mongodb aggregate cond $in

florinrelea answered on October 14, 2021 Popularity 10/10 Helpfulness 2/10

answer mongodb aggregate cond $in

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mongodb aggregate cond $in

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cond: {
  $and: [{
  	$in: ['$$viewsDoc.year', [1, 2, 3]]
  }]
}

Popularity 10/10 Helpfulness 2/10 Language typescript

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Tags: aggregate mongodb typescript

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Contributed on Oct 14 2021

florinrelea

0 Answers Avg Quality 2/10

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aggregate mongodb

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// MonogDB Aggregation Framework
// $project -> $match -> $group -> $sort -> ouput
// $count, $skip, Sunwind,
// $match
db.users.aggregate([
{ stage 1}, $project
 { stage 2}, $match
  { stage 3}, $group
   { stage 4}, $sort
]);

* match : 
db.users.aggregate([{$match:{'status':'inactive'}}]).pretty();

* group :
// group by department 
db.users.aggregate([{$group:{_id:'$department'}}]).pretty();
{"id" : "IT" } {
{"_id" : "TECH" }
{_"id" : "HR" }

- multiple group 
db.users.aggregate([{$group:{_id:{age:'Şage',gender:'$gender'}}}]).pretty();

{"id" : { "age" : "27", 'gender' : "female" }}
{"_id" : { "age" : "25", "gender" : "female" } }
{ "_id" : { "age" : "26" "gender" : "male" } }
{ "_id" : { "age" : "24", "gender" : "male" } }
{ "_id" : { "age" : "22", "gender" : "male" } }
{"_id" : { "age" : "27", "gender" : "male" } }

* sort 
db.users.aggregate([{$group:{_id:'$department'}},{$sort:{_id:1}}]);

* match 
db.users.aggregate([{$match:{status:'active'}},{$project:{name:1,email:1}}]).pretty();

* count 
db.users.aggregate([{$match:{status:'inactive'}},{$count:'totals'}]);

* limit 
db.users.aggregate([{$match:{status:'active'}},{$project:{name:1,email:1}},{Slimit:2}]).pretty();

* unwind 
- It is used for array attribute to show seprate result for that array.

db.users.aggregate([{$unwind:"$language"},{$match:{status:'inactive'}},{$project:{name:1,language:1}}]).pretty();
db.users.aggregate([{$unwind:'$language'},{$group:{_td:'$language'}}]);

* lookup
- Joining two tables
 
db.users.aggregate({
    $lookup:{
        from: "department",
        localField:"dept", 06
        foreignField: "name"
        as: "anything"
})

* out 
- It will create the new collection with that name and storing the data result into it. 
db.users.aggregate([{ $match:{status:'active'}},{$project:{email:1,name:1, id:0}},{'$out':'info'}]).pretty();

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Tags: aggregate aggr

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Contributed on Apr 16 2022

Abhishek

0 Answers Avg Quality 2/10

mongodb aggregate cond $in

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