Suppose I want to sort the data based on the current city first and then the remaining country data. Is there any way I achieve that in MongoDB?
Example
[
{ id: 2, name: 'sdf' },
{ id: 3, name: 'sfs' },
{ id: 3, name: 'aaa' },
{ id: 1, name: 'dsd' },
];
What I want as an outcome is the data with id 3 at first and the remaining other.
like
[
{ id: 3, name: 'sfs' },
{ id: 3, name: 'aaa' },
{ id: 1, name: 'dsd' },
{ id: 2, name: 'sdf' },
];
It's just a example,
My actual requirement is to sort the data based on certain category first and then the remaining one
It's not possible within mongodb but you could first fetch the documents from the db and then sort them in Javascript (or whatever other language you're using to present the data).
On a side note, having duplicate values in the "id" field is not a good practice and defies the definition of id itself.
There is no straight way to sort condationaly in MongoDB, as per your example you can try aggregation query,
$facet to separate result for both types of documents
first, to get id: 3 documents
second, to get id is not 3 documents and sort by id in ascending order
$project and $concatArrays to concat both arrays in siquance
$unwind deconstruct all array
$replaceRoot to replace all object to root
db.collection.aggregate([
{
$facet: {
first: [
{ $match: { id: 3 } }
],
second: [
{ $match: { id: { $ne: 3 } } },
{ $sort: { id: 1 } }
]
}
},
{
$project: {
all: { $concatArrays: ["$first", "$second"] }
}
},
{ $unwind: "$all" },
{ $replaceRoot: { newRoot: "$all" } }
])
Playground
Related
{
id: 1,
name: "sree",
userId: "001",
paymentData: {
user_Id: "001",
amount: 200
}
},
{
id: 1,
name: "sree",
userId: "001",
paymentData: {
user_Id: "002",
amount: 200
}
}
I got this result after unwind in aggregation any way to check user_Id equal to userId
Are you looking to only retrieve the results when they are equal (meaning you want to filter out documents where the values are not the same) or are you looking to add a field indicating whether the two are equal?
In either case, you append subsequent stage(s) to the aggregation pipeline to achieve your desired result. If you want to filter the documents, the new stage may be:
{
$match: {
$expr: {
$eq: [
"$userId",
"$paymentData.user_Id"
]
}
}
}
See how it works in this playground example.
If instead you want to add a field that compares the two values, then this stage may be what you are looking for:
{
$addFields: {
isEqual: {
$eq: [
"$userId",
"$paymentData.user_Id"
]
}
}
}
See how it works in this playground example.
You could also combine the two as in:
{
$addFields: {
isEqual: {
$eq: [
"$userId",
"$paymentData.user_Id"
]
}
}
},
{
$match: {
isEqual: true
}
}
Playground demonstration here
I have an issue that need to insert index number when get data. First i have this data for example:
[
{
_id : 616efd7e56c9530018e318ac
student : {
name: "Alpha"
email: null
nisn: "0408210001"
gender : "female"
}
},
{
_id : 616efd7e56c9530018e318af
student : {
name: "Beta"
email: null
nisn: "0408210001"
gender : "male"
}
}
]
and then i need the output like this one:
[
{
no:1,
id:616efd7e56c9530018e318ac,
name: "Alpha",
nisn: "0408210001"
},
{
no:2,
id:616efd7e56c9530018e318ac,
name: "Beta",
nisn: "0408210002"
}
]
i have tried this code but almost get what i expected.
{
'$project': {
'_id': 0,
'id': '$_id',
'name': '$student.name',
'nisn': '$student.nisn'
}
}
but still confuse how to add the number of index. Is it available to do it in $project or i have to do it other way? Thank you for the effort to answer.
You can use $unwind which can return an index, like this:
db.collection.aggregate([
{
$group: {
_id: 0,
data: {
$push: {
_id: "$_id",
student: "$student"
}
}
}
},
{
$unwind: {path: "$data", includeArrayIndex: "no"}
},
{
"$project": {
"_id": 0,
"id": "$data._id",
"name": "$data.student.name",
"nisn": "$data.student.nisn",
"no": {"$add": ["$no", 1] }
}
}
])
You can see it works here .
I strongly suggest to use a $match step before these steps, otherwise you will group your entire collection into one document.
You need to run a pipeline with a $setWindowFields stage that allows you to add a new field which returns the position of a document (known as the document number) within a partition. The position number creation is made possible by the $documentNumber operator only available in the $setWindowFields stage.
The partition could be an extra field (which is constant) that can act as the window partition.
The final stage in the pipeline is the $replaceWith step which will promote the student embedded document to the top-level as well as replacing all input documents with the specified document.
Running the following aggregation will yield the desired results:
db.collection.aggregate([
{ $addFields: { _partition: 'students' }},
{ $setWindowFields: {
partitionBy: '$_partition',
sortBy: { _id: -1 },
output: { no: { $documentNumber: {} } }
} },
{ $replaceWith: {
$mergeObjects: [
{ id: '$_id', no: '$no' },
'$student'
]
} }
])
I have a collection with the following data structure:
{
_id: ObjectId,
text: 'This contains some text',
type: 'one',
category: {
name: 'Testing',
slug: 'test'
},
state: 'active'
}
What I'm ultimately trying to do is get a list of categories and counts. I'm using the following:
const query = [
{
$match: {
state: 'active'
}
},
{
$project: {
_id: 0,
categories: 1
}
},
{
$unwind: '$categories'
},
{
$group: {
_id: { category: '$categories.name', slug: '$categories.slug' },
count: { $sum: 1 }
}
}
]
This returns all categories (that are active) and the total counts for documents matching each category.
The problem is that I want to introduce two additional $match that should still return all the unique categories, but only affect the counts. For example, I'm trying to add a text search (which is indexed on the text field) and also a match for type.
I can't do this at the top of the pipeline because it would then only return categories that match, not only affect the $sum. So basically it would be like being able to add a $match within the $group only for the $sum. Haven't been able to find a solution for this and any help would be greatly appreciated. Thank you!
You can use $cond inside of your $group statement:
{
$group: {
_id: { category: '$categories.name', slug: '$categories.slug' },
count: { $sum: { $cond: [ { $eq: [ "$categories.type", "one" ] }, 1, 0 ] } }
}
}
I have a mongodb collection with structure like that:
[
{
name: "name1",
instances: [{value:1, score:2}, {value:2, score:5}, {value:2.5, score:9}]
},
{
name: "name2",
instances: [{value:6, score:3}, {value:1, score:6}, {value:3.7, score:5.2}]
}
]
When I want to get all the data from a document, I use aggregate because I want each instance returned as a separate document:
db.myCollection.aggregate([{$match:{name:"name1"}}, {$unwind:"$instances"}, {$project:{name:1, value:"$instances.value", score:"$instances.score"}}])
And everything works like I want it to.
Now for my question: I want to filter the returned data by score or by value. For example, I want an array of all the subdocuments of name1 which have a value greater or equal to 2.
I tried to add to the $match object 'instances.value':{$gte:2}, but it didn't filter anything, and I still get all 3 documents for this query.
Any ideas?
After unwinding instances then again used $match as below
db.collectionName.aggregate({
"$match": {
"name": "name1"
}
}, {
"$unwind": "$instances"
}, {
"$match": {
"instances.value": {
"$gte": 2
}
}
}, {
$project: {
name: 1,
value: "$instances.value",
score: "$instances.score"
}
})
Or if you tried $match after project then used as below
db.collectionName.aggregate([{
$match: {
name: "name1"
}
}, {
$unwind: "$instances"
}, {
$project: {
name: 1,
value: "$instances.value",
score: "$instances.score"
}
}, {
"$match": {
"value": {
"$gte": 2
}
}
}])
I am implementing a small application using mongodb as a backend. In this application I have a data structure where the documents will contain a field that contains an array of subdocuments.
