Mysql 使用sequelizejs在第二个表上使用条件进行SQL联接

Mysql 使用sequelizejs在第二个表上使用条件进行SQL联接,mysql,join,sequelize.js,Mysql,Join,Sequelize.js,我有以下型号: const User = sequelize.define('user', { name: Sequelize.TEXT, avatar: Sequelize.TEXT, }); const Event = sequelize.define('event', { title: Sequelize.STRING, description: Sequelize.TEXT, timestamp: Sequelize.DATE, }); con

我有以下型号:

const User = sequelize.define('user', {
    name: Sequelize.TEXT,
    avatar: Sequelize.TEXT,
});

const Event = sequelize.define('event', {
    title: Sequelize.STRING,
    description: Sequelize.TEXT,
    timestamp: Sequelize.DATE,
});

const EventUser = sequelize.define('event_user', {});

Event.hasMany(EventUser);
EventUser.belongsTo(User);
因此,
EventUser
持有我的特定事件参与者列表

我正在尝试查询用户X参与的所有事件:

let events = await Event.findAll({
    include: [
        {model: EventUser, include: [{model: User, where: {id: 1}}]}
    ],
});

但它不会获取该活动的其他参与者名单。不知道如何才能做到,谢谢

正确的方法是定义用户和事件之间的多对多关系

User.belongsToMany(Event, {through: 'EventUser'});
Event.belongsToMany(User, {through: 'EventUser'});
然后,您可以获得某些用户的所有事件,如下所示:

User.findById(userId).then(function(user){
    return user.getEvents();
}).then(function(events){
    //here's events
});
或通过一个查询:

Event.findAll({
    include: [{
        model: User,
        where: {id: userId}
    }]
}).then(function(events){
    //here's events
});

“一个查询”选项不起作用(它只返回该用户作为参与者,而不是所有参与者),但第一个选项非常有效,谢谢!