Question

我正在尝试将现有配方从我的数据库添加到用户，这是我的功能：

exports.addRecipeToUser =  (user,recipeName) => {

  let recipe =  this.findRecipe(recipeName);
  console.log("i am in model",recipe)

  User.updateOne({username: user},{ $push: { recipes: recipe} },function (err) {
    if (err) console.log(err);
  });

}

exports.findRecipe = async (recipeName) => {

  Recipe.findOne({name: recipeName}, function (err, docs) {
    if (err){
        return err;
    }
    else{
      console.log("testing 1111111111",docs);
        return docs;
    }
  });
  }

当我这样调用这个函数时：

model.addRecipeToUser("a@b.com","pancake");

这是我得到的：

i am in model Promise { undefined }
testing 1111111111 {
  categories: [ 'Dessert' ],
  _id: 60cb54b80790970dab918bbc,
  name: 'pancake',
  img: 'https://www.auxdelicesdupalais.net/wp-content/uploads/2020/06/pancakes-fluffy-2.jpg',
  description: 'e2e2e2 dewf; ewk  kks  lel f ',
  ingredients: [
    {
      _id: 60cb54bc0790970dab918bbe,
      name: 'cheese',
      quantity: 2,
      unit: ''
    }
  ],
  steps: [
    {
      _id: 60cb54bf0790970dab918bc0,
      description: 'e2e2e2 dewf; ewk  kks  lel f ',
      img: 'https://www.auxdelicesdupalais.net/wp-content/uploads/2020/06/pancakes-fluffy-2.jpg'
    }
  ],
  __v: 0
}

在robo 3T中，保存的值为Null，我在没有异步的情况下尝试过，但得到了相同的结果，我查看了其他问题并尝试了答案，但仍然未定义。为什么 findRecipe 在 console.log 之后被调用？我该怎么做才能得到正确的对象？

Answer 1

您将回调与 async/await 混合使用，并且未正确使用 async/await。

exports.addRecipeToUser = async (user,recipeName) => {
 try {
    const recipe = await Recipe.findOne({name: recipeName});
    await User.updateOne({username: user}, { $push: { recipes: 
     recipe} });
 } catch(error) {
        console.error(error)
 }
}

Answer 2

我真的想通了，我希望这会帮助某人：

exports.addRecipeToUser = async  (user,recipeName) => {

  let recipe = await  Recipe.findOne({name: recipeName}, function (err, docs) {
    if (err){
        return err;
    }
    else{
      console.log("testing 1111111111",docs);
        return docs;
    }
  });
  console.log("i am in model",recipe)

  User.updateOne({username: user},{ $push: { recipes: recipe} },function (err) {
    if (err) console.log(err);
  });

}

猫鼬中的 findOne 返回空对象，然后返回正确的对象

2 个答案: