typescript - Sequelize 动态选择属性？作品？

Question

面临一个问题 - 是否真的可以将动态选择属性传递给 FindAll 方法？它不起作用，因为它没有得到认可。

selectAttributes= "'x_name','y_name','z_name','start_z'"; //prepare this dynamically
groupBy = ['x_name','y_name','z_name','start_z','start_z']       
const results =  await XX.findAll({
        group:groupBy,           
        attributes:['x_name','y_name','z_name','start_z','start_z',
                                    [Sequelize.fn(Constants.SEQUELIZE_SUM_FUNCTION, 
                                      Sequelize.col(obj.metrics)),obj.metrics]],

        where:result.where , //I have this built dynamically which works great
        order: result.order,//I have this built dynamically which works great

    });

//上面的代码有效并获取select中的所有列。但我想与 group 和 where 以及 order by 一起动态传递列；我在 where/orderby/group by 上取得了成功，但没有选择属性。我正在制作一个字符串并将其作为选择属性传递，但它无法识别它并在上面的代码中仅返回 SUM 列。

如何概括 FindALL 的 SELECT 属性？

Answer 1

可能这会帮助其他希望构建动态选择属性的人。
selectAttributes= build your array here dynamically //prepare this dynamically
groupBy = ['x_name','y_name','z_name','start_z','start_z']

const results = await XX.findAll({
group:groupBy,

attributes:[...selectAttributes, [Sequelize.fn(Constants.SEQUELIZE_SUM_FUNCTION, Sequelize.col(obj.tasks)),obj.tasks]],
where:result.where , //I have this built dynamically which works great
order: result.order,//I have this built dynamically which works great
});