Question

我正在尝试在一个集合中查找find（）并将该值连接到随后的find（）返回的正确对象。但是我不能保证在第一个find（）中检索到的值将与index匹配。如何确保将第一个值（布尔值）附加到第二个find（）的正确实例上？

到目前为止，我所使用的是第一个条件的索引，但是如果实例被删除，它可能不匹配

我的模特：

let Instance= new Schema({
  imgName: String,
  offer: String,
  brand: String,
  desc: String,
  keywords: Array,
  loc: String,
  location: {
    type: {
      type: String, // Don't do `{ location: { type: String } }`
      enum: ['Point'], // 'location.type' must be 'Point'
      required: true
    },
    coordinates: {
      type: [Number],
      required: true
    }
  },
  categories: Array,
  userId: String,
  active: Boolean,
  stockPic: String,
  startTime: Number,
  endTime: Number,
  range: Boolean
});

mongoose.model('Beams2', Instance);

let LikesSchema = new Schema({
  userId: String,
  likeId: String,
  categories: Array,
  public: Boolean
});

mongoose.model('Likes', LikesSchema);

//My query:

exports.findAllLikes = function(req, res){
  Likes.find({'userId': req.body.userId}, function(err, results) {
    let likeIds = results.map((currentValue, index, array) => {
        return currentValue.likeId;
    });

    let statusArr = results.map((currentValue, index, array) => {
        return currentValue.public;
    });

    Instances.find({'_id': { $in: likeIds }}, function(err, favs){
      if (err) return console.log(err);

      let newArr = [];
      favs.forEach(function(element, i) { 
        //console.log(statusArr[i]);
        let post = element.toObject();
        post.public = statusArr[i]; //need to guarantee correct value
        newArr.push(post);
        console.log(post);
      });

      return res.send(newArr);
    });

  });
};

Answer 1

我认为，这是您可以实现想要的目标的方法：

exports.findAllLikes = function(req, res) {
    Likes.find({
        'userId': req.body.userId
    }, (err, results) => {
        if (err) {
            console.log('Error while fetching likes: ', err);
            return res.status(500).json({
                Error: 'Error while fetching likes.'
            });
        }
        const likeIds = results.map(({
            likeId
        }) => likeId);

        Instances.find({
            '_id': {
                $in: likeIds
            }
        }, (err, favs) => {
            if (err) {
                console.log('Error while fetching likes: ', err);
                return res.status(500).json({
                    Error: 'Error while fetching Instances.'
                });
            }
            let newArr = [];
            favs.forEach((elem) => {
                let post = elem.toObject();
                const likeDoc = results.find(({
                    likeId
                }) => likeId === post._id.toString());
                post.public = (likeDoc && likeDoc.public) ? likeDoc.public : '';
                newArr.push(post);
            });
            console.log(post);
            return res.status(200).send(newArr);
        });
    });
};

除非您确定自己具有相同的长度数组，否则不要使用索引来映射内容。

希望这会有所帮助：）

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1 个答案: