javascript - 从 SequelizeJS 中的表中选择特定属性

Question

那些是我的表(不包括所有列)和关系

var client = schema.define('client', {
    name: {
        type: Sequelize.STRING,
        allowNull: false
    },
}

var task = schema.define('task', {
    name: {
        type: Sequelize.STRING,
        unique: true,
        allowNull: false
    },
    description: {
        type: Sequelize.STRING,
    }
}

var clientTask = schema.define('clientTask', {
    value: {
        type: Sequelize.STRING,
        allowNull: false,
        defaultValue: false
    },
}

client.belongsToMany(task, { through: clientTask });
task.belongsToMany(client, { through: clientTask });

我只想从任务中获取 name 并从 clientTask 中获取 value，我通过客户端 ID 搜索，这是我目前尝试的方法。

client.findAll({ 
    attributes: [], 
    where: {id: clientId}, 
    include: [{ 
        model: task, 
        attributes: ['name'] 
    }]
}).then(function (clients) { 
    //client.tasks is array with task objects(models) with only name attribute
    //client.tasks[0].clientTask is object(models) with all attributes but I want only `value`
}

基本上我想要的是这个查询

Select
  tasks.name,
  clienttasks.value
From
  clients Inner Join
  clienttasks
    On clienttasks.clientId = clients.id Inner Join
  tasks
    On clienttasks.taskId = tasks.id
Where clients.id = ?

Answer 1

你可以这样查询

clients.findById(1, {
    attributes: ['id'],
    include: [{
        model: tasks,
        attributes: ['name'],
        required: false
    }]
}).then(function(client) {
    return client.tasks.map(function(task) {
        return {
            name: task.name,
            value: task.clients_tasks.value
        };
    });
}).then(function(result) {
    console.log(result);
    // The rest of you logic here...
});