Which value pass to param select all records in mongo db collections using meteor.
For Example :
var filterResult = auth_user.find({staff:0,issuper:0}).fetch();
The above one to pass either 0 or 1 this time not a problem.But some times pass only staff value and issuper is to select all records and vice versa.If not given two values select all records in a collection.
I need the above one in a single query? is it possible or suggest me what to do for this?
Its hard to know what you're asking but in general you can get everything in the collection without any query at all:
var filterResult = auth_user.find({}).fetch();
Hi I found that the output of the two queries was same, but I want to know is there any difference in executing the query.
First:
db.collectionname.find({}).pretty()
Second:
db.offers.find().pretty()
There is no difference between these two queries.
db.collectionname.find({}).pretty()
You are not giving any query params here so the results are same like
db.collectionname.find().pretty()
In short: Both function in the same way.
1) db.collectionname.find({})
Here you are not specifying any query parameters and its just an empty document {} so it will return you all the documents present in the collection
2) db.offers.find()
Here you didn't specify any queries. So it need not even look at the parameters, it will just print you all the documents in the collection. find() short form of find({})
I have the following problem: I can't limit number of results when using distinct. Exemple :
$stores = $this->dm->createQueryBuilder('Application\Document\Item')
->distinct('storeName')
->limit(10)
->getQuery()
->execute();
This query render 100 entries but I want only 10 results.
With query builder class in ORM you need to use:
->setMaxResults(10);
As per #Siol and #john Smith said, in ODM you could use limit:
->limit(10);
I don't think distinct will work with limit as suggested in the Jira mongodb issue ticket Ability to use Limit() with Distinct():
The current Distinct() implementation only allows for bringing back
ALL distinct values in the collection or matching a query, but there
is no way to limit these results. This would be very convenient and
there are many use cases.
I have over 300k records in one collection in Mongo.
When I run this very simple query:
db.myCollection.find().limit(5);
It takes only few miliseconds.
But when I use skip in the query:
db.myCollection.find().skip(200000).limit(5)
It won't return anything... it runs for minutes and returns nothing.
How to make it better?
One approach to this problem, if you have large quantities of documents and you are displaying them in sorted order (I'm not sure how useful skip is if you're not) would be to use the key you're sorting on to select the next page of results.
So if you start with
db.myCollection.find().limit(100).sort({created_date:true});
and then extract the created date of the last document returned by the cursor into a variable max_created_date_from_last_result, you can get the next page with the far more efficient (presuming you have an index on created_date) query
db.myCollection.find({created_date : { $gt : max_created_date_from_last_result } }).limit(100).sort({created_date:true});
From MongoDB documentation:
Paging Costs
Unfortunately skip can be (very) costly and requires the server to walk from the beginning of the collection, or index, to get to the offset/skip position before it can start returning the page of data (limit). As the page number increases skip will become slower and more cpu intensive, and possibly IO bound, with larger collections.
Range based paging provides better use of indexes but does not allow you to easily jump to a specific page.
You have to ask yourself a question: how often do you need 40000th page? Also see this article;
I found it performant to combine the two concepts together (both a skip+limit and a find+limit). The problem with skip+limit is poor performance when you have a lot of docs (especially larger docs). The problem with find+limit is you can't jump to an arbitrary page. I want to be able to paginate without doing it sequentially.
The steps I take are:
Create an index based on how you want to sort your docs, or just use the default _id index (which is what I used)
Know the starting value, page size and the page you want to jump to
Project + skip + limit the value you should start from
Find + limit the page's results
It looks roughly like this if I want to get page 5432 of 16 records (in javascript):
let page = 5432;
let page_size = 16;
let skip_size = page * page_size;
let retval = await db.collection(...).find().sort({ "_id": 1 }).project({ "_id": 1 }).skip(skip_size).limit(1).toArray();
let start_id = retval[0].id;
retval = await db.collection(...).find({ "_id": { "$gte": new mongo.ObjectID(start_id) } }).sort({ "_id": 1 }).project(...).limit(page_size).toArray();
This works because a skip on a projected index is very fast even if you are skipping millions of records (which is what I'm doing). if you run explain("executionStats"), it still has a large number for totalDocsExamined but because of the projection on an index, it's extremely fast (essentially, the data blobs are never examined). Then with the value for the start of the page in hand, you can fetch the next page very quickly.
i connected two answer.
the problem is when you using skip and limit, without sort, it just pagination by order of table in the same sequence as you write data to table so engine needs make first temporary index. is better using ready _id index :) You need use sort by _id. Than is very quickly with large tables like.
db.myCollection.find().skip(4000000).limit(1).sort({ "_id": 1 });
In PHP it will be
$manager = new \MongoDB\Driver\Manager("mongodb://localhost:27017", []);
$options = [
'sort' => array('_id' => 1),
'limit' => $limit,
'skip' => $skip,
];
$where = [];
$query = new \MongoDB\Driver\Query($where, $options );
$get = $manager->executeQuery("namedb.namecollection", $query);
I'm going to suggest a more radical approach. Combine skip/limit (as an edge case really) with sort range based buckets and base the pages not on a fixed number of documents, but a range of time (or whatever your sort is). So you have top-level pages that are each range of time and you have sub-pages within that range of time if you need to skip/limit, but I suspect the buckets can be made small enough to not need skip/limit at all. By using the sort index this avoids the cursor traversing the entire inventory to reach the final page.
