Question 169678 in Mongodb Question posted in
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Can MongoDB filter after populate?

I’d like filter all records with populated attributes in a query. This is TypeScript code:

router.get("/slice", async (req, res) => {
    if (req.query.first && req.query.rowcount) {
        const first: number = parseInt(req.query.first as string);
        const rowcount: number = parseInt(req.query.rowcount as string);
        const question: string = req.query.question as string;
        if (question) {
            let result = await QuestionAnswer.QuestionAnswer.find().populate('question_id').exec();
            let filtered: any[] = [];
            result.forEach((record) => {
                let q = record.get("question_id.question");
                if (q.includes(question))
                    filtered.push(record);
            });
            filtered = filtered.slice(first, first + rowcount);
            res.send(filtered);
        } else {
            let result = await QuestionAnswer.QuestionAnswer.find().skip(first).limit(rowcount).populate('question_id');
            res.send(result);
        }
    } else {
        res.status(404).send();
    }
});

Schema is the following:

const questionSchema = new Schema({
    question: {
        type: String,
        required: true,
    },
    topic_id: {
        type: Types.ObjectId,
        required: true,
        ref: 'topic',
    },
    explanation: {
        type: String,
        required: true,
    },
});

const questionAnswerSchema = new Schema({
    question_id: {
        type: Types.ObjectId,
        required: true,
        ref: 'question',
    },
    answer: {
        type: String,
        required: true,
    },
    isCorrect: {
        type: Boolean,
        required: true,
    },
});

How can I do it in a query instead array filtering?
I found a solution with aggregation and lookup. Is there any way with find?

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    With aggregation I can filter the result with the following code:

        const query = {
            'question_id.question': { $regex: question, $options: 'i' },
    };
    QuestionAnswer.QuestionAnswer.aggregate([
        {
            $lookup: {
                from: 'questions',
                localField: 'question_id',
                foreignField: '_id',
                as: 'question',
            },
        },
        {
            $project: {
                _id: 1,
                answer: 1,
                isCorrect: 1,
                question_id: { $arrayElemAt: ['$question', 0] },
            },
        }
    ]).match(query).exec().then((result) => {
        res.send(result);
    });

  2. 0

    Something like this may work using regex:

    router.get("/slice", async (req, res) => {
  if (req.query.first && req.query.rowcount) {
    const first: number = parseInt(req.query.first as string);
    const rowcount: number = parseInt(req.query.rowcount as string);
    const question: string = req.query.question as string;

    if (question) {
      const filtered = await QuestionAnswer.QuestionAnswer.find({
        'question_id.question': { $regex: question, $options: 'i' }
      }).populate('question_id').skip(first).limit(rowcount).exec();

      res.send(filtered);
    } else {
      const result = await QuestionAnswer.QuestionAnswer.find()
        .skip(first)
        .limit(rowcount)
        .populate('question_id')
        .exec();

      res.send(result);
    }
  } else {
    res.status(404).send();
  }
});

    Specifying "i" in options is for case insensitive filtering.

    Haven’t tested it, but should work as expected.

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