Match documents if a value in an array of sub-documents is greater than some value only if the same document contains a field that is equal to some value
I have a collection that contains documents with an array of sub-documents. This array of sub-documents contains a field that dictates whether or not I can filter the documents in the collection based on another field in the sub-document. This'll make more sense when you see an example of the document.
{
"_id":"ObjectId('XXX')",
"Data":{
"A":"",
"B":"-25.78562 ; 28.35629",
"C":"165"
},
"SubDocuments":[
{
"_id":"ObjectId('XXX')",
"Data":{
"Value":"XXX",
"DataFieldId":"B"
}
},
{
"_id":"ObjectId('XXX')",
"Data":{
"Value":"",
"DataFieldId":"A"
}
},
{
"_id":"ObjectId('XXX')",
"Data":{
"Value":"105",
"DataFieldId":"Z"
}
}
]
}
I only want to match documents that contain sub-documents with a DataFieldId that is equal to Z but also filter for Values that are greater than 105 only if Data Field Id is equal to Z.
Try as below:
db.collection.aggregate([
{
$project: {
_id:1,
Data:1,
filteredSubDocuments: {
$filter: {
input: "$SubDocuments",
as: "subDoc",
cond: {
$and: [
{ $eq: ["$$subDoc.Data.DataFieldId", "Z"] },
{ $gte: ["$$subDoc.Data.Value", 105] }
]
}
}
}
}
}
])
Resulted response will be:
{
"_id" : ObjectId("5cb09659952e3a179190d998"),
"Data" : {
"A" : "",
"B" : "-25.78562 ; 28.35629",
"C" : "165"
},
"filteredSubDocuments" : [
{
"_id" : "ObjectId('XXX')",
"Data" : {
"Value" : 105,
"DataFieldId" : "Z"
}
}
]
}
This can be done by using the $elemMatch operator on sub-documents, for details you can click on provided link. For your problem you can try below query by using $elemMatch which is match simpler than aggregation:
db.collectionName.find({
"SubDocuments": {
$elemMatch: {
"Data.DataFieldId": "Z" ,
"Data.Value" : {$gte: 105}
}
} })
Its working fine, I have verified it locally, one modification you required is that you have to put the value of SubDocuments.Data.Value as Number or Long as per your requirements.
Related
We are now using mongo-db to store data from tests. I am using Mongo-Shell
The document is structured like this:
{
static1:"abc",
static2:"xyz",
static3:"asd", [...],
nested:[
{
data1: "d1",
data2: "d2",
},
{
data1: "dx",
data4: "d4",
data5: "d5",
data6: "d6",
},
{
data1: "ds",
data8:"data8"
}, [...]
]
}
So the static-data is always in the same structure, but for every measurement the object can look different. It could be voltage with upper, lower and actual value. Or just comparison of a target value and an actual value. Highly dynamic.
Same data1-name represent same attributes.
Now e.g. I want to display only some static data and one (or several) attributes of a nested document.
I am using this query:
find({}, {_id:0, data1:1, "nested.data8":1}).pretty()
As expected, only the static-data1 is displayed, but the dynamic measurements has lots of empty objects in the shell-output
Example-Output:
{
"static1" : "123",
"nested" : [
{ },
{ },
{ }, [...] ,
{ "data8" : "OK" }
] }
The desired output would be:
{
"static1" : "123",
"nested" : [
{ "data8" : "d8" }
]
}
I also tried this query on the mongo shell:
aggregate( { $addFields: {"static":"$static1", "data8":"$nested.data8"} }, { $project:{"static1":1, "nested.data8":1} } ).pretty()
But the result is the same.
I hope there is a ways to get rid of the empty documents in the output.
Thanks
You can try $filter operator to filter result of nested array by checking condition not equal to empty object {}
db.collection.aggregate([
{
$project: {
_id: 0,
data1: 1,
"nested.data8": 1
}
},
{
$set: {
nested: {
$filter: {
input: "$nested",
cond: { $ne: ["$$this", {}] }
}
}
}
}
])
Playground
I need help in aggregate this query, I need aggregate values of debito
{
"_id" : ObjectId("5a088f6584ccb0a665900726"),
"usuario" : "tamura",
"creditos" : [
{
"nome_do_credito" : "credito inicial",
"credito" : 0
}
],
"debitos" : [
{
"nome_do_debito" : "debito inicial",
"debito" : 0
},
{
"nome_do_debito" : "Faculdade",
"debito" : "150.00"
}
]
}
I need the output
debito : 150
(0+150)
You will first need to turn all your debito fields into a numerical type (as in 150.00) since you cannot do Maths on strings (as in "150.00"). And then the following query should do the trick:
db.collection.aggregate({
$project: {
"debitos": {
$sum: "$debitos.debito"
}
}
})
In case you have more than one document in your collection and you want the total sum over all documents you can run this:
db.collection.aggregate({
$unwind: "$debitos" // flatten the "debitos" array
}, {
$group: {
"_id": null, // do not really group, just throw all documents in the same group
"debitos": {
$sum: "$debitos.debito" // sum up all debito fields
}
}
})
I have a mongoDB orders collection, the documents of which look as follows:
[{
"_id" : ObjectId("59537df80ab10c0001ba8767"),
"shipments" : {
"products" : [
{
"orderDetails" : {
"id" : ObjectId("59537df80ab10c0001ba8767")
}
},
{
"orderDetails" : {
"id" : ObjectId("59537df80ab10c0001ba8767")
}
}
]
},
}
{
"_id" : ObjectId("5953831367ae0c0001bc87e1"),
"shipments" : {
"products" : [
{
"orderDetails" : {
"id" : ObjectId("5953831367ae0c0001bc87e1")
}
}
]
},
}]
Now, from this collection, I want to filter out the elements in which, any of the values at shipments.products.orderDetails.id path is same as value at _id path.
