Consider this JSON content:
{
students: [
{
student_name: aaa,
classes: [
{
name: class1,
properties: {
id: 1
}
},
{
name: class2,
properties: {
id: 2
}
}
},
{
student_name: bbb,
classes: [
{
name: class2,
properties: {
id: 1
}
}
}
]
}
students is not nested. classes is not nested. I only want to match student_name aaa document by this below query:
{
"query": {
"bool": {
"must": [
{ "term": { "classes.name": "class1" }},
{ "term": { "classes.properties.id": "1" }}
]
}
}
}
But my query is matching both aaa and bbb, because it is considering my term statements as separate queries. How can I just match student aaa?
You have to make the student as a nested type. Otherwise it gets flattened and your query will match both the documents.
From the same ElasticSearch documentation:
When ingesting key-value pairs with a large, arbitrary set of keys,
you might consider modeling each key-value pair as its own nested
document with key and value fields. Instead, consider using the
flattened data type, which maps an entire object as a single field and
allows for simple searches over its contents. Nested documents and
queries are typically expensive, so using the flattened data type for
this use case is a better option.
Please refer to examples given in the same documentation and it will be clear to you. When student is changed to nested you should be able to get your expected results.
Related
I am trying to query my collection of matches (games) and find if a certain user has already sent data to the 'reportMessages' array of Objects.
const results = await Match.findOne({ 'users': req.params.userIdOfReportSender, '_id': req.params.matchId, 'reportMessages.sentBy': req.params.userIdOfReportSender }, 'reportMessages' )
However, the above query returns the following:
{
_id: 5fd382c65d5395e0778f2f8a,
reportMessages: [
{
_id: 5fd610f27ae587189c45b6ca,
content: 'jajatest',
timeStamp: 2020-12-13T13:02:42.102Z,
sentBy: 'XbVvm6g3nsRmPg3P1pBvVl84h6C2'
},
{ sentBy: "'anotheruser123" }
]
}
How can I get it to only return the first reportMessage, i.e. the one sent by XbVvm6g3nsRmPg3P1pBvVl84h6C2?
Mongoose findOne docs (https://mongoosejs.com/docs/api.html#model_Model.findOne) show that you can provide arguments to say which fields to select (in their case 'name length' but don't show a way to only select the fields in case they match a certain condition.
Is this even possible? Tried googling this seemingly easy question for quite some time without success
Kind regards
You can get only the subdocument you want with this aggregation query:
Match.aggregate([
{
$match: { _id: req.params.matchId }
},
{
$project: {
reportMessages: {
$filter: {
input: '$reportMessages',
as: 'msg',
cond: { $eq: ['$$msg.sentBy', req.params.userIdOfReportSender] }
}
}
}
},
{
$project: {
reportMessage: { $arrayElemAt: [ '$reportMessages', 0 ] },
}
},
{ $replaceWith: '$reportMessage' }
]);
Note that you only need to specify the document _id to get a single result, since _ids are unique.
I have a document structure called session which contains an array called seances in which there is an array called student.
I would like to retrieve all the documents where a given value is in the array students; for example if the value is Alice, and I have the following 3 documents :
Doc 1
{
name: "doc1"
seances: [
{
date: "02/01",
students: ["Alice","John","Brahim"]
}
]
}
Doc 2
{
name: "doc2"
seances: [
{
date: "12/04",
students: ["Alice","John","Brahim"]
}
]
}
Doc 3
{
name: "doc3"
seances: [
{
date: "21/03",
students: ["Arysse","John","Brahim"]
},
{
date: "22/05",
students: ["Steward","John","Brahim"]
}
]
}
In this case, it should return Doc 1 and Doc 2 because it contains the value "Alice" inside the array Students which contains the array sessions despite other fields like date.
Is it possible to do that using only where clause with firestore?
It's not possible. You can't query for the contents of arrays inside other object arrays. In order to satisfy this query, you'll have to keep another field with the array you want to search, and use an array-contains query on that field.
I'm trying to find all users except for a few, like this:
// get special user IDs
var special = db.special.find({}, { _id: 1 }).toArray();
// get all users except for the special ones
var users = db.users.find({_id: {$nin: special}});
This doesn't work because the array that I'm passing to $nin is not and array of ObjectId but an array of { _id: ObjectId() }
Variable special looks like this after the first query:
[ { _id: ObjectId(###) }, { _id: ObjectId(###) } ]
But $nin in the second query needs this:
[ ObjectId(###), ObjectId(###) ]
How can I get just the ObjectId() in an array from the first query so that I can use them in the second query?
