I have tried and tried on Meteor and on Robomongo (Mongodb) to select objects with dot notation.
I would like to be able to filter team.0.wageringStats.wageringStraightSpread objects (sometimes subjects can be fields or arrays - thats another issue)
In the first image I can select team.wageringStats.wageringStraightSpread and get back all the subOjects of team (team has siblings not shown in images)
The second image I tried team.0.wageringStats.wageringStraightSpread and I get no fields.
Lastly i tried team.[0].wageringStats.wageringStraightSpread and
team[0].wageringStats.wageringStraightSpread and get the same result : 0 fields
I am at a loss and would like some help. Thank you
I am not sure what you are trying to do now? Because in your first command, you already have a list of team that match your criteria and then, put it into the loop of meteor to process. Why do you need to find only the first one ? By the way, in order to select the nth of the result set in mongodb, you will need something like skip and limit
db.collections.find({'team.wageringStats.wageringStraightSpread':1}).limit(1).skip(0)
(in skip, you need to pass the offset you need to reach to)
Also, if you only care about the first one, findOne is the one you need to do the query
db.collections.findOne({'team.wageringStats.wageringStraightSpread':1})
Be aware that the syntax of mongodb and meteor for querying is a bit different
Related
Looking for a way to view the results of a query where the displayed fields in the doc are ordered (lexicographically in my case).
Example:
I'm getting back from a query one document, which is what I need. This document has 30 fields and I'm looking to see the value in one of them. My issue is that the order of the fields is, well, kinda random. Not sorted in any way I'm aware of.
Background:
I am writing a mobile application which has lazy loading page. My backend is using go and mongodb with mongo-go driver. There are 10 elements on that page and i want to get next ten when i scroll to the bottom. I am planning to send ObjectID (_id) as request query parameter and get next ten elements starting from the index of id + 1.
I write what i want in mongo shell "language" so that more people understand what i want and can help in shell syntax.
Is there a way to get index of the document by it's _id or may be i can get skip until it in skip().
something like db.collection.find().skip(idOfDocument+1).limit(10)
I found the answer here.
nextDocuments = db.collection.find({'_id'> last_id}).limit(10)
I need to get the users whose ids are contained in an array. For this i'm using the $in operator, however being this inside an aggregate operation, i'd like to get back a specific user all the time it's id is present in the array, not just one. For example:
The ids array is A=[a,b,c,b] and U(x) is user with id x
with users.find({_id:{$in:A}}) i get these users as result: U(a),U(b),U(c)
instead i'd like to get back the result: U(a),U(b),U(c),U(b)
so get the user back every time it's id appears.
I understand that $in is working as expected but does anyone have an idea on how can i achieve this?
Thanks
This isn't possible using a MongoDB query.
MongoDB's query engine iterates over the documents in a collection (or over an index if there's a useful one) and returns to you any documents that match your query, in the order it finds them. Whether b appears once, twice, or a hundred times in your query makes no difference: the document with _id of b matches the query and is returned once, when MongoDB finds it.
You can do a post-processing step in your programming language to repeat documents as many times as you want.
This seems like it should be very simple but I can't get it to work. I want to select all documents A where there are one or more B elements in a sub collection.
Like if a Store document had a collection of Employees. I just want to find Stores with 1 or more Employees in it.
I tried something like:
{Store.Employees:{$size:{$ne:0}}}
or
{Store.Employees:{$size:{$gt:0}}}
Just can't get it to work.
This isn't supported. You basically only can get documents in which array size is equal to the value. Range searches you can't do.
What people normally do is that they cache array length in a separate field in the same document. Then they index that field and make very efficient queries.
Of course, this requires a little bit more work from you (not forgetting to keep that length field current).
I have collections with huge amount of Documents on which I need to do custom search with various different queries.
Each Document have boolean property. Let's call it "isInTop".
I need to show Documents which have this property first in all queries.
Yes. I can easy do sort in this field like:
.sort( { isInTop: -1 } );
And create proper index with field "isInTop" as last field in it. But this will be work slowly, as indexes in mongo works best with unique fields.
So is there is solution to show Documents with field "isInTop" on top of each query?
I see two solutions here.
First: set Documents wich need to be in top the _id from "future". As you know, ObjectId contains timestamp. So I can create ObjectId with timestamp from future and use natural order
Second: create separate collection for Ducuments wich need to be in top. And do queries in it first.
Is there is any other solutions for this problem? Which will work fater?
UPDATE
I have done this issue with sorting on custom field which represent rank.
Using the _id field trick you mention has the problem that at some point in time you will reach the special time, and you can't change the _id field (without inserting a new document and removing the old one).
Creating a special collection which just holds the ones you care about is probably the best option. It gives you the ability to logically (and to some extent, physically) separate the documents.
Newly introduced in mongodb there is also support for a "sparse" index which may fulfill your needs as well. You could only set the "isInTop" field when you want it to be special, and then create a sparse index on it which would not have the problems you would normally have with a single indexed boolean field (in btrees).