未处理的拒绝 SequelizeDatabaseError：无法添加外键约束

Posted 2023-02-23

【中文标题】未处理的拒绝 SequelizeDatabaseError：无法添加外键约束

【英文标题】：Unhandled rejection SequelizeDatabaseError: Cannot add foreign key constraint

【发布时间】：2018-12-15 17:36:14

【问题描述】：

我有 2 个表：admin_user 和 admin_account。

const Sequelize = require('sequelize');
module.exports = (sequelize, DataTypes) => 
  var admin_user = sequelize.define('admin_user', 
    id: 
      autoIncrement: true,
      type: Sequelize.INTEGER,
      primaryKey: true
    ,
    name: 
      type: Sequelize.STRING
    ,
    email: 
      type: Sequelize.STRING,
      allowNull: false,
      unique: true,
      validate: 
        isEmail: true,
      
    ,
    user_name:
      type: Sequelize.STRING,
      allowNull: false,
      unique: true
    
  );

  admin_user.associate = (models) =>
    admin_user.hasOne(models.admin_account,  foreignKey: 'admin_user_id' );
  ;

  return admin_user;
;


'use strict';
const Sequelize = require('sequelize');
module.exports = (sequelize, DataTypes) => 
  var admin_account = sequelize.define('admin_account', 
    admin_user_id:
      type: Sequelize.INTEGER,
      primaryKey: true
    ,
    password: 
      type: Sequelize.STRING
    
  );
  return admin_account;
;

并在创建表时出现此错误：

执行（默认）：CREATE TABLE IF NOT EXISTS admin_users (id INTEGER auto_increment , name VARCHAR(255), email VARCHAR(255) NOT NULL UNIQUE, user_name VARCHAR(255) NOT NULL UNIQUE, createdAt DATETIME NOT NULL, updatedAt DATETIME NOT NULL, 主键 (id)) ENGINE=InnoDB; 执行（默认）：SHOW INDEX FROM admin_users FROM milkman_prod1 执行（默认）： CREATE TABLE IF NOT EXISTS admin_accounts (admin_user_id INTEGER , password VARCHAR(255), createdAt DATETIME NOT NULL, updatedAt DATETIME NOT NULL, PRIMARY KEY (admin_user_id), FOREIGN KEY (admin_user_id) REFERENCES admin_users (id) ON DELETE SET NULL ON UPDATE CASCADE) ENGINE=InnoDB; 未处理的拒绝 SequelizeDatabaseError: 无法添加外键约束

【参考方案1】：

请尝试以下代码：

    'use strict';
    const Sequelize = require('sequelize');
    module.exports = (sequelize, DataTypes) => 
      var admin_account = sequelize.define('admin_user', 
        admin_user_id:
          type: Sequelize.INTEGER,
          primaryKey: true
        ,
        password: 
          type: Sequelize.STRING
        
      );
admin_account.associate = (models) =>
    admin_account.belongsTo(models.admin_user);
  ;
      return admin_account;
    ;

【讨论】：

谢谢里亚朱尔。它将消除错误，但不会在表中输入任何内容。我想出了另一个解决方案： admin_user.associate = (models) => admin_user.hasOne(models.admin_account, foreignKey: name: 'admin_user_id', allowNull: false , onDelete: 'restrict', onUpdate: 'restrict ' ); ;