I'm trying to find a better performing solution maybe where you don't take the whole string array from collection and instead just return the count. Basically the countDocuments equivalent for a string array in one user collection.
I have a solution that works, but if that list gets really long then it won't be as good performance-wise.
User.findById(
{ _id: req.params.id }, { _id: 0 })
.select("blacklistGroup").then((result) => {
let output = JSON.parse(JSON.stringify(result.blacklistGroup));
console.log(output.length + " + people");
});
I was hoping to get an example of a more efficient solution more aligned with countDocuments, but for string array. I appreciate the help!
Here is the model definition of blacklistGroup
blacklistGroup: [String]
collection example
"blacklistGroup": [
"5e98e8e785e69146f841d239",
"5e9867ff5e72550988820dd3",
"5e98e90d85e69146f841d23a",
"5e98c950f4fb3f63b4634e30",
"5e99fcf3a506cf570056898b",
"5e99fd15a506cf570056898c",
"5e99fd3aa506cf570056898d",
"5e99fd64a506cf570056898e",
"5e99fda5a506cf570056898f"
]
Try This One:
User.aggregate([
{
$match: { _id: req.params.id }
},
{
$project: { blacklistGroupLength: { $size: '$blacklistGroup' } }
}
])
I've found something from MongoDB docs that might help you. Here is the link
Related
I'm trying to return all documents which satisfy document.field.value > document.array.length. I'm trying to do this using MongoClient in ExpressJS and was not able to find answers on SO.
Here is what I have tried but this gives an empty list.
const db = ref()
const c = db.collection(t)
const docs = await c.find({
"stop.time":{$gt: new Date().toISOString()},
"stop.expect": {$gt: { $size: "stop.riders"}}
})
console.log(docs)
Also tried replacing
"stop.expect": {$gt: { $size: "stop.riders"}}
with the below code which does not compile
$expr: {$lt: [{$size: "stop.riders"}, "stop.expect"]}
Sample data:
{ "stop": { "expect":3, "riders": ["asd", "erw", "wer"] } },
{ "stop": { "expect":4, "riders": ["asq", "frw", "wbr"] } }
To filter the query with a complex query involving the calculation, you need to use the $expr operator, which allows the aggregation operators.
Next, within the $expr operator, to refer to the field, you need to add the prefix $.
db.collection.find({
$expr: {
$lt: [
{
$size: "$stop.riders"
},
"$stop.expect"
]
}
})
Demo # Mongo Playground
To compare fields within a document to each other, you must use $expr. It would look like this:
const docs = await c.find({
$expr: { $gt: ["$stop.expect", { $size: "$stop.riders" }] },
});
I omitted the stop.time condition because it's not in your sample data and it's unclear what type of field it is, if it actually is in your data (whether it's a Date or String would be very important).
Note that this sort of query is unable to use an index.
This is a Controller in which I'm trying to catch multiple candidates id(ObjectId) and try to store it in the database in the array Candidates. But data is not getting pushed in Candidates column of Array type.
routes.post('/Job/:id',checkAuthenticated,function(req,res){
var candidates=req.body.candidate;
console.log(candidates);
Job.update({_id:req.params.id},{$push:{Appliedby : req.user.username}},{$push:{Candidates:{$each:
candidates}}}
});
Console screens output
[ '5eb257119f2b2f0b4883558b', '5eb2ae1cff3ae7106019ad7e' ] //candidates
you have to do all the update operations ($set, $push, $pull, ...) in one object, and this object should be the second argument passed to the update method after the filter object
{$push:{Appliedby : req.user.username}},{$push:{Candidates:{$each: candidates}}
this will update the Appliedby array only, as the third object in update is reserved for the options (like upsert, new, ....)
you have to do something like that
{ $push: { Appliedby: req.user.username, Candidates: { $each: candidates } } }
then the whole query should be something like that
routes.post('/Job/:id', checkAuthenticated, function (req, res) {
var candidates = req.body.candidate;
console.log(candidates);
Job.update(
{ _id: req.params.id }, // filter part
{ $push: { Appliedby: req.user.username, Candidates: { $each: candidates } } } // update part in one object
)
});
this could do the trick I guess, hope it helps
If I have this schema...
person = {
name : String,
favoriteFoods : Array
}
... where the favoriteFoods array is populated with strings. How can I find all persons that have "sushi" as their favorite food using mongoose?
