How find in mongoose by an array property - mongodb

Question

I have defined my Conversation scheme like this:
const { Schema, model } = require("mongoose");
const ConversationSchema = Schema(
{
members: {
type: Array,
},
},
{ timestamps: true }
);
module.exports = model("Conversation", ConversationSchema);
My problem is that when I want to create a conversation model I search first if there is already a conversation.
const newConversation = async (req, res = response) => {
try {
const { senderId, receiverId } = req.body;
const conversation = await Conversation.find({
members: { $in: [senderId, receiverId] },
});
if (conversation.length === 0) {
const dbConversation = new Conversation({
members: [senderId, receiverId],
});
await dbConversation.save();
return res.status(201).json({
ok: true,
conversation: dbConversation
});
} else {
return res.status(403).json({
ok: false,
msg: "Conversation already exist",
});
}
} catch (err) {
return res.status(500).json({
ok: false,
msg: "Please contact with administrator",
});
}
};
senderId and receivedId are the ids of the users that are in that conversation, but it doesn't work.
How can I make it check if there is already a conversation with both ids?

Answer 1

Per the comments, we came to understand that the thing that wasn't working about the current code was always taking the code path that returned the message that the "Conversation already exist". This meant that the following query was always returning data:
const conversation = await Conversation.find({
members: { $in: [senderId, receiverId] },
});
The logic here does not match the logic implied in the question. This syntax uses the $in operator to find documents whose members array has at least one of the values passed to it (here the senderId and the receiverId).
To instead find documents where both of those people are present in the members array, you want to use the $all operator instead:
const conversation = await Conversation.find({
members: { $all: [senderId, receiverId] },
});
Working Mongo Playground example here.