I use the following use case as a basis:
http://docs.mongodb.org/manual/use-cases/inventory-management/
As you can see from the example, each document have a field called carted, which is an array of subdocuments.
{
_id: 42,
last_modified: ISODate("2012-03-09T20:55:36Z"),
status: 'active',
items: [
{ sku: '00e8da9b', qty: 1, item_details: {...} },
{ sku: '0ab42f88', qty: 4, item_details: {...} }
]
}
This fits me perfect, except for one problem:
I want to count each unique item (with "sku" as the unique identifier key) in the entire collection where each document adds the count by 1 (multiple instances of the same "sku" in the same document will still just count 1). E.g. I would like this result:
{ sku: '00e8da9b', doc_count: 1 },
{ sku: '0ab42f88', doc_count: 9 }
After reading up on MongoDB, I am quite confused about how to do this (fast) when you have a complex schema as described above. If I have understood the otherwise excellent documentation correct, such operation may perhaps be achieved using either the aggregation framework or the map/reduce framework, but this is where I need some input:
Which framework would be better suited to achieve the result I am looking for, given the complexity of the structure?
What kind of indexes would be preferred in order to gain the best possible performance out of the chosen framework?
MapReduce is slow, but it can handle very large data sets. The Aggregation framework on the other hand is a little quicker, but will struggle with large data volumes.
The trouble with your structure shown is that you need to "$unwind" the arrays to crack open the data. This means creating a new document for every array item and with the aggregation framework it needs to do this in memory. So if you have 1000 documents with 100 array elements it will need to build a stream of 100,000 documents in order to groupBy and count them.
You might want to consider seeing if there's a schema layout that will server your queries better, but if you want to do it with the Aggregation framework here's how you could do it (with some sample data so the whole script will drop into the shell);
db.so.remove();
db.so.ensureIndex({ "items.sku": 1}, {unique:false});
db.so.insert([
{
_id: 42,
last_modified: ISODate("2012-03-09T20:55:36Z"),
status: 'active',
items: [
{ sku: '00e8da9b', qty: 1, item_details: {} },
{ sku: '0ab42f88', qty: 4, item_details: {} },
{ sku: '0ab42f88', qty: 4, item_details: {} },
{ sku: '0ab42f88', qty: 4, item_details: {} },
]
},
{
_id: 43,
last_modified: ISODate("2012-03-09T20:55:36Z"),
status: 'active',
items: [
{ sku: '00e8da9b', qty: 1, item_details: {} },
{ sku: '0ab42f88', qty: 4, item_details: {} },
]
},
]);
db.so.runCommand("aggregate", {
pipeline: [
{ // optional filter to exclude inactive elements - can be removed
// you'll want an index on this if you use it too
$match: { status: "active" }
},
// unwind creates a doc for every array element
{ $unwind: "$items" },
{
$group: {
// group by unique SKU, but you only wanted to count a SKU once per doc id
_id: { _id: "$_id", sku: "$items.sku" },
}
},
{
$group: {
// group by unique SKU, and count them
_id: { sku:"$_id.sku" },
doc_count: { $sum: 1 },
}
}
]
//,explain:true
})
Note that I've $group'd twice, because you said that an SKU can only count once per document, so we need to first sort out the unique doc/sku pairs and then count them up.
If you want the output a little different (in other words, EXACTLY like in your sample) we can $project them.
With the latest mongo build (it may be true for other builds too), I've found that slightly different version of cirrus's answer performs faster and consumes less memory. I don't know the details why, seems like with this version mongo somehow have more possibility to optimize the pipeline.
db.so.runCommand("aggregate", {
pipeline: [
{ $unwind: "$items" },
{
$group: {
// create array of unique sku's (or set) per id
_id: { id: "$_id"},
sku: {$addToSet: "$items.sku"}
}
},
// unroll all sets
{ $unwind: "$sku" },
{
$group: {
// then count unique values per each Id
_id: { id: "$_id.id", sku:"$sku" },
count: { $sum: 1 },
}
}
]
})
to match exactly the same format as asked in question, grouping by "_id" should be skipped