【发布时间】:2017-02-17 22:16:14
【问题描述】:
我有一个没有任何主键的续集模型:
module.exports = (sequelize, DataTypes) => {
const usersDoors = sequelize.define('usersDoors',
{
user_uid: {
type: DataTypes.UUID,
allowNull: false,
},
door_uid: {
type: DataTypes.UUID,
allowNull: false,
},
property_manager_uid: {
type: DataTypes.UUID,
allowNull: true,
},
tenant_uid: {
type: DataTypes.UUID,
allowNull: true,
},
created_at: {
type: DataTypes.INTEGER,
allowNull: false,
},
},
{
tableName: 'users_doors',
indexes: [
{
name: 'doors_users_indexes',
unique: true,
fields: ['user_uid', 'door_uid', 'property_manager_uid', 'tenant_uid'],
},
],
classMethods: {
associate: (models) => {
usersDoors.belongsTo(models.users, { foreignKey: 'user_uid' });
usersDoors.belongsTo(models.doors, { foreignKey: 'door_uid' });
usersDoors.belongsTo(models.propertyManagers, { foreignKey: 'property_manager_uid' });
usersDoors.belongsTo(models.tenants, { foreignKey: 'tenant_uid' });
},
},
});
return usersDoors;
};
当我插入数据(通过 sequelize)时,sequelize 会向表中添加一个复合主键,其中包括前 2 列(user_uid 和 door_uid)。这破坏了我的东西,我不想要它。
如何在插入数据时阻止 sequelize 创建不需要的主键?
更多细节:
- 方言:postgresql
- sequelize(节点包)版本:3.30.2
- postgresql 版本:psql(9.6.1,服务器 9.5.5)
- 节点版本(这与任何事情无关):6.9.1
最后,这是此模型的 sequelize 迁移:
const tableName = 'users_doors';
module.exports = {
up: (queryInterface, Sequelize) => {
return queryInterface.createTable(tableName, {
user_uid: {
type: Sequelize.UUID,
allowNull: false,
references: {
model: 'users',
key: 'uid',
},
},
door_uid: {
type: Sequelize.UUID,
allowNull: false,
references: {
model: 'doors',
key: 'uid',
},
},
property_manager_uid: {
type: Sequelize.UUID,
allowNull: true,
references: {
model: 'property_managers',
key: 'uid',
},
},
tenant_uid: {
type: Sequelize.UUID,
allowNull: true,
references: {
model: 'tenants',
key: 'uid',
},
},
created_at: {
type: Sequelize.INTEGER,
allowNull: false,
},
})
.then(() => {
return queryInterface.addIndex('users_doors',
['user_uid', 'door_uid', 'property_manager_uid', 'tenant_uid'],
{
indexName: 'doors_users_indexes',
indicesType: 'UNIQUE',
});
});
},
down: (queryInterface) => {
return queryInterface.dropTable(tableName)
.then(() => {
return queryInterface.removeIndex('users_doors', ['user_uid', 'door_uid', 'property_manager_uid', 'tenant_uid']);
});
},
};
【问题讨论】:
标签: node.js postgresql sequelize.js