How can I query array of embedded document in mongodb.
{
_id: 1,
zipcode: "63109",
students: [
{ name: "john", school: 102, age: 14 },
{ name: "jess", school: 102, age: 11 },
{ name: "jeff", school: 108, age: 15 }
]
}
{
_id: 2,
zipcode: "63110",
students: [
{ name: "ajax", school: 100, age: 7 },
{ name: "achilles", school: 100, age: 8 },
]
}
{
_id: 3,
zipcode: "63109"
}
Let's say for the above data how will retrieve only those lines where age>=7 and age<=10. For id_1 only the row with age 10 should be returned. Both the rows in id_2 and id_3. Also, id_4 has no field called students. So, I DO NOT, want to see it in the output.
Edit:-
The output I get looks something like this when I do a filter. But I DO NOT want the row that has "None".
The last document _id: 3 does not have the field students but filtering is done on students, so the output has a "None" corresponding to it.
I wish to handle 2 cases:
No filtering should be applied where the array "students" don't exist under an id.
Array students exists but is empty []
Both these cases end up in the output if simply filtered.
{'_id': ObjectId('5cdaefd393436906b016ddb4'),
'students': [{'name': ajax,
'school': '100',
'age': '7'},
{'name': achilles,
'school': '100',
'age': '8'}],
{'_id': ObjectId('5cdaefd393436906b016ddb3'), 'students': None}
Note: I am using a different but identical data set, so the output is not exact, but I hope you get the idea. The 'student' is an embedded document which may or may not be present in all documents.
Edit : Finally I used unwind to achieve what I wanted.
You can use below aggregation
db.collection.aggregate([
{ "$match": { "$expr": { "$gte": [{ "$size": { "$ifNull": ["$students", []] } }, 1] }}},
{ "$addFields": {
"students": {
"$filter": {
"input": { "$ifNull": ["$students", []] },
"cond": {
"$and": [
{ "$gte": ["$$this.age", 7] },
{ "$lte": ["$$this.age", 10] }
]
}
}
}
}}
])
Related
Considering the following document structure:
{_id: 1, name: 'joe', snapshot: null, age: 30}
{_id: 2, name: 'joe', snapshot: 'snapshot1', age: 30}
{_id: 3, name: 'joe', snapshot: 'snapshot15', age: 30}
{_id: 4, name: 'joe', snapshot: 'snapshot23', age: 30}
How would I perform a query that groups on the name field and adds an additional field that is a count of the remaining records containing subtree: 'additionalinfo'. It would look like this:
{_id: 1, name: 'joe', snapcount: 3, age: 30}
I've been able to group using aggregations but I can't quite get it like this.
My own solution:
I ultimately restructured my data to look like this instead:
{
_id: 1,
name: 'joe',
snapshots: [
{name: 'snap17', id: 1},
{name: 'snap15', id: 2},
{name: 'snap14', id: 3}
],
age: 30
}
This allows me to just check snapshots.length to solve my original problem. However; the answers in this post where very helpful and answered the original question.
Adding another aggregation query to do it: playground link: try it
db.collection.aggregate([
{
$match: {
"snapshot": {
$exists: true,
$ne: null
}
}
},
{
$group: {
_id: "$name",
snapcount: {
$sum: 1
},
age: {
"$first": "$age"
},
name: {
"$first": "$name"
}
}
},
{
"$unset": "_id"
}
])
Based on the comments, the query worked for OP:
db.collection.aggregate([
{
$match: {
"snapshot": {
$exists: true,
$ne: null
}
}
},
{
$group: {
_id: "$name",
snapcount: {
$sum: 1
},
age: {
"$first": "$age"
},
name: {
"$first": "$name"
},
id: {
"$first": "$_id"
}
}
},
{
"$unset": "_id"
}
])
Here's one way you could do it.
db.collection.aggregate([
{
"$group": {
"_id": "$name",
"name": {"$first": "$name"},
"age": {"$first": "$age"},
"snapcount": {
"$sum": {
"$cond": [
{"$eq": [{"$type": "$snapshot"}, "string"]},
1,
0
]
}
}
}
},
{"$unset": "_id"}
])
Try it on mongoplayground.net.
I'm trying to find the last element of an array that matches a condition, for example if I had the data from the $elemMatch page:
{
_id: 1,
students: [
{ name: "john", school: 102, age: 10 },
{ name: "jess", school: 102, age: 11 },
{ name: "jeff", school: 108, age: 15 }
]
}
I want to find the last element that has school: 102 in it, in this case I will get { name: "jess", school: 102, age: 11 }.
But $elemMatch will return john instead.
I feel like there supposed to be an easy way to do it, but couldn't find any solution without using $unwind and complex aggregations.
Any ideas?
You can use $filter to get only students from school 102 and then $slice with -1 to get last element from filtered array. To make it as short as possible you can use $let to define temporary variable, try:
db.col.aggregate([
{
$addFields: {
students: {
$let: {
vars: {
filtered: { $filter: { input: "$students", as: "s", cond: { $eq: [ "$$s.school", 102 ] } } }
},
in: { $slice: [ "$$filtered", -1 ] }
}
}
}
}
])
Given the following data, I'm trying to get an average of all their ages, at the same time I want to return an array of their names. Ideally, I want to do this in just one query but can't seem to figure it out.
