Is there a way to push all the documents of a given collection in a array?
I did this but is there any quicker way?
var ops = [];
db.getCollection('stock').find({}).forEach(function (stock) {
ops.push(stock);
})
PS: I use Mongo 3.4
You can just use the toArray function on the cursor that's returned from find, like this:
var ops = db.getCollection('stock').find({}).toArray();
Note: As with your original solution, this might suffer with performance if the stock collection contains millions of documents.
As an aside, you can use db.stock directly to shorten the query a little bit:
var ops = db.stock.find({}).toArray();
Try using lean query option. in your case:
db.getCollection('stock').find({}).lean()
You could as well use $facet which will allow you to create the array on the server side - provided the resulting document array is no bigger than 16MB in which case you'll get an exception:
db.stock.aggregate({
$facet: {
ops: [ { $match: {} } ]
}
})
In order to reduce the amount of data returned you could limit the number of returned fields in the above pipeline (instead of an empty $match stage - which is a hack anyway - you would then use $project).
Related
I'm not sure this is possible, but i'd like to create a single view or at least a single query that looks in different collections based on what's being queried.
for example, if the first character is an "A" look in the "Aresults" collection, if it's a "B" look in the "Bresults" collection, etc.
I could potentially create a "A-Z" collection with just those letters, and do a $lookup from there based on a condition, but i'm not sure how to do that either.
I am aware that i could create a view with a $unionWith having all the "*results" collections, but that seems very inefficient.
Any other ideas? Is there perhaps some type of dynamic query structure within mongodb like in MySQL (couldn't find any)?
Thanks
Something like this?
const prefix = db.meta_data.findOne({field: condition}).prefix ;
db.createView('view_name', prefix + 'results', [<your aggregation pipeline>]);
or this?
const pipeline = [];
db.meta_data.find({ field: condition }).forEach(x => {
pipeline.push({ $unionWith: { coll: prefix + 'results' } });
});
db.collection.aggregate([pipeline]);
I would like to count the number of documents returned by an aggregation.
I'm sure my initial aggregation works, because I use it later in my programm. To do so I created a pipeline variable (here called pipelineTest, ask me if you want to see it in detail, but it's quite long, that's why I don't give the lines here).
To count the number of documents returned, I push my pipeline with :
{$count: "totalCount"}
Now I would like to get (or log) totalCount value. What should I do ?
Here is the aggregation :
pipelineTest.push({$count: "totalCount"});
cursorTest = collection.aggregate(pipelineTest, options)
console.log(cursorTest.?)
Thanks for your help, I read lot and lot doc about aggregation and I still don't understand how to read the result of an aggregation...
Assuming you're using async/await syntax - you need to await on the result of the aggregation.
You can convert the cursor to an array, get the first element of that array and access totalCount.
pipelineTest.push({$count: "totalCount"});
cursorTest = await collection.aggregate(pipelineTest, options).toArray();
console.log(cursorTest[0].totalCount);
Aggregation
db.mycollection.aggregate([
{
$count: "totalCount"
}
])
Result
[ { totalCount: 3 } ]
Your Particulars
Try the following:
pipelineTest.push({$count: "totalCount"});
cursorTest = collection.aggregate(pipelineTest, options)
console.log(cursorTest.totalCount)
I need to search for the existence of a property that is within another object.
the collection contains documents that look like:
"properties": {
"source": {
"a/name": 12837,
"a/different/name": 76129
}
}
As you can see below, part of the query string is from a variable.
With some help from JohnnyHK (see mongo query - does property exist? for more info), I've got a query that works by doing the following:
var name = 'a/name';
var query = {};
query['properties.source.' + name] = {$exists: true};
collection.find(query).toArray(function...
Now I need to see if I can index the collection to improve the performance of this query.
I don't have a clue how to do this or if it is even possible to index for this.
Suggestions?
2 things happening in here.