I was hoping for something along the lines of:
PersonModel.find({ favoriteFoods : { $contains : "sushi" }, function(...) {...});
(I know that there is no $contains in mongodb, just explaining what I was expecting to find before knowing the solution)
As favouriteFoods is a simple array of strings, you can just query that field directly:
PersonModel.find({ favouriteFoods: "sushi" }, ...); // favouriteFoods contains "sushi"
But I'd also recommend making the string array explicit in your schema:
person = {
name : String,
favouriteFoods : [String]
}
The relevant documentation can be found here: https://docs.mongodb.com/manual/tutorial/query-arrays/
There is no $contains operator in mongodb.
You can use the answer from JohnnyHK as that works. The closest analogy to contains that mongo has is $in, using this your query would look like:
PersonModel.find({ favouriteFoods: { "$in" : ["sushi"]} }, ...);
I feel like $all would be more appropriate in this situation. If you are looking for person that is into sushi you do :
PersonModel.find({ favoriteFood : { $all : ["sushi"] }, ...})
As you might want to filter more your search, like so :
PersonModel.find({ favoriteFood : { $all : ["sushi", "bananas"] }, ...})
$in is like OR and $all like AND. Check this : https://docs.mongodb.com/manual/reference/operator/query/all/
In case that the array contains objects for example if favouriteFoods is an array of objects of the following:
{
name: 'Sushi',
type: 'Japanese'
}
you can use the following query:
PersonModel.find({"favouriteFoods.name": "Sushi"});
In case you need to find documents which contain NULL elements inside an array of sub-documents, I've found this query which works pretty well:
db.collection.find({"keyWithArray":{$elemMatch:{"$in":[null], "$exists":true}}})
This query is taken from this post: MongoDb query array with null values
It was a great find and it works much better than my own initial and wrong version (which turned out to work fine only for arrays with one element):
.find({
'MyArrayOfSubDocuments': { $not: { $size: 0 } },
'MyArrayOfSubDocuments._id': { $exists: false }
})
Incase of lookup_food_array is array.
match_stage["favoriteFoods"] = {'$elemMatch': {'$in': lookup_food_array}}
Incase of lookup_food_array is string.
match_stage["favoriteFoods"] = {'$elemMatch': lookup_food_string}
Though agree with find() is most effective in your usecase. Still there is $match of aggregation framework, to ease the query of a big number of entries and generate a low number of results that hold value to you especially for grouping and creating new files.
PersonModel.aggregate([
{
"$match": {
$and : [{ 'favouriteFoods' : { $exists: true, $in: [ 'sushi']}}, ........ ] }
},
{ $project : {"_id": 0, "name" : 1} }
]);
There are some ways to achieve this. First one is by $elemMatch operator:
const docs = await Documents.find({category: { $elemMatch: {$eq: 'yourCategory'} }});
// you may need to convert 'yourCategory' to ObjectId
Second one is by $in or $all operators:
const docs = await Documents.find({category: { $in: [yourCategory] }});
or
const docs = await Documents.find({category: { $all: [yourCategory] }});
// you can give more categories with these two approaches
//and again you may need to convert yourCategory to ObjectId
$in is like OR and $all like AND. For further details check this link : https://docs.mongodb.com/manual/reference/operator/query/all/
Third one is by aggregate() function:
const docs = await Documents.aggregate([
{ $unwind: '$category' },
{ $match: { 'category': mongoose.Types.ObjectId(yourCategory) } }
]};
with aggregate() you get only one category id in your category array.