Or, is there a better way of achieving what I'm trying to do?
Use the cursor.map() method returned by the find() function to transform the list of { _id: ObjectId(###) } documents to an array of ObjectId's as in the following
var special = db.special.find({}, { _id: 1 }).map(function(doc){
return doc._id;
});
Another approach you can consider is using the $lookup operator in the aggregation framework to do a "left outer join" on the special collection and filtering the documents on the new "joined" array field. The filter should match on documents whose array field is empty.
The following example demonstrates this:
db.users.aggregate([
{
"$lookup": {
"from": "special",
"localField": "_id",
"foreignField": "_id",
"as": "specialUsers" // <-- this will produce an arry of "joined" docs
}
},
{ "$match": { "specialUsers.0": { "$exists": false } } } // <-- match on empty array
])
I have users' collection whose schema is like:
{
_id: unique number,
name: 'asdf',
age: '12',
gender: 'm',
address: [
{area: 'sdf',
city: 'sdq',
state: 'wfw'},
{area: 'asdf',
city: 'sdfs',
state: 'vfdwd'}
]
}
I want to find out the users for whom all the values of state in address should be the value I pass. If even one of the state value doesn't match with the value I pass the user shouldn't be returned.
I tried simple find, aggregation framework with $unwind, $match but nothing seemed to get solution. Can you please help me out...
Thanks
P.S. please bear with multiple addresses for the sake of question. :)
To find out if all array entries match the state "wfw", do an aggregation like the following:
db.users.aggregate([
{ "$project" : {
"test" : {
"$allElementsTrue" : [{
"$map" : {
"input" : "$address",
"as" : "a",
"in" : { "$eq" : ["wfw", "$$a.state"] }
}
}]
}
} },
{ "$match" : { "test" : true } }
])
This aggregation takes each document, maps "state equals 'wfw'" over the address array to get a boolean array, and tests if the entire array is true, storing the result in `test, and then filtering the results based on test. You will need MongoDB 2.6 for support of some of the operators.
I don't know if I understand.
I replicated your document. When you want to retrieve an user by state you can do in many ways
If you search with single value you can do
db.g.find({ "address.state": "wfw" })
and retrieve an user
You can use $all
db.g.find( { "address.state": { $all: ["wfw","vfdwd"] } } ) // retrieve User
db.g.find( { "address.state": { $all: ["wfw","vfdwd","foo"] } } ) // don't retrieve User
or you can use $and
db.g.find( { $and: [ { "address.state":"wfw" },{ "address.state":"vfdwd" }] } )
But I don't know if I understand your question
Update and the correct answer
db.g.find( { "address.state": { $nin: ["wfw"] } } )
Let me Know
I need to rename indentifier in this:
{ "general" :
{ "files" :
{ "file" :
[
{ "version" :
{ "software_program" : "MonkeyPlus",
"indentifier" : "6.0.0"
}
}
]
}
}
}
I've tried
db.nrel.component.update(
{},
{ $rename: {
"general.files.file.$.version.indentifier" : "general.files.file.$.version.identifier"
} },
false, true
)
but it returns: $rename source may not be dynamic array.
For what it's worth, while it sounds awful to have to do, the solution is actually pretty easy. This of course depends on how many records you have. But here's my example:
db.Setting.find({ 'Value.Tiers.0.AssetsUnderManagement': { $exists: 1 } }).snapshot().forEach(function(item)
{
for(i = 0; i != item.Value.Tiers.length; ++i)
{
item.Value.Tiers[i].Aum = item.Value.Tiers[i].AssetsUnderManagement;
delete item.Value.Tiers[i].AssetsUnderManagement;
}
db.Setting.update({_id: item._id}, item);
});
I iterate over my collection where the array is found and the "wrong" name is found. I then iterate over the sub collection, set the new value, delete the old, and update the whole document. It was relatively painless. Granted I only have a few tens of thousands of rows to search through, of which only a few dozen meet the criteria.
Still, I hope this answer helps someone!
Edit: Added snapshot() to the query. See why in the comments.
You must apply snapshot() to the cursor before retrieving any documents from the database.
You can only use snapshot() with unsharded collections.
From MongoDB 3.4, snapshot() function was removed. So if using Mongo 3.4+ ,the example above should remove snapshot() function.
