'How to retreive an object from an array of Objects in mongodb given a objectid
I have an array of reviews, I want to retrieve only a review from an array of objects inside a schema.
Here is my schema:
const sellerSchema = new mongoose.Schema({
user: {
type: mongoose.Schema.Types.ObjectId,
ref: "User",
unique: true,
},
reviews: [
{
by: {
type: mongoose.Schema.Types.ObjectId,
ref: "User",
unique: true,
},
title: {
type: String,
},
message: {
type: String,
},
rating: Number,
imagesUri: [{ String }],
timestamp: {
type: Date,
default: Date.now,
},
},
],
});
How can I get a single review from the array if 'by' will be req.user._id. I have the previous code, but it is not working to retrieve the review only that satisfies the query.
try {
const seller_id = mongoose.Types.ObjectId(req.params._id);
const review = await Seller.findOne(
{ _id: seller_id },
{ reviews: { $elemMatch: { by: req.user._id } } } //get the review that matches to the user_id
);
res.status(200).send(review);
} catch (err) {
//sends the status code with error code message to the user
res.status(502).send({
error: "Error retreiving review.",
});
}
This retrieves the whole seller document, but I just want to retrieve the object review with the given a user_id === by: ObjectID
Solution 1:[1]
give it a try
try {
const seller_id = mongoose.Types.ObjectId(req.params._id);
const review = await Seller.findOne(
{ _id: seller_id , "reviews.by": req.user._id}, { "reviews.$": 1 }
);
res.status(200).send(review);
}
Solution 2:[2]
Try to cast also the user _id as ObjectId and unify your conditions into a single object:
try {
const seller_id = mongoose.Types.ObjectId(req.params._id);
const user_id = mongoose.Types.ObjectId(req.user._id);
const review = await Seller.findOne({
_id: seller_id,
reviews: { $elemMatch: { by: user_id } },
});
res.status(200).send(review);
} catch (err) {
//sends the status code with error code message to the user
res.status(502).send({
error: 'Error retreiving review.',
});
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Muhammad Umar |
| Solution 2 | lpizzinidev |
