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Replica of "for" loop in mongoose aggregation

I have some code as this

const quizzes = await Quiz.find({});
        const filtered = [];
        for (const quiz of quizzes){
            const response = await Response.aggregate([
                {
                    $match: {
                        userID: `${userID}`, 
                        quizID: `${quiz._id}`
                    }
                }
            ]);
            if (response.length === 0){
                filtered.push(quiz._id);
            }
        }
       console.log(filtered);

I want to write a aggregate method to replace my for loop, how can I achieve this? I assume there is no requirement to post my schema's, all I need is help with how to traverse a whole collection items and perform some query on them with some other collection.

over 4 years ago · Santiago Trujillo
2 Respuestas
Responde la pregunta

0

  • $lookup to join response collection and match quizID and userID not equal to input userID
  • $match if returned response from lookup is not empty
  • $group by null and construct array of quizID in filtered
const response = await Quiz.aggregate([
  {
    $lookup: {
      from: "response",
      let: { quizID: "$_id" },
      pipeline: [
        {
          $match: {
            userID: { $ne: userID },
            $expr: { $eq: ["$$quizID", "$quizID"] }
          }
        }
      ],
      as: "response"
    }
  },
  { $match: { response: { $ne: [] } } },
  {
    $group: {
      _id: null,
      filtered: { $push: "$_id" }
    }
  }
]);

const filtered = response.length ? response[0].filtered : [];
console.log(filtered);

Playground

over 4 years ago · Santiago Trujillo Denunciar

0

I suppose, it would be great if you have provided models as well, but anyway if you want to remove for loop here what I suggest:

quizz_ids = quizzes.map(q => q._id);
const responses = await Response.aggregate([
            {
                $match: {
                    userID: `${userID}`, 
                    quizID: { $in: quizz_ids}
                }
            }
        ]);
over 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
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