I have a mongo database with a collection of countries.
One property (currencies) contains an array of currencies.
A currency has multiple properties:
"currencies": [{
"code": "EUR",
"name": "Euro",
"symbol": "€"
}],
I wish to select all countries who use Euro's besides other currencies.
I'm using the following statement:
db.countries.find({currencies: { $in: [{code: "EUR"}]}})
Unfortunately I'm getting an empty result set.
When I use:
db.countries.find({"currencies.code": "EUR"})
I do get results. Why is the first query not working and the second one succesfull?
The first query is not working as it checks whether the whole currency array is in the array, which is never true.
It is true when:
currencies: {
$in: [
[{
"code": "EUR",
"name": "Euro",
"symbol": "€"
}],
...
]
}
I believe that $elemMatch is what you need besides the dot notation.
db.collection.find({
currencies: {
$elemMatch: {
code: "EUR"
}
}
})
Sample Mongo Playground
MongoDB works in the same way if the query field is an array or a single value, that's why the second one works.
So why the first one doesn't work? The problem here is that you are looking for an object that is exactly defined as {code: "EUR"}: no name or symbol field are specified. To make it work, you should change it to:
db.getCollection('countries').find({currencies: { $in: [{
"code" : "EUR",
"name" : "Euro",
"symbol" : "€"
}]}})
or query the subfield directly:
db.getCollection('stuff').find({"currencies.code": { $in: ["EUR"]}})
Related
Hi I am trying to increment the count of the matching requirement in an array. My sample collection looks like the following:
{
"_id": ObjectId("60760ba2e870fa518f2ae48b"),
"userId": "6075f7289822d94dca8066b4",
"requirements": [
{
"searchText": "zee5",
"planType": "basic",
"mode": "PRIVATE",
"count": 32.0
},
{
"searchText": "sony",
"planType": "standard",
"mode": "PUBLIC",
"count": 12.0
},
{
"searchText": "prime",
"planType": "premium",
"mode": "PRIVATE",
"count": 2
}
]
}
If a user searches for prime, with filter premium and PRIVATE, then the count of the last requirement should be updated. If he searches for prime, with filter standard and PRIVATE, then the new requirement will be inserted with count 1.
I am doing in two steps. First I fire an update with the following query and then if no update, I fire a push query with count 1:
db.getCollection('userProfile').update({ "$and" : [{ "requirements.searchText" : {$eq:"prime"}}, {"requirements.mode" : {$eq: "PUBLIC"}}, {"requirements.planType": {$eq: "standard"}}, { "userId" : "6075f7289822d94dca8066b4"}]}, {$inc: {"requirements.$.count" : 1}})
I was expecting that the above query will not update any requirement, since there is no exact match. Interestingly, it increments the count of the second requirement with (sony, standard, public). What is wrong with the query? How can I get it right?
Demo - with Update - https://mongoplayground.net/p/-ISXaAayxxv
Demo No update - https://mongoplayground.net/p/88bTj3lz7U_
Use $elemMatch to make sure all properties are present in the same object inside the array
The $elemMatch operator matches documents that contain an array field with at least one element that matches all the specified query criteria.
db.collection.update(
{
"requirements": {
$elemMatch: { "searchText": "prime","mode": "PUBLIC", "planType": "standard" }
},
"userId": "6075f7289822d94dca8066b4"
},
{ $inc: { "requirements.$.count": 1 } }
)
Problem -
Your current query will match any document with all these fields in
requirements array in any object, if they match 1 property in 1 index of the array and another match in the next index query will find the document valid.
"searchText": "prime",
"mode": "PUBLIC",
"planType": "standard"
I am beginner in MongoDB and struck at a place I am trying to fetch data from nested array but is it taking so long time as data is around 50K data, also it is not much accurate data, below is schema structure please see once -
{
"_id": {
"$oid": "6001df3312ac8b33c9d26b86"
},
"City": "Los Angeles",
"State":"California",
"Details": [
{
"Name": "Shawn",
"age": "55",
"Gender": "Male",
"profession": " A science teacher with STEM",
"inDate": "2021-01-15 23:12:17",
"Cars": [
"BMW","Ford","Opel"
],
"language": "English"
},
{
"Name": "Nicole",
"age": "21",
"Gender": "Female",
"profession": "Law student",
"inDate": "2021-01-16 13:45:00",
"Cars": [
"Opel"
],
"language": "English"
}
],
"date": "2021-01-16"
}
Here I am trying to filter date with date and Details.Cars like
db.getCollection('news').find({"Details.Cars":"BMW","date":"2021-01-16"}
it is returning details of other persons too which do not have cars- BMW , Only trying to display details of person like - Shawn which have BMW or special array value and date too not - Nicole, rest should not appear but is it not happening.
Any help is appreciated. :)
A combination of $match on the top-level fields and $filter on the array elements will do what you seek.
db.foo.aggregate([
{$match: {"date":"2021-01-16"}}
,{$addFields: {"Details": {$filter: {
input: "$Details",
as: "zz",
cond: { $in: ['BMW','$$zz.Cars'] }
}}
}}
,{$match: {$expr: { $gt:[{$size:"$Details"},0] } }}
]);
Notes:
$unwind is overly expensive for what is needed here and it likely means "reassembling" the data shape later.
We use $addFields where the new field to add (Details) already exists. This effectively means "overwrite in place" and is a common idiom when filtering an array.
The second $match will eliminate docs where the date matches but not a single entry in Details.Cars is a BMW i.e. the array has been filtered down to zero length. Sometimes you want to know this info so if this is the case, do not add the final $match.