Data:
users:[
{user:{
id: 1,
name: “Bob”,
age: 23
}},
{user:{
id: 1,
name: “Susan”,
age: 32
}},
{user:{
id: 2,
name: “Jeff”,
age: 45
}
}]
Query:
var dbmatch = db.users.aggregate([
{$match: {"id" : 1}},
{$group: {_id: null, avg_age: { $avg: "$age" }}},
{$group: {_id : { name: "$name"}}}
)]
Running the above groups one at a time outputs the results I expect, either an _id of null and an average of 27.5, or an array of the names.
When I combine them as you see above using a comma, I get:
Issue Generated Code:
[ { _id: {name: null } } ]
Expected Generated Code:
[
{name:"Bob"},
{name:"Susan"},
avg_age: 27.5
]
Any help would be greatly appreciated!
Not sure if this is exactly what you want, but this query
db.users.aggregate([
{
$match: {
id: 1
}
},
{
$group: {
_id: "$id",
avg_age: {
$avg: "$age"
},
names: {
$push: {
name: "$name"
}
}
}
},
{
$project: {
_id: 0
}
}
])
Results in this result:
[
{
"avg_age": 27.5,
"names": [
{
"name": "Bob"
},
{
"name": "Susan"
}
]
}
]
This will duplicate names, so if there are two documents with the name Bob, it will be two times in the array. If you don't want duplicates, change $push to $addToSet.
Also, if you want names to be just an array of names instead of objects, change names query to
names: {
$push: "$name"
}
This will result in
[
{
"avg_age": 27.5,
"names": ["Bob", "Susan"]
}
]
Hope it helps,
Tomas :)
You can use $facet aggregation to run the multiple queries at once
db.collection.aggregate([
{ "$facet": {
"firstQuery": [
{ "$match": { "id": 1 }},
{ "$group": {
"_id": null,
"avg_age": { "$avg": "$age" }
}}
],
"secondQuery": [
{ "$match": { "id": 1 }},
{ "$group": { "_id": "$name" }}
]
}}
])
Is there an quick efficient way to duplicate elements in a mongo db collections based on a property. In the example below, I am trying to duplicate the elements based on a jobId.
I am using Spring boot, so any example using Spring boot API would be even more helpful.
Original Collection
{ _id: 1, jobId: 1, product: "A"},
{ _id: 2, jobId: 1, product: "B"},
{ _id: 3, jobId: 1, product: "C"},
After duplication
{ _id: 1, jobId: 1, product: "A"},
{ _id: 2, jobId: 1, product: "B"},
{ _id: 3, jobId: 1, product: "C"},
{ _id: 4, jobId: 2, product: "A"},
{ _id: 5, jobId: 2, product: "B"},
{ _id: 6, jobId: 2, product: "C"},
You can use following aggregation:
db.col.aggregate([
{
$group: {
_id: null,
values: { $push: "$$ROOT" }
}
},
{
$addFields: {
size: { $size: "$values" },
range: { $range: [ 0, 3 ] }
}
},
{
$unwind: "$range"
},
{
$unwind: "$values"
},
{
$project: {
_id: { $add: [ "$values._id", { $multiply: [ "$range", "$size" ] } ] },
jobId: { $add: [ "$values.jobId", "$range" ] },
product: "$values.product",
}
},
{
$sort: {
_id: 1
}
},
{
$out: "outCollection"
}
])
The algorithm is quite simple here: we want to iterate over two sets:
first one defined by all items from your source collection (that's why I'm grouping by null)
second one defined artificially by $range operator. It will define how many times we want to multiply our collection (3 times in this example)
Double unwind generates as much documents as we need. Then the formula for each _id is following: _id = _id + range * size. Last step is just to redirect the aggregation output to your collection.
I have a collection like below
{ _id: 1, zipcode: "63109", students: [
{ name: "john", school: 102, age: 10 },
{ name: "jess", school: 102, age: 11 },
{ name: "jeff", school: 108, age: 15 }
] }
{ _id: 2, zipcode: "63110", students: [
{ name: "ajax", school: 100, age: 7 },
{ name: "achilles", school: 100, age: 8 },
] }
{ _id: 3, zipcode: "63109", students: [
{ school: 100, age: 7 },
{ school: 100, age: 8 },
] }
{ _id: 4, zipcode: "63109", students: [
{ name: "barney", school: 102, age: 7 },
{ name: "ruth", school: 102, age: 16 },
] }
Note:Some document doesnot have name field.
I want to findout the count of name field.
Kindly suggest query to find count in mongo
You can do this with aggregation framework with $unwind. Let say your collection name is test:
db.test.aggregate(
{$unwind: '$students'},
{$match:{'students.name':{$exists:1}}},
{$group:{_id: '$students.name', count:{$sum:1}}},
{$project:{tmp:{name:'$_id', count:'$count'}}},
{$group:{_id:'Total Names', total:{$sum:1}, data:{$addToSet:'$tmp'}}}
)
Hope this is what you want!
You need to use the aggregation framework in this case.
The first aggregation step would be to use $unwind to turn the arrays into streams of documents.
Then you can use $group as a second aggregation step with $sum:1 to get the count of every name.
db.collection.aggregate([
{
$unwind: "$students"
},
{
$group: {
_id:"$students.name",
count: { $sum:1 }
}
}
]
This is doable through the aggregation framework
I'll assume your collection is called mycoll:
db.mycoll.aggregate([
{$unwind:'$students'},
{$group:{_id:'$name', 'count':{$sum:1}},
{$match:{'students.name':{$exists:1}}}
]);
References:
Aggregations http://docs.mongodb.org/manual/tutorial/aggregation-zip-code-data-set/
Exists: http://docs.mongodb.org/manual/reference/operator/query/exists/
Summing '1' essentially turns it into a count.