My collection has around 1.3M documents (not that big), properly indexed, but still takes a big performance hit by the issue.
After reading other answers, the solution forward is clear; the paginated collection must be sorted by a counting integer similar to the auto-incremental value of SQL instead of the time-based value.
The problem is with skip; there is no other way around it; if you use skip, you are bound to hit with the issue when your collection grows.
Using a counting integer with an index allows you to jump using the index instead of skip. This won't work with time-based value because you can't calculate where to jump based on time, so skipping is the only option in the latter case.
On the other hand,
by assigning a counting number for each document, the write performance would take a hit; because all documents must be inserted sequentially. This is fine with my use case, but I know the solution is not for everyone.
The most upvoted answer doesn't seem applicable to my situation, but this one does. (I need to be able to seek forward by arbitrary page number, not just one at a time.)
Plus, it is also hard if you are dealing with delete, but still possible because MongoDB support $inc with a minus value for batch updating. Luckily I don't have to deal with the deletion in the app I am maintaining.
Just write this down as a note to my future self. It is probably too much hassle to fix this issue with the current application I am dealing with, but next time, I'll build a better one if I were to encounter a similar situation.
If you have mongos default id that is ObjectId, use it instead. This is probably the most viable option for most projects anyway.
As stated from the official mongo docs:
The skip() method requires the server to scan from the beginning of
the input results set before beginning to return results. As the
offset increases, skip() will become slower.
Range queries can use indexes to avoid scanning unwanted documents,
typically yielding better performance as the offset grows compared to
using skip() for pagination.
Descending order (example):
function printStudents(startValue, nPerPage) {
let endValue = null;
db.students.find( { _id: { $lt: startValue } } )
.sort( { _id: -1 } )
.limit( nPerPage )
.forEach( student => {
print( student.name );
endValue = student._id;
} );
return endValue;
}
Ascending order example here.
If you know the ID of the element from which you want to limit.
db.myCollection.find({_id: {$gt: id}}).limit(5)
This is a lil genious solution which works like charm
For faster pagination don't use the skip() function. Use limit() and find() where you query over the last id of the precedent page.
Here is an example where I'm querying over tons of documents using spring boot:
Long totalElements = mongockTemplate.count(new Query(),"product");
int page =0;
Long pageSize = 20L;
String lastId = "5f71a7fe1b961449094a30aa"; //this is the last id of the precedent page
for(int i=0; i<(totalElements/pageSize); i++) {
page +=1;
Aggregation aggregation = Aggregation.newAggregation(
Aggregation.match(Criteria.where("_id").gt(new ObjectId(lastId))),
Aggregation.sort(Sort.Direction.ASC,"_id"),
new CustomAggregationOperation(queryOffersByProduct),
Aggregation.limit((long)pageSize)
);
List<ProductGroupedOfferDTO> productGroupedOfferDTOS = mongockTemplate.aggregate(aggregation,"product",ProductGroupedOfferDTO.class).getMappedResults();
lastId = productGroupedOfferDTOS.get(productGroupedOfferDTOS.size()-1).getId();
}
If my collection has 10 records.
my $records = $collection->find;
while (my $record = $records->next){
do something;
}
Are there ten roundtrips to the mongodb server?
If so, is there any way to limit it to one roundtrip?
Thanks.
The answer is it's just one query per batch of records/documents returned in groups of 100 by default.
If your result set is 250 docs, the first access of the cursor to get doc 1 will load docs 1-100 in memory, when doc 101 is accessed this causes another 100 docs to be loaded from the server, and finally one more query for the last 50 docs.
See the mongodb docs about cursors and "getmore" command.
It's a single query, just like querying a RDBMS.
As per the documentation:
my $cursor = $collection->find({ i => { '$gt' => 42 } });
Executes the given $query and returns a MongoDB::Cursor with the results
my $cursor = $collection->query({ }, { limit => 10, skip => 10 });
Valid query attributes are:
limit - Limit the number of results.
skip -Skip a number of results.
sort_by - Order results.
No, i am absolutely sure that in above code only one roundtrip to the server. For example in c# the same code will load all data only once, when you start iteration.
while (my $record = $records->next){
^^^
here on first iteration driver load all 10 records
It seems to me logical have only one request to the server.
From documentation:
The shell find() method returns a
cursor object which we can then
iterate to retrieve specific documents
from the result
You can use the "mongosniff" tool to figure out the operations over the wire. Apart from that: you basically have no other option then iterating over the cursor....so why do you care?