I recommend you look into using real dates i.e. ISODate instead of strings so that you can easily take advantage of MongoDB date math and date formatting functions.
Is a common mistake think that find({nested.array:value}) will return only the nested object but actually, this query return the whole object which has a nested object with desired value.
The query is returning the whole document where value BMW exists in the array Details.Cars. So, Nicole is returned too.
To solve this problem:
To get multiple elements that match the criteria you can do an aggregation stage using $unwind to separate the different objects into array and match by the criteria you want.
db.collection.aggregate([
{
"$match": { "Details.Cars": "BMW", "date": "2021-01-26" }
},
{
"$unwind": "$Details"
},
{
"$match": { "Details.Cars": "BMW" }
}
])
This query first match by the criteria to avoid $unwind over all collection.
Then $unwind to get every document and $match again to get only the documents you want.
Example here
To get only one element (for example, if you match by _id and its unique) you can use $elemMatch in this way:
db.collection.find({
"Details.Cars": "BMW",
"date": "2021-01-16"
},
{
"Details": {
"$elemMatch": {
"Cars": "BMW"
}
}
})
Example here
You can use $elemenMatch into query or projection stage. Docs here and here
Using $elemMatch into query the way is this:
db.collection.find({
"Details": {
"$elemMatch": {
"Cars": "BMW"
}
},
"date": "2021-01-16"
},
{
"Details.$": 1
})
Example here
The result is the same. In the second case you are using positional operator to return, as docs says:
The first element that matches the query condition on the array.
That is, the first element where "Cars": "BMW".
You can choose the way you want.
Hello Good Developers,
I am facing a situation in MongoDB where I've JSON Data like this
[{
"id": "GLOBAL_EDUCATION",
"general_name": "GLOBAL_EDUCATION",
"display_name": "GLOBAL_EDUCATION",
"profile_section_id": 0,
"translated": [
{
"con_lang": "US-EN",
"country_code": "US",
"language_code": "EN",
"text": "What is the highest level of education you have completed?",
"hint": null
},
{
"con_lang": "US-ES",
"country_code": "US",
"language_code": "ES",
"text": "\u00bfCu\u00e1l es su nivel de educaci\u00f3n?",
"hint": null
}...
{
....
}
]
I am projecting result using the following query :
db.collection.find({
},
{
_id: 0,
id: 1,
general_name: 1,
translated: {
$elemMatch: {
con_lang: "US-EN"
}
}
})
here's a fiddle for the same: https://mongoplayground.net/p/I99ZXBfXIut
I want those records who don't match $elemMatch don't get returned at all.
In the fiddle output, you can see that the second item doesn't have translated attribute, In this case, I don't want the second Item at all to be returned.
I am using Laravel as Backend Tech, I can filter out those records using PHP, but there are lots of records returned, and I think filtering using PHP is not the best option.
You need to use $elemMatch in the first parameter
db.collection.find({
translated: {
$elemMatch: {
con_lang: "IT-EN"
}
}
})
MongoPlayground
In my database I have a field of name. In some records it is an empty string, in others it has a name in it.
In my query, I'm currently doing:
db.users.find({}).sort({'name': 1})
However, this returns results with an empty name field first, then alphabetically returns results. As expected, doing .sort({'name': -1}) returns results with a name and then results with an empty string, but it's in reverse-alphabetical order.
Is there an elegant way to achieve this type of sorting?
How about:
db.users.find({ "name": { "$exists": true } }).sort({'name': 1})
Because after all when a field you want to sort on is not actually present then the returned value is null and therefor "lower" in the order than any positive result. So it makes sense to exclude those results if you really are only looking for something with a matching value.
If you really want all the results in there and regarless of a null content, then I suggest you "weight" them via .aggregate():
db.users.aggregate([
{ "$project": {
"name": 1,
"score": {
"$cond": [
{ "$ifNull": [ "$name", false ] },
1,
10
]
}
}},
{ "$sort": { "score": 1, "name": 1 } }
])
And that moves all null results to the "end of the chain" by assigning a value as such.
If you want to filter out documents with an empty "name" field, change your query: db.users.find({"name": {"$ne": ""}}).sort({"name": 1})
I have mongodb document with data as following :
[{
"name": "Robin singh",
"developer": "java",
"address": "Robin Singh,Mohali",
"search_string": ["Robin", "Singh", "java"]
}, {
"name": "Rohan singh",
"developer": "java",
"address": "Rohan Singh,Mohali",
"search_string": ["Rohan", "Singh", "java"]
}]
I want to search document with developer java with name Rohan singh and I used this query:
{"search_string":{"$all":["Robin","Singh","java"]}}
but I am getting both results.
If you have both name and developer then you can simply use find query -
db.collection.find({"name":"Rohan singh", "developer":"java"})
Additionally, If you want to check when search_string contains exact parameters which you are passing in $all then - You need to use combination of $size and $all operator to get desired result. size must be the number of parameters that you are using in $all. Here you must check size of search_string to number of params in $all so that it will search given parameters ($all) only in those search_string which size matches to number of params.
Following query might be helpful to you.
db.collection.find({
"name":"Rohan singh",
"search_string": {
$all: ["Rohan", "singh","java"]
},
"search_string": {
$size: 3 // This is the number of param you pass in $all
}
}).pretty()
Try:
db.testcollection.aggregate([
{ $match: {"search_string":{"$all":["Robin","Singh","java"]}} },
{ $group: { _id: "$name"} }
])