I get this code snippets from my projects where I had to find docs with specific category/categories, so you can easily customize it according to your needs.
For Loopback3 all the examples given did not work for me, or as fast as using REST API anyway. But it helped me to figure out the exact answer I needed.
{"where":{"arrayAttribute":{ "all" :[String]}}}
In case You are searching in an Array of objects, you can use $elemMatch. For example:
PersonModel.find({ favoriteFoods : { $elemMatch: { name: "sushiOrAnytthing" }}});
With populate & $in this code will be useful.
ServiceCategory.find().populate({
path: "services",
match: { zipCodes: {$in: "10400"}},
populate: [
{
path: "offers",
},
],
});
If you'd want to use something like a "contains" operator through javascript, you can always use a Regular expression for that...
eg.
Say you want to retrieve a customer having "Bartolomew" as name
async function getBartolomew() {
const custStartWith_Bart = await Customers.find({name: /^Bart/ }); // Starts with Bart
const custEndWith_lomew = await Customers.find({name: /lomew$/ }); // Ends with lomew
const custContains_rtol = await Customers.find({name: /.*rtol.*/ }); // Contains rtol
console.log(custStartWith_Bart);
console.log(custEndWith_lomew);
console.log(custContains_rtol);
}
I know this topic is old, but for future people who could wonder the same question, another incredibly inefficient solution could be to do:
PersonModel.find({$where : 'this.favouriteFoods.indexOf("sushi") != -1'});
This avoids all optimisations by MongoDB so do not use in production code.
I now use mongooses to pull and pull subdocuments to the array, and now I want to change the contents of the detail field of that subdocument with the _id of the subdocument.
{
subDocument: [{
_id: ObjectId('123'),
detail: 'I want update this part'
}]
}
I tried to use the $set method as shown below but it did not work as expected.
Model.findByIdAndUpdate(uid, { $Set: {subDocument: {_id: _id}}});
Looking at the for statement as shown below is likely to have a bad effect on performance. So I want to avoid this method.
const data = findById(uid);
for(...) {
if(data.subDocument[i]._id==_id) {
data.subDocument[i].detail = detail
}
}
Can you tell me some mongodb queries that I can implement?
And, Is it not better to use the 'for(;;)' statement shown above than to search using mongodb's query?
This should work:
Model.findOneAndUpdate({"subdocument._id": uid},
{
$set: {
"subdocument.$.detail ": "detail here"
}
},
).exec(function(err, doc) {
//code
});
To find subdocument by id, I am using something like this :
var subDocument = data.subDocument.id(subDocumentId);
if (subDocument) {
// Do some stuff
}
else {
// No subDocument found
}
Hope it helps.
I have this collection :
[
{
_id: ObjectId('myId1'),
probes: ['id_probe_1', 'id_probe_2']
},
{
_id: ObjectId('myId2'),
probes: ['id_probe_1', 'id_probe_3']
}
]
I want to get an array like this :
['id_probe_1', 'id_probe_2', 'id_probe_3']
So I try this request (from nodeJS driver) :
let find = [
{
$match: {
_id: {
$in: [new ObjectId('miId1'), new ObjectId('myId2')]
}
}
},
{
$group: {
_id: null,
probes: {
$addToSet: {
$each: '$probes'
}
}
}
}
];
This doesn't work, give me this error :
invalid operator '$each'
From the doc, they mention that it will appends the whole array as a single element.
If the value of the expression is an array, $addToSet appends the whole array as a single element.
But they don't say how to have an unique array. So I use the $each operator like this page indicates (I don't really know what's the difference...)
Is there a way to make this work ?
Thanks !
insert $unwind before $group
{$unwind:"$probes"},
then remove $each
Why don't you try distinct operation? In mongo shell, db.col.distinct('probs'); you can try the distinct function in nodejs mongo driver.