I have a collection where i have objects like:
{
"_id" : ObjectId("5ab212249a639865c58b744e"),
"levels" : [
{
"levelId" : 0,
"siteId" : "5a0ff11dc7bd083ea6a706b1",
"title" : "Hospital Services"
},
{
"levelId" : 1,
"siteId" : "5a0ff220c7bd083ea6a706d0",
"title" : "Reference Testing"
},
{
"levelId" : 2,
"siteId" : "5a0ff24fc7bd083ea6a706da",
"title" : "Des Moines(Reference Testing)"
}
]
}
I want to sort on the title field of 2nd object of levels array e.g. levels.2.title
Currently my mongo query looks like:
db.getCollection('5aaf63a69a639865c58b2ab9').aggregate([
{$sort : {'levels.2.title':1}}
])
But it is not giving desired results.
Please help.
You can try below query in 3.6.
db.col.aggregate({$sort:{"levels.2.title":1}});
This aggregation and find semantics are different in 3.4. More on jira here
So
db.col.find().sort({"levels.2.title":1})
works as expected and aggregation sort is not working as expected.
Use below aggregation in 3.4.
Use $arrayElemAt to project the second element in $addFields to keep the computed value as the extra field in the document followed by $sort sort on field.
$project with exclusion to drop the sort field to get expected output.
db.col.aggregate([
{"$addFields":{ "sort_element":{"$arrayElemAt":["$levels", 2]}}},
{"$sort":{"sort_element.title":-1}},
{"$project":{"sort_element":0}}
])
Also, You can use $let expression to output the title field directly in $addFields stage.
db.col.aggregate([
{"$addFields":{ "sort_field":{"$let:{"vars":{"ele":{$arrayElemAt":["$levels", 2]}}, in:"$$ele.title"}}}},
{"$sort":{"sort_field":-1}},
{"$project":{"sort_field":0}}
])
Related
I have this collection in MongoDB that contains the following entries. I'm using Robo3T to run the query.
{
"_id" : ObjectId("xxx1"),
"Evaluation Date" : "2021-09-09",
"Results" : [
{
"Name" : "ABCD",
"Version" : "3.2.x"
}
]
"_id" : ObjectId("xxx2"),
"Evaluation Date" : "2022-09-09",
"Results" : [
{
"Name" : "ABxD",
"Version" : "5.2.x"
}
]
}
This document contains multiple entries of similar format. Now, I need to extract the latest value for "Version".
Expected output:
5.2.x
Measures I've taken so far:
(1) I've only tried findOne() and while I was able to extract the value of "Version": db.getCollection('TestCollectionName').findOne().Results[0].Version
...only the oldest entry was returned.
3.2.x
(2) Using the find().sort().limit() like below, returns the entire document for the latest entry and not just the data value that I wanted; db.getCollection('TestCollectionName').find({}).sort({"Results.Version":-1}).limit(1)
Results below:
"_id" : ObjectId("xxx2"),
"Evaluation Date" : "2022-09-09",
"Results" : [
{
"Name" : "ABxD",
"Version" : "5.2.x"
}
]
(3) I've tried to use sort() and limit() alongside findOne() but I've read that findOne is maybe deprecated and also not compatible with sort. And thus, resulting to an error.
(4) Finally, if I try to use sort and limit on find like this: db.getCollection('LD_exit_Evaluation_Result_MFC525').find({"Results.New"}).sort({_id:-1}).limit(1) I would get an unexpected token error.
What would be a good measure for this?
Did I simply mistake to/remove a bracket or need to reorder the syntax?
Thanks in advance.
I'm not sure if I understood well, but maybe this could be what are you looking for:
db.collection.aggregate([
{
"$project": {
lastResult: {
"$last": "$Results"
},
},
},
{
"$project": {
version: "$lastResult.Version",
_id: 0
}
}
])
It uses aggregate with some operators: the first $project calculate a new field called lastResult with the last element of each array using $last operator. The second $project is just to clean the output. If you need the _id reference, just remove _id: 0 or change its value to 1.
You can check how it works here: https://mongoplayground.net/p/jwqulFtCh6b
Hope I helped
I have following students collection
{ "_id" : ObjectId("5f282eb2c5891296d8824130"), "name" : "Rajib", "mark" : "1000" }
{ "_id" : ObjectId("5f282eb2c5891296d8824131"), "name" : "Rahul", "mark" : "1200" }
{ "_id" : ObjectId("5f282eb2c5891296d8824132"), "name" : "Manoj", "mark" : "1000" }
{ "_id" : ObjectId("5f282eb2c5891296d8824133"), "name" : "Saroj", "mark" : "1400" }
My requirement is to sort the collection basing on 'mark' field in descending order. But it should not display 'mark' field in final result. Result should come as:
{ "name" : "Saroj" }
{ "name" : "Rahul" }
{ "name" : "Rajib" }
{ "name" : "Manoj" }
Following query I tried and it works fine.
db.students.find({},{"_id":0,"name":1}).sort({"mark":-1})
My MongoDB version is v4.2.8. Now question is what is the equivalent Aggregation Query of the above query. I tried following two queries. But both didn't give me desired result.
db.students.aggregate([{"$project":{"name":1,"_id":0}},{"$sort":{"mark":-1}}])
db.students.aggregate([{"$project":{"name":1,"_id":0,"mark":1}},{"$sort":{"mark":-1}}])
Why it is working in find()?
