When I run the following query I am getting the document that matches as normal which is "LON" in this case.
But is there any way that I can make the response seperately return the values that didn't match or found, which is "BUJ" in this case. Instead of running a for loop for individual values.
ports = [
"LON",
"BUJ"
];
findDatas = async(coll, values, key) => {
let datas = await coll.find({[key] : values});
// let datas = await coll.find().where(key).in(values);
console.log(datas)
}
findDatas(airportsModel, ports, "iata_code")
In my DB I only have "LON" which mean "BUJ" is not found. So is there any way to make mongo to tell that the given values haven't been found? along with the found ones.
This code dynamically creates a $match stage to find the documents, the uses $facet to split into 2 pipelines, the first returns the documents, the second uses a $group stage created from the input array to count how many documents match each element. The result should be a document with 2 fields: documents and counts
matchdata={};
matchdata[key]={"$in":ports};
groupdata = {_id:null};
ports.forEach(function(p){
groupdata[p] = {"$sum":{"$cond":[{"$eq":["$" + key, p ]},1,0]}}
});
db.coll.aggregate([
{$match:matchdata},
{$facet:{
documents:[{$match:{}}],
counts:[{$group: groupdata},{$project:{_id:0}}]
}}
])
Related
Hi I have two mongodb collections. The first one returns json data (array) and with the output of this, I want to return documents that match.
When I run Console.log (req.bidder.myBids) I get the following output:
[{"productId":"3798b537-9c7b-4395-9e41-fd0ba39aa984","price":3010},{"productId":"3798b537-9c7b-4395-9e41-fd0ba39aa984","price":3020},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1040},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1050},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1060},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1070},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1090},{"productId":"4c4bd71c-6664-4d56-b5d3-6428fe1bed19","price":1100}]
The productId has duplicates, I want to remove duplicates and then call a routine that finds all the products that match and output as json.
So far I have this code that only outputs one document, but cant figure out how to add the array of productId's and then fetch all corresponding products.
var agencyId = req.body.agencyId;
var productId = req.body.productId;
if (!validate.STRING(agencyId)) {
res.apiError(messages.server.invalid_request);
} else {
dbProduct.find({productId:{$in:['3798b537-9c7b-4395-9e41-fd0ba39aa984','4c4bd71c-6664-4d56-b5d3-6428fe1bed19']}
}).then(dbRes => {
console.log(dbRes);
Updated code and works with hard-wired productId and updated above code. Looking at how to get the array data and transpose replacing the hard-wired productId's
The $in operator is what you want. See the docs here: https://docs.mongodb.com/manual/reference/operator/query/in/
Really don't know how to describe my problem. I tried my best. I have a collection and I want anyone to search through multiple parameters. I used the $match filter for this. here is my code.
db.product.aggregate([{$match:{lastStatus:req.query.status,name:req.query.name}}])
This is working fine. but I am taking the match values from the user via API and there is a chance that the user didn't pass these values, in that case, I return the products based on the applied match. suppose the user only passes the name, then the result comes on the basis of the name, not on lastStatus. right now this query does not work if any of the provided values is empty or undefined.
what can be the solutions
You can check the condition if any of the field has value then set the value in the match variable.
let match = {};
// MATCH STATUS
if (req.query.status) match["lastStatus"] = req.query.status;
// MATCH NAME
if (req.query.name && req.query.name.trim() != "") match["name"] = req.query.name.trim();
db.product.aggregate([{ $match: match }]);
I have a Post model that has a publisher property defined in its schema (I'm using Mongoose). The publisher property is a string that refers to a publisher's name.
I also have an array called sourceNames that holds all the different publisher names. I want to query my database for ALL the posts whose publisher matches any one of the array elements in sourceName. My current query looks like this:
const query = postModel
.find({ publisher: { $all: sourceNames } })
.limit(limit)
.skip(startIndex);
My query isn't returning anything when I exec it. Does anyone know if what I'm trying to do is possible in a single query (Rather than loop over sourceNames and make a query for each individual element?
