等待来自链状sequelize函数异步响应

Question

我是新来的异步和Web应用程序开发，我无法想出一个办法，我的网页呈现之前从一个函数返回的数据。此外新作sequelize和格式化或最佳做法的任何反馈意见表示赞赏。

我曾尝试设置此功能为这实际上不会返回数据，如果我把它作为响应res.send（配方）的路线。但我想它作为我的网页被呈现之前，我可以调用一个函数

getRecipes.js

const sequelized = require('sequelize'); 
const op = sequelized.Op;

async function getRecipes(){
    //SELECT * FROM ingredients
    ingredients.findAll({ 
        where: {}, 
        raw : true 
    }) 
    .then(ingredients_result =>{ 
        //Get ingredient that expires soon
        //Find recipes of the ingredient that expires soon
        recipe_ingredient.findAll({ 
            where: {}, 
            raw: true 
        }) 
        .then(recipe_ingrdient_result =>{ 
            //If we have all ingredients for a recipe then find name of that recipe by ID
            recipes.findAll({ 
                where: {recipe_id: {[op.in]: suggested_recipes}} 
            }) 
            .then(recipes =>{
                someinfo = JSON.stringify(recipes);
                // This is where i need the date to be returned from
                return someinfo; // But every time i load a page this returns undefined
            }) 
        })
    })
}

module.exports.getRecipes = getRecipes;

路线/ user.js的

const getStuff = require('./getRecipes');
router.get('/dashboard', async function(req, res){
    //This returns undefined
    var r_name = await getStuff.getRecipes();
    res.render('dashboard',{
            title:"Dashboard",
        });
    })

我可能误解了如何异步工作，所以任何帮助，将不胜感激！我知道希望能够通过运行getRecipes功能的页面获取呈现之前检索结果。

Answer 1

首先，如果你所做的功能async然后用它（DO READ）：

这就是你应该在你的代码中使用async/await：

async function getRecipes() {
    //SELECT * FROM ingredients
    let ingredients_result = await ingredients.findAll({ // <------- CHANGE IS HERE
        where: {},
        raw: true
    });

    //Get ingredient that expires soon
    //Find recipes of the ingredient that expires soon
    let recipe_ingrdient_result = await recipe_ingredient.findAll({ // <------- CHANGE IS HERE
        where: {},
        raw: true
    });

    //If we have all ingredients for a recipe then find name of that recipe by ID
    let recipes_result = await recipes.findAll({ // <------- CHANGE IS HERE
        where: {
            recipe_id: {
                [op.in]: suggested_recipes
            }
        }
    })

    let someinfo = JSON.stringify(recipes_result);
    return someinfo; 
}