As per Cursor.Sort, When a set of results are both sorted and projected, the MongoDB query engine will always apply the sorting first.
Why it isn't working in aggregate()?
As per Aggregation Pipeline, The MongoDB aggregation pipeline consists of stages. Each stage transforms the documents as they pass through the pipeline. Pipeline stages do not need to produce one output document for every input document; e.g., some stages may generate new documents or filter out documents.
You need to correct:
You should change pipeline order, because if you have not selected mark field in $project then it will no longer available in further pipelines and it will not affect $sort operation.
db.students.aggregate([
{ "$sort": { "mark": -1 } },
{ "$project": { "name": 1, "_id": 0 } }
])
Playground: https://mongoplayground.net/p/xtgGl8AReeH
db.rohit1.find({{$add:["a_image","b_image"]}}:{gt:10})
E QUERY [thread1] SyntaxError: invalid property id #(shell):1:16
You can not use $add with find in the mongodb. Instead i will suggest you to use aggregate.
First you need to create another field which is the sum of two fields a_image and b_image using $project.And, then you need to use $match to check your condition.
try this:
db.rohit1.aggregate([{
$project :{
add : {$add : ["$a_image","$b_image"]},
field1 : "$field1"
... //include all the fields you want to have in the result
}
},{
$match : {add : {$gt : 10}}
}]);
For more information on $project and $match, please read linked mongodb documentation.
EDIT:
you can use $pow inside $add in the same projection.
Try this code:
db.rohit1.aggregate([{
$project :{
add : {$add : [{ $pow : ["$a_image",2]},{$pow : ["$b_image",2]}]},
field1 : "$field1"
... //include all the fields you want to have in the result
}
},{
$match : {add : {$gt : 10}}
}]);
I need few values to pass through the pipeline(without changes) to the subsequent pipeline aggregator while using a group
For eg if my input is
{ "_id" : 1, "tags" : [ "a", "b", "b", "c" ], "text" : "a" },
{ "_id" : 2, "tags" : [ "a", "c" ], "text" : "b" }
and I like the output to be
{ _id: 1, tags : [a,b,c], text : a} , { _id: 2, tags : [a,c], text : b}
I did an unwind and group and do $addToSet to remove the dups. My question is how do I make
text appear in the output as is. the text key and values need to just pass through this pipeline, However i am forced to use accumulators
I tried this code so far
use test
var unwind = { $unwind : "$tags"};
var group = { $group : { _id : "$_id" , tags : { $addToSet : "$tags" }
, text : { $first : "$text" }}};
db.test.aggregate(unwind,group)
and it works fine but that is not the intention of $first and it is suggested to be used with sort. What is the right way to do this ?
There's nothing wrong with using $first for this. As an alternative, you could add text to your _id, but that's hardly better as then you'd need to add a $project to the end of your pipeline to move it back out of _id.
You have to do the same sort of mildly awkward thing with SQL group bys.
All fields but the _id field must use an accumulator with $group, as specified in the $group documentation page. The problem is that it's unclear what it would mean to "pass" a value through the pipeline with $group, since MongoDB has no way of knowing that the value of that field is going to be the same for all documents you are grouping together. You have to choose what value you want for that field with the accumulator.
I hate this kind of questions but maybe you can point me to obvious. I'm using Mongo 2.2.2.
I have a collection (in replica set) with 6M documents which has string field called username on which I have index. The index was non-unique but recently I made it unique. Suddenly following query gives me false alarms that I have duplicates.
db.users.aggregate(
{ $group : {_id : "$username", total : { $sum : 1 } } },
{ $match : { total : { $gte : 2 } } },
{ $sort : {total : -1} } );
which returns
{
"result" : [
{
"_id" : "davidbeges",
"total" : 2
},
{
"_id" : "jesusantonio",
"total" : 2
},
{
"_id" : "elesitasweet",
"total" : 2
},
{
"_id" : "theschoolofbmx",
"total" : 2
},
{
"_id" : "longflight",
"total" : 2
},
{
"_id" : "thenotoriouscma",
"total" : 2
}
],
"ok" : 1
}
I tested this query on sample collection with few documents and it works as expected.
One of 10gen responded in their JIRA.
Are there any updates on this collection? If so, I'd try adding {$sort: {username:1}} to the front of the pipeline. That will ensure that you only see each username once if it is unique.
If there are updates going on, it is possible that aggregation would see a document twice if it moves due to growth. Another possibility is that a document was deleted after being seen by the aggregation and a new one was inserted with the same username.
So sorting by username before grouping helped.
I think the answer may lie in the fact that your $group is not using an index, it's just doing a scan over the entire collection. These operators can use and index currently in the aggregation framework:
$match $sort $limit $skip
And they will work if placed before:
$project $unwind $group
However, $group by itself will not use an index. When you do your find() test I am betting you are using the index, possibly as a covered index (you can verify by looking at an explain() for that query), rather than scanning the collection. Basically my theory is that your index has no dupes, but your collection does.
Edit: This likely happens because a document is updated/moved during the aggregation operation and hence is seen twice, not because of dupes in the collection as originally thought.
If you add an operator earlier in the pipeline that can use the index but not alter the results fed into $group, then you can avoid the issue.