Example: Mongo db collection:
[
{
"user_id": 1,
"lead_id": 901,
...
...
},
{
"user_id": 2,
"lead_id": 902,
...
...
},
{
"user_id": 2,
"lead_id": 903,
...
...
},
{
"user_id": 3,
"lead_id": 999,
...
...
},
]
I have an array of objects:
const reqArr = [
{
"user_id": 1,
"lead_id": 901,
},
{
"user_id": 2,
"lead_id": 999,
}
]
I want to query the DB based on both user_id and lead_id present in each object in the array reqArr. something like this or with aggregate pipeline
Model.find(reqArr)
My desired output would be something like this:
[{
"user_id": 1,
"lead_id": 901,
...
...
}]
I only know how to query based on the array of one field like Model.find({_id: {"$in": [array of ids ]}}) but not with more than one field in an array of object.
Any idea or workaround to achieve something like the above output or negation of the above output? I only need which objects in reqArr are present in db collection or which are not. Thank you.
Santiago Trujillo
The simplest way to do the above would be this:
const results = await Promise.all(reqArr.map(req => {
const { user_id, lead_id } = req;
return Model.findOne({ user_id, lead_id });
}));
We map over each item in reqArr. For each item (req), we destructure into two variables user_id and lead_id.
We then query for a single document (I assume there is only one document for each user_id and lead_id pair, otherwise this will need to be slightly adjusted.
We then return the findOne promise, resulting in an array of promises which we pass into Promise.all to await them all asynchronously.
This should result flat array of results that match the lead_iq/user_id pairs in reqArr.
If some of these combos aren't present in the database, those elements of the array will be undefined and you can go on to handle those however you need to.
simple $or
db.collection.aggregate([
{
"$match": {
"$or": [
{
"user_id": 1,
"lead_id": 901
},
{
"user_id": 2,
"lead_id": 902
}
]
}
}
])
node
const reqArr = [
{
"user_id": 1,
"lead_id": 901,
},
{
"user_id": 2,
"lead_id": 999,
}
]
Model.find({ "$or": reqArr})