Short
Just replace $all with $in
Expl
$all is trying to match an array with all elements in your array.
$in instead, tries to match a string or array with one in the array.
This is the concerned part from the schema
`
var CandidateSchema = new Schema({
calculateScore:[{
jobname:{type:Schema.ObjectId,ref: 'Job'}
,Score:{type:Number,default:0}
}]
})
`
A candidate can apply to multiple jobs and get scored differently for different jobs. I want to sort the candidates depending on the specific job's Score. Any Idea?
Assuming the variable objectId holds the ObjectId of the referred Job, you can aggregate the records to get the records sorted by the score of that particular Job.
Since the stage operator $project does not support the $elemeMatch operation, we cannot use it to directly get the Job sub document that we want and sort based on it.
$project a separate field named temp_score to have a copy of the original calculateScore array.
$redact other sub documents from calculateScore other than whose jobname contains the
id we are looking for. Now calculateScore will contain only one
element in the array, i.e the element whose jobname is the id
that we have specified.
Based on this sub document's score sort the records in descending
order.
Once the sorting is done, project our original calculatescore
field, which is in temp_score.
The code:
var objectId = ObjectId("ObjectId of the referred Job"); // Needs to be fetched
// from the Job collection.
model.aggregate(
{$project:{"temp_score":{"level":{$literal:1},
"calculateScore":"$calculateScore"},
"calculateScore":1}},
{$redact:{$cond:[
{$and:[
{$eq:[{$ifNull:["$jobname",objectId]},objectId]},
{$ne:["$level",1]}
]
},
"$$DESCEND",
{$cond:[{$eq:["$level",1]},
"$$KEEP","$$PRUNE"]}]}},
{$sort:{"calculateScore.Score":-1}},
{$project:{"_id":1,
"calculateScore":"$temp_score.calculateScore"}},
function(err,res)
{
console.log(res);
}
);
I'm performing this mongo query:
db.mytable.find({id:{$in:["1","2", "3", "4" ]}});
It returns all results in a strange order, as it follows:
4,3,2,1
I need to retrieve all results in same order as it was defined in the query array.
1,2,3,4
Is it possible ?
There is indeed no guarantee about the order of results returned from your query, but you could do a sort afterwards with the result. Two examples, the first one with the order you wanted, the second one reversed.
const arr = ["1", "2", "3", "4" ];
db.collection.find({ id: { $in: arr }})
.then(result => {
let sorted = arr.map(i => result.find(j => j.id === i));
console.log(sorted) // 1, 2, 3, 4
let reversed = arr.reverse().map(i => result.find(j => j.id === i));
console.log(reversed) // 4, 3, 2, 1
});
In case you want to do real MongoDB ID lookups, use db.collection.find({ _id: { $in: arr }}) and .map(i => result.find(j => j._id == i)) (Notice the two equal signs instead of three)
A couple of things to note:
1.) MongoDB, like most databases, makes no guarantees about the order of results returned from your query unless you use a call to sort(). If you really want to guarantee that your query result is returned in a a specific order, you'll need to specify that specific sort order.
2.) In general, the most recently updated/moved document will show up at the end of your result set but there are still no guarantees. MongoDB uses "natural order" for its native ordering of objects and although this is very close to the order of insertion, it is not guaranteed to be the same.
3.) Indexed fields will behave differently. It's worth pointing out that it looks like your query is using id and not _id. The former, _id would be indexed by default and id would not be indexed unless you've explicitly added an index to that field.
You can read more about MongoDB's sorting and ordering here:
http://www.mongodb.org/display/DOCS/Sorting+and+Natural+Order
you can write a query like this :
db.mytable.find({id:{$in:["1","2", "3", "4" ]}}).sort({id:1})
To have your results ORDER BY id ASC
Source : MongoDB Advanced Queries
But if you just want to order the results based on your $in array, try to sort your $in array in the reverse order, the result regarding to the first element of your $in array is likely to appear as the last element of the results