First probably you are looking for sparse indexes.
http://docs.mongodb.org/manual/core/index-sparse/
In your case it could be a sparse index on "properties.source.a/name" field. Making indexes on field will dramatically improve your query lookup time.
db.yourCollectionName.createIndex( { "properties.source.a/name": 1 }, { sparse: true } )
Second thing. Always when you want to know whether your query is fast/slow, use mongo console, run your query and on its result call explain method.
db.yourCollectionName.find(query).explain();
Thanks to it you will know whether your query uses indexes or not, how many documents it had to check in order to complete query and some others useful information.
I have a mongoDB document that has the following structure:
{
user:user_name,
streams:[
{user:user_a, name:name_a},
{user:user_b, name:name_b},
{user:user_c, name:name_c}
]
}
I want to use $pullAll to remove from the streams array, passing it an array of streams (the size of the array varies from 1 to N):
var streamsA = [{user:"user_a", name:"name_a"},{user:"user_b", name:"name_b"}]
var streamsB = [{name:"name_a", user:"user_a"},{name:"name_b", user:"user_b"}]
I use the following mongoDB command to perform the update operation:
db.streams.update({name:"user_name", {"$pullAll:{streams:streamsA}})
db.streams.update({name:"user_name", {"$pullAll:{streams:streamsB}})
Removing streamsA succeeds, whereas removing streamsB fails. After digging through the mongoDB manuals, I saw that the order of fields in streamsA and streamsB records has to match the order of fields in the database. For streamsB the order does not match, that's why it was not removed.
I can reorder the streams to the database document order prior to performing an update operation, but is there an easier and cleaner way to do this? Is there some flag that can be set to update and/or pullAll to ignore the order?
Thank You,
Gary
The $pullAll operator is really a "special case" that was mostly intended for single "scalar" array elements and not for sub-documents in the way you are using it.
Instead use $pull which will inspect each element and use an $or condition for the document lists:
db.streams.update(
{ "user": "user_name" },
{ "$pull": { "streams": { "$or": streamsB } }}
)
That way it does not matter which order the fields are in or indeed look for an "exact match" as the current $pullAll operation is actually doing.
I am trying to retrieve one element from a mongo collection, the one with the greatest _id field. I know this can be done by querying:
db.collection.find().sort({_id: -1}).limit(1)
But it kind of seems unelegant and I was wondering whether there is a way to get that specific element using findOne()
Note: I want to do this because, from what I've read in ObjectId, the first bytes correspond to the miliseconds since the Epoch and thus, the last element being inserted will have the greatest _id. Is there any other way to retrieve the last element inserted in a collection?
You should use find, like you already are, and not aggregation which will be slower since it needs to scan all the values of _id fields to figure out the max.
As comments pointed out there is no difference between using find() and findOne() - functionally or elegance-wise. In fact, findOne in the shell (and in the drivers which implement it) is defined in terms of find (with limit -1 and with pretty print in the shell).
If you really want to do the equivalent of
db.collection.find().sort({_id:-1}).limit(1).pretty()
as findOne you can do it with this syntax:
db.collection.findOne({$query:{},$orderby:{_id:-1}})
You can get max _id using aggregation of mongodb. Find and sort may overkill's.
db.myCollection.aggregate({
$group: {
_id: '',
last: {
$max: "$_id"
}
}
});
with PHP driver (mongodb)
using findOne()
$filter=[];
$options = ['sort' => ['_id' => -1]]; // -1 is for DESC
$result = $collection->findOne(filter, $options);
$maxAge = $result['age']
import pymongo
tonystark = pymongo.MongoClient("mongodb://localhost:27017/")
mydb = tonystark["tonystark_db"]
savings = mydb["customers"]
x = savings.find().sort("_id")
for s in x:
print(s)
$maxId="";
$Cursor =$collection->find();
foreach($cursor as $document) {
$maxid =max($arr=(array($document['id'])));
}
print_r($maxid+1);