I am unable to add the following object:
[
{ 'option1':'opt1','option2':'opt2','option3':'opt3'},
]
as a sub filed to:
const question_list=await Questions.find({ $and: [{categoryid:categoryId},{ isDeleted: false }, { status: 0 }] }, { name: 1 });
question_list=[{"_id":"5eb167fb222a6e11fc6fe579","name":"q1"},{"_id":"5eb1680abb913f2810774c2a","name":"q2"},{"_id":"5eb16b5686068831f07c65c3","name":"q5"}]
I want the final Object to be as:
[{"_id":"5eb167fb222a6e11fc6fe579","name":"q1","options":[
{ 'option1':'opt1','option2':'opt2','option3':'opt3'},
]},{"_id":"5eb1680abb913f2810774c2a","name":"q2","options":[
{ 'option1':'opt1','option2':'opt2','option3':'opt3'},
]},{"_id":"5eb16b5686068831f07c65c3","name":"q5","options":[
{ 'option1':'opt1','option2':'opt2','option3':'opt3'},
]}]
what is the best possible solution?
You need to use aggregation-pipeline to do this, instead of .find(). As projection in .find() can only accept $elemMatch, $slice, and $ on existing fields : project-fields-from-query-results. So to add a new field with new data to documents, use $project in aggregation framework.
const question_list = await Questions.aggregate([
{
$match: {
$and: [{ categoryid: categoryId }, { isDeleted: false }, { status: 0 }]
}
},
{
$project: {
_id: 0,
name: 1,
options: [{ option1: "opt1", option2: "opt2", option3: "opt3" }]
}
}
]);
Test : mongoplayground
Related
This is the schema model
const tokoSchema = new mongoose.Schema({
name: String
product: [{display: Boolean}]
}
So, what I want is filtering the data by product.display. When I filtered it using toko.find({"product.display": true}) it only shows list of toko that has product.display: true but what I want is if the product display is false it keep showing the _id and name of toko with empty array of product []. Example query expected
// showing all tokos
[
{
_id: blabla
name: "test",
product: [{_id: blabla, display: true}], // filter display:true
},
{
_id: blabla,
name: "test",
product: [] // no product with display:true
}
]
example of toko.find({"product.display": true}) query
// not showing all tokos
[
{
_id: blabla
name: "test",
product: [{_id: blabla, display: true}]
},
]
any solution for this? should I use aggregate?
You need $filter for an array:
db.toko.aggregate([
{
$addFields: {
product: {
$filter: {
input: "$product",
cond: {
$eq: [ "$$this.display", true ]
}
}
}
}
}
])
Mongo Playground
I have a collection with the following data structure:
{
_id: ObjectId,
text: 'This contains some text',
type: 'one',
category: {
name: 'Testing',
slug: 'test'
},
state: 'active'
}
What I'm ultimately trying to do is get a list of categories and counts. I'm using the following:
const query = [
{
$match: {
state: 'active'
}
},
{
$project: {
_id: 0,
categories: 1
}
},
{
$unwind: '$categories'
},
{
$group: {
_id: { category: '$categories.name', slug: '$categories.slug' },
count: { $sum: 1 }
}
}
]
This returns all categories (that are active) and the total counts for documents matching each category.
The problem is that I want to introduce two additional $match that should still return all the unique categories, but only affect the counts. For example, I'm trying to add a text search (which is indexed on the text field) and also a match for type.
I can't do this at the top of the pipeline because it would then only return categories that match, not only affect the $sum. So basically it would be like being able to add a $match within the $group only for the $sum. Haven't been able to find a solution for this and any help would be greatly appreciated. Thank you!
You can use $cond inside of your $group statement:
{
$group: {
_id: { category: '$categories.name', slug: '$categories.slug' },
count: { $sum: { $cond: [ { $eq: [ "$categories.type", "one" ] }, 1, 0 ] } }
}
}
I am having some issue writing a find/aggregate mongo query where my requirement is to get all the documents but having condition like:
Suppose I have 2 documents:
{
_id: 5ccaa76939d95d395791efd2,
name: 'John Doe',
email: 'john.doe#foobar.com',
private: true
}
{
_id: 5ccaa76939d95d395791efd2,
name: 'Jane Doe',
email: 'jane.doe#foobar.com',
private: false
}
Now the query I am trying to get my head around is if the field private is true then when I query I must get all documents except email fields not included if private is true, like this:
{
_id: 5ccaa76939d95d395791efd2,
name: 'John Doe',
private: true
}
{
_id: 5ccaa76939d95d395791efd2,
name: 'Jane Doe',
email: 'jane.doe#foobar.com',
private: false
}
Tried $redact, $cond, $$PRUNE, $$DESCEND in aggregate() as well as came across $$REMOVE (looks like it is newest feature) but unable to get the required output. Please help me out with the Query
You can use $$REMOVE to remove a field from returned documents.
db.collection.aggregate([
{ "$addFields": {
"email": {
"$cond": [
{ "$eq": ["$private", true] },
"$$REMOVE",
"$email"
]
}
}}
])
MongoPlayground
Thank you Anthony Winzlet, his solution worked like a charm.