I tried:
db.orders.aggregate([{
"$addFields": {
"same": {
"$eq": ["$shipments.products.orderDetails.id", "$_id"]
}
}
}])
to add a field same as a flag to decide whether the values are equal, but the value of same comes as false for all documents.
EDIT
What I want to do is compare the _id field the the documents with all shipments.products.orderDetails.id values in the array.
If even 1 of the shipments.products.orderDetails.ids match the value of the _id field, I want that document to be present in the final result.
PS I am using MongoDB 3.4, and have to use the aggregation pipeline.
Your current attempt fails because the notation returns an "array" in comparison with a "single value".
So instead either use $in where available, which can compare to see if one value is "in" an array:
db.orders.aggregate([
{ "$addFields": {
"same": {
"$in": [ "$_id", "$shipments.products.orderDetails.id" ]
}
}}
])
Or notate both as arrays using $setIsSubset
db.orders.aggregate([
{ "$addFields": {
"same": {
"$setIsSubset": [ "$shipments.products.orderDetails.id", ["$_id"] ]
}
}}
])
Where in that case it's doing a comparison to see if the "sets" have an "intersection" that makes _id the "subset" of the array of values.
Either case will return true when "any" of the id properties within the array entries at the specified path are a match for the _id property of the document.
There is three documents in collection test:
// document 1
{
"id": 1,
"score": [3,2,5,4,5]
}
// document 2
{
"id": 2,
"score": [5,5]
}
// document 3
{
"id": 3,
"score": [5,3,3]
}
I want to fetch documents that score field contains [5,5].
query:
db.test.find( {"score": {"$all": [5,5]}} )
will return document 1, 2 and 3, but I only want to fetch document 1 and 2.
How can I do this?
After reading your problem I personally think mongodb not supported yet this kind of query. If any one knows about how to find this using mongo query they defiantly post answers here.
But I think this will possible using mongo forEach method, so below code will match your criteria
db.collectionName.find().forEach(function(myDoc) {
var scoreCounts = {};
var arr = myDoc.score;
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
scoreCounts[num] = scoreCounts[num] ? scoreCounts[num] + 1 : 1;
}
if (scoreCounts[5] >= 2) { //scoreCounts[5] this find occurrence of 5
printjsononeline(myDoc);
}
});
Changed in version 2.6.
The $all is equivalent to an $and operation of the specified values; i.e. the following statement:
{ tags: { $all: [ "ssl" , "security" ] } }
is equivalent to:
{ $and: [ { tags: "ssl" }, { tags: "security" } ] }
I think you need to pass in a nested array -
So try
db.test.find( {"score": {"$all": [[5,5]]}} )
Source
Changed in version 2.6.
When passed an array of a nested array (e.g. [ [ "A" ] ] ), $all can now match documents where the field contains the nested array as an element (e.g. field: [ [ "A" ], ... ]), or the field equals the nested array (e.g. field: [ "A" ]).
http://docs.mongodb.org/manual/reference/operator/query/all/
You can do it with an aggregation. The first step can use an index on { "score" : 1 } but the rest is hard work.
db.test.aggregate([
{ "$match" : { "score" : 5 } },
{ "$unwind" : "$score" },
{ "$match" : { "score" : 5 } },
{ "$group" : { "_id" : "$_id", "sz" : { "$sum" : 1 } } }, // use $first here to include other fields in the results
{ "$match" : { "sz" : { "$gte" : 2 } } }
])
I have a sample document like shown below
{
"_id" : "docID",
"ARRAY" : [
{
"k" : "value",
"T" : "20:15:35",
"I" : "Hai"
},
{
"K" : "some value",
"T" : "20:16:35",
"I" : "Hello"
},
{
"K" : "some other value",
"T" : "20:15:35",
"I" : "Update"
}
]
}
I am trying to update the last element in the "ARRAY" based on field "ARRAY.T"(which is only field i know at the point of update), but what my problem is first element in the array matches the query and its ARRAY.I field is updated.
Query used to update:
db.collection.update( { _id: "docID","ARRAY.T" : "20:15:35"},
{ $set: { "ARRAY.$.I": "Updated value" }
})
Actually i don't know index of the array where to update so i have to use ARRAY.I in the query, is there any way to to tell Mongodb to update the first element matched the query from last of the array.
I understand what you are saying in that you want to match the last element in this case or in fact process the match in reverse order. There is no way to modify this and the index stored in the positional $ operator will always be the "first" match.
But you can change your approach to this, as the default behavior of $push is to "append" to the end of the array. But MongoDB 2.6 introduced a $position modifier so you can in fact always "pre-pend" to the array meaning your "oldest" item is at the end.
Take this for example:
db.artest.update(
{ "array": { "$in": [5] } },
{ "$push": { "array": { "$each": [5], "$position": 0 } }},
{ "upsert": true }
)
db.artest.update(
{ "array": { "$in": [5] } },
{ "$push": { "array": { "$each": [6], "$position": 0 } }},
{ "upsert": true }
)
This results in a document that is the "reverse" of the normal $push behavior:
{ "_id" : ObjectId("53eaf4517d0dc314962c93f4"), "array" : [ 6, 5 ] }
Alternately you could apply the $sort modifier when updating your documents in order to "order" the elements so they were reversed. But that may not be the best option if duplicate values are stored.
So look into storing your arrays in "reverse" if you intend to match the "newest" items "first". Currently that is your only way of getting your "match from last" behavior.