMP,1);
[1] Unhandled rejection SequelizeForeignKeyConstraintError:
Cannot add or update a child row: a foreign key constraint fails
(`9uhfggfhf2`.`projectimages`,
CONSTRAINT `projectimages_ibfk_1`
FOREIGN KEY (`ProjectId`)
REFERENCES `Projects` (`id`)
ON DELETE SET NULL
ON UPDATE CASCADE)
此外,如果我在同步时不向ProjectsImages添加行,也不会出现任何错误,并且如果我尝试通过以下方式在DB中手动插入行:
INSERT INTO `ProjectImages` (`id`,`img`,`createdAt`,`updatedAt`,`ProjectsIdA`)
VALUES (1,'https://fakeimg.pl/500x500/',CURRENT_TIMESTAMP,CURRENT_TIMESTAMP,1);
它可以正常工作。
有人知道那里发生了什么吗?
答案 0 :(得分:0)
正如我所说,我正在使用类似于this的东西来与数据库同步。
在代码段的末尾,我必须播种数据:
seed([
articles,
projectImages,
projects,
tags,
tagsProjects,
users,
usersLogins
]);
您会注意到,我在创建projectImages
之前先插入projects
。按正确的顺序放置东西可以解决问题。
seed([
articles,
projects,
projectImages,
tags,
tagsProjects,
users,
usersLogins
]);
:)