如何在javascript中运行异步inside.map？

我有一个数据库选择函数，我希望在其中具有以下输出格式：

{
  "recipe_id" : 1,
  "recipe_name": "Spaghetti Bolognese",
  "created_at": "2021-01-01 08:23:19.120",
  "steps": [
    {
      "step_id": 11,
      "step_number": 1,
      "step_instructions": "Put a large saucepan on a medium heat",
      "ingredients": []
    },
    {
      "step_id": 12,
      "step_number": 2,
      "step_instructions": "Add 1 tbsp olive oil",
      "ingredients": [
        { "ingredient_id": 27, "ingredient_name": "olive oil", "quantity": 0.014 }
      ]
    },
  ]
}

我创建了以下数据库函数来执行步骤部分：

const getIngredientsStepById = async (step_id) =>{
  const data = await db('step_ingredient_info as sii')
    .join('steps', 'steps.step_id', 'sii.step_id')
    .join('ingredients', 'ingredients.ingredient_id', 'sii.ingredient_id')
    .where('steps.step_id', step_id)

  let returnObj = data[0]? data.map(ingredient => {
     return {
      ingredient_id: ingredient.ingredient_id,
      ingredient_name: ingredient.ingredient_name,
      quantity: ingredient.quantity
    }
  }) : []

  return returnObj
}

它提供以下输出

[
    {
        "ingredient_id": 1,
        "ingredient_name": "egg",
        "quantity": "2 eggs"
    },
    {
        "ingredient_id": 2,
        "ingredient_name": "salt",
        "quantity": "a pinch of salt"
    }
]

我试图将其与当前代码结合起来：

const getRecipeById = async recipe_id => {

  let data = await db('recipes as r')
    .join('steps as st', 'st.recipe_id', 'r.recipe_id')
    .where('r.recipe_id', recipe_id)
    .orderBy('st.step_number')

  let recipeObj = {
    recipe_name: data[0].recipe_name,
    steps:
      data[0].step_id ? data.map(step =>{
        return {
          step_id: step.step_id,
          step_number: step.step_number,
          instructions: step.step_instructions
        }
      }) : []
  };

  return recipeObj
}

它目前有这个输出，这是好的：

{
    "recipe_name": "fried egg",
    "steps": [
        {
            "step_id": 1,
            "step_number": 1,
            "instructions": "add oil and heat pan"
        },
        {
            "step_id": 2,
            "step_number": 2,
            "instructions": "crack egg into pan in medium heat for 2-3 mins and sprinkle salt"
        },
        {
            "step_id": 3,
            "step_number": 3,
            "instructions": "serve"
        }
    ]
}

但现在，我尝试使用async获取成分，以便在步骤中添加成分数组，代码在其中停止工作，它显示在控制台中，但没有得到输出

使用async的代码是：

const getRecipeById = async recipe_id => {

  let data = await db('recipes as r')
    .join('steps as st', 'st.recipe_id', 'r.recipe_id')
    .where('r.recipe_id', recipe_id)
    .orderBy('st.step_number')

  let recipeObj = {
    recipe_name: data[0].recipe_name,
    steps:
      data[0].step_id ? data.map(async step =>{
        let ingredients = await getIngredientsStepById(step.step_id);
        console.log(ingredients);
        return {
          step_id: step.step_id,
          step_number: step.step_number,
          instructions: step.step_instructions,
          ingredients: ingredients
        }
      }) : []
  };

  return recipeObj
}

---

map

，

forEach

，

reduce

等，

async

将无效

但是看看你的代码，你可以把第二阶段变成一个简单的

const getRecipeById = async recipe_id => {

  let data = await db('recipes as r')
    .join('steps as st', 'st.recipe_id', 'r.recipe_id')
    .where('r.recipe_id', recipe_id)
    .orderBy('st.step_number')

  let recipeObj = {
    recipe_name: data[0].recipe_name,
    steps:
      data[0].step_id ? data.map(step =>{
        return {
          step_id: step.step_id,
          step_number: step.step_number,
          instructions: step.step_instructions,
        }
      }) : []
  };

  for (const step of recipeObj.steps) {
     step.ingredients = await getIngredientsStepById(step.step_id);
  }

  return recipeObj
}

---

这回答了你的问题吗？

async

函数返回一个承诺，因此

map（）

生成一个承诺数组。你需要调用

Promise。所有的
都会等待结果。你的
map
调用中仍然有一个
async
函数，正如Bergi指出的那样，删除了map调用中的async，但剩下的都是天赐的。非常感谢。我甚至没想过要走这条路。@Bergi干杯，好的老拷贝粘贴…）