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).
Related
The chat app schema that I have is something like below.
1. conversations {participants[user_1, user_2], convsersation_id}
2. messages {sender: user_1, sonversation_id, timestamps}
I want to map this relationship using existing _id:ObjectId which is already indexed.
But if I want to get all conversation of user_1 I have to first search in which conversation that user is involed and get that conversation's _id and again search for the messages in messages using that conversation _id.
So my questions are -
Does length of indexed field (here _id) matters while searching?
Should I create another shorter indexed fields?.
Also if there is any better alternative schema please suggest.
I would suggest you to maintain the data as sub documents instead of array. The advantage you have is you can build another index (only) on conversation_id field, which you want to query to know the user's involvement
When you maintain it as array, you cannot index the converstaion_id field separately, instead you will have to build a multi key index, which indexes all the elements of the array (sender and timestamps fields) which you are never going to use for querying and it also increases the index size
Answering you questions:
Does length of indexed field (here _id) matters while searching? - Not really
Should I create another shorter indexed fields? - Create sub-document and index converstaion_id
Also if there is any better alternative schema please suggest. - Maintain the array fields as sub-documents
How costly is it to index some fields in MongoDB,
I have a table where i want uniqueness combining two fields, Every where i search they suggested compound index with unique set to true. But what i was doing is " Appending both field1_field2 and making it a key, so that field2 will be always unique for field1.(and add Application logic) As i thought indexing is costly.
And also as MongoDB documentation advices us not to use Custom Object ID like auto incrementing number, I end up giving big numbers to Models like Classes, Students etc, (where i could have used easily used 1,2,3 in sql lite), I didn't think to add a new field for numbering and index that field for querying.
What are the best practices advice for production
The advantage of using compound indexes vs your own indexed field system is that compound indexes allows sorting quicker than regular indexed fields. It also lowers the size of every documents.
In your case, if you want to get the documents sorted with values in field1 ascending and in field2 descending, it is better to use a compound index. If you only want to get the documents that have some specific value contained in field1_field2, it does not really matter if you use compound indexes or a regular indexed field.
However, if you already have field1 and field2 in seperate fields in the documents, and you also have a field containing field1_field2, it could be better to use a compound index on field1 and field2, and simply delete the field containing field1_field2. This could lower the size of every document and ultimately reduce the size of your database.
Regarding the cost of the indexing, you almost have to index field1_field2 if you want to go down that route anyways. Queries based on unindexed fields in MongoDB are really slow. And it does not take much more time adding a document to a database when the document has an indexed field (we're talking 1 millisecond or so). Note that adding an index on many existing documents can take a few minutes. This is why you usually plan the indexing strategy before adding any documents.
TL;DR:
If you have limited disk space or need to sort the results, go with a compound index and delete field1_field2. Otherwise, use field1_field2, but it has to be indexed!
I'm new to MongoDB and I'm not how to best solve my fairly basic problem.
I have a Collection of "emoji" Documents in my database. At any given time, there is one (and only one) "selected" emoji Document. This is determined and updated by the application. How can I persist the information of which one is selected to the database?
Approach 1:
Add a new Collection to hold this kind of metadata of the emoji collection? I'm thinking it would hold a single document with a reference to the currently selected emoji document. This seems to hurt the OO design. A whole collection, with a single document, to hold a single property. But it does have flexibility to add more metadata.
Approach 2:
Add a new boolean field to each emoji Document indicating whether or not it is the current selected emoji. This seems like a lot of extra info to track for each Document, when only one should have a true value. I would also be concerned with maintaining consistency.
I know I'm not the first person to have this issue, but I couldn't find a solution this is as a general case. Thanks!
MongoDB is schemaless so you can just add the boolean field to the currently selected emoji and remove it when the selection changes. You should add a parse unique index to make querying this field faster. You could set the field using this syntax:
db.emojis.update({name:"b"},{$set:{selected:true}})
And simply unset it like this:
db.emojis.update({name:"b"},{$unset:{selected:""}})
You could create the following parse unique index to ensure there is only ever one field with selected:true
db.emojis.createIndex( { selected: 1 } , { sparse: true, unique: true } )
I have a class/object that have a guid and i want to use that field as the _id object when it is saved to Mongodb. Is it possible to use other value instead of the ObjectId?
Is there any performance consideration when doing bulk insert when there is an _id field? Is _id an index? If i set the _id to different field, would it slow down the bulk insert? I'm inserting about 10 million records.
1) Yes you can use that field as the id. There is no mention of what API (if any) you are using for inserting the documents. So if you would do the insertion at the command line, the command would be:
db.collection.insert({_id : <BSONString_version_of_your_guid_value>, field1 : value1, ...});
It doesn't have to be BsonString. Change it to whatever Bson value is closest matching to your guid's original type (except the array type. Arrays aren't allowed as the value of _id field).
2) As far as i know, there IS effect on performance when db.collection.insert when you provide your own ids, especially in bulk, BUT if the id's are sorted etc., there shouldn't be a performance loss. The reason, i am quoting:
The structure of index is a B-tree. ObjectIds have an excellent
insertion order as far as the index tree is concerned: they are always
increasing, meaning they are always inserted at the right edge of
B-tree. This, in turn, means that MongoDB only has to keep the right
edge of the B-Tree in memory.
Conversely, a random value in the _id field means that _ids will be
inserted all over the tree. Then the machine must move a page of the
index into memory, update a tiny piece of it, then probably ignore it
until it slides out of memory again. This is less efficient.
:from the book `50 Tips and Tricks for MongoDB Developers`
The tip's title says - "Override _id when you have your own simple, unique id." Clearly it is better to use your id if you have one and you don't need the properties of an ObjectId. And it is best if your ids are increasing for the reason stated above.
3) There is a default index on _id field by MongoDB.
So...
Yes. It is possible to use other types than ObjectId, including GUID that will be saved as BinData.
Yes, there are considerations. It's better if your _id is always increasing (like a growing number, or ObjectId) otherwise the index needs to rebuild itself more often. If you plan on using sharding, the _id should also be hashed evenly.
_id indeed has an index automatically.
It depends on the type you choose. See section 2.
Conclusion: It's better to keep using ObjectId unless you have a good reason not to.
I'm planning to add a Collection to a mongodb database that will have a text field that should be unique for each Document. Lookups from this Collection will almost always be based on this field. This field can contain as many as 100+ chars.
My question is, should this field be the _id field, or should I just add an index for it? What would the performance impact for either approach be?
I suggest you to use your unique text as _id.
It will reduce data size and eliminate an index. Here is the reference. 9th page will guide you.