If anyone faces same problem and requires to include more than 1 fields, I am doing so by writing this method:
function condition(privateFieldLimitationArray, publicFieldLimitationArray) {
const condition = {};
privateFieldLimitationArray.map((d, i) => {
condition[d] = {
"$cond": [
{ "$eq": ["$private", true] },
"$$REMOVE",
publicFieldLimitationArray.includes(d) ? '$$REMOVE' : '$'+d
]
}
});
return condition;
}
Then, you can use the above function like:
const privateLimitationArray = ['updatedAt', 'createdAt', 'email', 'lname', 'friendslist', '__v'];
const publicLimitationArray = ['updatedAt', 'createdAt', '__v'];
YourModel.aggregate([
{
$match: {
// Your query to find by
}
}, {
"$addFields": condition(privateLimitationArray, publicLimitationArray)
}
])
.then(result => {
// handle result
})
.catch(error => {
// handle error
});
is it possible in MongoDB to find some objects that match a query and then to modify the result without modifying the persistent data?
For example, let
students = [
{ name: "Alice", age: 25 },
{ name: "Bob", age: 22 },
{ name: "Carol", age: 19 },
{ name: "Dave", age: 18}
]
Now, I want to query all students that are younger than 20 and in the search result, I just want to replace "age: X" with "under20: 1" resulting in the following:
result = [
{ name: "Carol", under20: 1 },
{ name: "Dave", under20: 1}
]
without changing anything in the database.
Sure, it is possible to get the result and then call a forEach on it, but that sounds so inefficient because I have to rerun every object again, so I'm searching for an alternative. Or is there no one?
A possible solution would be to use an aggregation pipline with a $match followed by a $project:
db.students.aggregate(
[
{
$match: { age: { $lt: 20 } }
},
{
$project:
{
_id: false,
name: true,
under20: { $literal: 1 }
}
}
]);
The $literal: 1 is required as just using under20: 1 is the same as under20: true, requesting that field under20 be included in the result: which would fail as under20 does not exist in the document produced by the match.
Or to return all documents in students and conditionally generate the value for under20 a possible solution would be to use $cond:
db.students.aggregate(
[
{
$project:
{
_id: false,
name: true,
under20:
{
$cond: { if: { $lt: [ "$age", 20 ] }, then: 1, else: 0 }
}
}
}
]);
This is how I check if a document exists:
var query = {};
if (req.body.id) {
query._id = {
$ne: new require('mongodb').ObjectID.createFromHexString(req.body.id)
};
}
Creditor.native(function(err, collection) {
collection.find({
$or: [{
name: req.body.name
}, {
sapId: req.body.sapId
},
query
]
}).limit(-1).toArray(function(err, creditors) {
if (creditors.length > 0) {
return res.send(JSON.stringify({
'message': 'creditor_exists'
}), 409);
} else {
return next();
}
})
});
To avoid that multiple documents exist with the same name or/and the same sapID I do this check on every creation/update of a document.
E.g. I want to update this document and give it a different name
{
name: 'Foo',
sapId: 123456,
id: '541ab60f07a955f447a315e4'
}
But when I log the creditors variable I get this:
[{
_id: 541a89a9bcf55f5a45c6b648,
name: 'Bar',
sapId: 3454345
}]
But the query should only match the same sapID/name. However there totally not the same. Is my query wrong?
You're currently finding docs where name matches OR sapId matches OR _id doesn't match. So that last clause is what's pulling in the doc you're seeing.
You probably mean to find docs where (name matches OR sapId matches) AND _id doesn't match.
collection.find({ $and: [
query,
{ $or: [{
name: req.body.name
}, {
sapId: req.body.sapId
}
] } ]
})
Or more simply:
collection.find({
_id: { $ne: require('mongodb').ObjectID.createFromHexString(req.body.id) },
$or: [{
name: req.body.name
}, {
sapId: req.body.sapId
}
]
})