As mentioned in the documentation there is no way to directly rename fields within arrays with a single command. Your only option is to iterate over your collection documents, read them and update each with $unset old/$set new operations.
I had a similar problem. In my situation I found the following was much easier:
I exported the collection to json:
mongoexport --db mydb --collection modules --out modules.json
I did a find and replace on the json using my favoured text editing utility.
I reimported the edited file, dropping the old collection along the way:
mongoimport --db mydb --collection modules --drop --file modules.json
Starting Mongo 4.2, db.collection.update() can accept an aggregation pipeline, finally allowing the update of a field based on its own value:
// { general: { files: { file: [
// { version: { software_program: "MonkeyPlus", indentifier: "6.0.0" } }
// ] } } }
db.collection.updateMany(
{},
[{ $set: { "general.files.file": {
$map: {
input: "$general.files.file",
as: "file",
in: {
version: {
software_program: "$$file.version.software_program",
identifier: "$$file.version.indentifier" // fixing the typo here
}
}
}
}}}]
)
// { general: { files: { file: [
// { version: { software_program: "MonkeyPlus", identifier: "6.0.0" } }
// ] } } }
Literally, this updates documents by (re)$setting the "general.files.file" array by $mapping its "file" elements in a "version" object containing the same "software_program" field and the renamed "identifier" field which contains what used to be the value of "indentifier".
A couple additional details:
The first part {} is the match query, filtering which documents to update (in this case all documents).
The second part [{ $set: { "general.files.file": { ... }}}] is the update aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline):
$set is a new aggregation operator which in this case replaces the value of the "general.files.file" array.
Using a $map operation, we replace all elements from the "general.files.file" array by basically the same elements, but with an "identifier" field rather than "indentifier":
input is the array to map.
as is the variable name given to looped elements
in is the actual transformation applied on elements. In this case, it replaces elements by a "version" object composed by a "software_program" and a "identifier" fields. These fields are populated by extracting their previous values using the $$file.xxxx notation (where file is the name given to elements from the as part).
I had to face the issue with the same schema. So this query will helpful for someone who wants to rename the field in an embedded array.
db.getCollection("sampledocument").updateMany({}, [
{
$set: {
"general.files.file": {
$map: {
input: "$general.files.file",
in: {
version: {
$mergeObjects: [
"$$this.version",
{ identifer: "$$this.version.indentifier" },
],
},
},
},
},
},
},
{ $unset: "general.files.file.version.indentifier" },
]);
Another Solution
I also would like rename a property in array: and I used thaht
db.getCollection('YourCollectionName').find({}).snapshot().forEach(function(a){
a.Array1.forEach(function(b){
b.Array2.forEach(function(c){
c.NewPropertyName = c.OldPropertyName;
delete c["OldPropertyName"];
});
});
db.getCollection('YourCollectionName').save(a)
});
The easiest and shortest solution using aggregate (Mongo 4.0+).
db.myCollection.aggregate([
{
$addFields: {
"myArray.newField": {$arrayElemAt: ["$myArray.oldField", 0] }
}
},
{$project: { "myArray.oldField": false}},
{$out: {db: "myDb", coll: "myCollection"}}
])
The problem using forEach loop as mention above is the very bad performance when the collection is huge.
My proposal would be this one:
db.nrel.component.aggregate([
{ $unwind: "$general.files.file" },
{
$set: {
"general.files.file.version.identifier": {
$ifNull: ["$general.files.file.version.indentifier", "$general.files.file.version.identifier"]
}
}
},
{ $unset: "general.files.file.version.indentifier" },
{ $set: { "general.files.file": ["$general.files.file"] } },
{ $out: "nrel.component" } // carefully - it replaces entire collection.
])
However, this works only when array general.files.file has a single document only. Most likely this will not always be the case, then you can use this one:
db.nrel.componen.aggregate([
{ $unwind: "$general.files.file" },
{
$set: {
"general.files.file.version.identifier": {
$ifNull: ["$general.files.file.version.indentifier", "$general.files.file.version.identifier"]
}
}
},
{ $unset: "general.files.file.version.indentifier" },
{ $group: { _id: "$_id", general_new: { $addToSet: "$general.files.file" } } },
{ $set: { "general.files.file": "$general_new" } },
{ $unset: "general_new" },
{ $out: "nrel.component" } // carefully - it replaces entire collection.
])