I actually want To display Full documents Without duplication
When I use distinct i can only display one particular field
How to display all fields without selecting that document which is repeated
Each document in MongoDB contains an _id field which must be unique in the collection. Hence, it is not possible to have two identical documents in a collection. Ergo, iterating a collection with no conditions will return all unique documents (which are all documents in the collection).
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I have application which has solr search facility. When i query for search result comes only fields which has the values. How to get the result with empty field values in solr? How to get all the fields in result even though the value of a field is empty?
Only the fields present is stored with each document, so if you want to keep empty fields actually available, you'll have to explicitly index empty content into the field. How you do that depends on how you're indexing documents to Solr today.
You can work around this by fetching the schema for the core / collection you're querying first, retrieve the field names and adding empty fields for each field present in the schema, if necessary.
It'll probably be easier to assume that missing fields are empty, and wrap your code in a check to see if the field is present, and if not, return an empty value instead.
The reason for this is that while Solr uses a schema, the underlying Lucene library does not - any document can contain a set of field names unrelated to the other documents present in the index. Since the schema can change independent of the content of the index as well (a schema change will not update existing documents), returning non-existing fields isn't straight forward - and for most cases the document returned should be as close to possible to what was actually indexed.
I have a collection which stores document that can have multiple subdocument (e.g. a group collection with multiple votings). My goal is to delete certain subdocuments on a certain date. I thought about using a package like "agenda" to cron-like delete certain subdocuments. Are there any more efficient or easier ways to do that
For a dating site I have a db of hundreds of different Users.
Fields such as "matches", "heartedMe" contain the ObjectId of different users in Array form.
When I delete a user I need to search every collection and delete any references to the deleted user by $pull ing their ObjectId from all relevant fields.
Any thoughts on the sort of query I would need to run? Currently I've just found All Users!
Thanks a lot!
I'm new to Firestore so I have a simple question:
Suppose I have a collection of users and a collection of groups. users have an array field of groupReferences to mark what groups it is a part of.
Now I want to query a subset of the collection of groups that matches those that a particular user is within.
Do I:
Pull all groups and somehow compare the references in the user, filtering out any that match?
Iterate through the user's references and pull each group one at a time?
Thanks for your help!
i need to develop a search application , where many documents are indexed with different fields and a id field which is unique for each of the document . Fields are not stored just indexed except for id field
i need to find out for each document , the documents similar to this, here all i have is unique id field of current document , i dont have any other fields of current document to form Terms and query the index for finding similar documents like current one.
How do i do this ? any help greatly appreciated .
I believe the simplest way to do this is to use Solr, and use Solr's MoreLikeThisHandler.
You can use a query likehttp://localhost:8983/solr/select?q=unique_id:2722&mlt=true&mlt.fl=manu,cat&mlt.mindf=1&mlt.mintf=1&fl=id,score
Do you have any control over how these documents are indexed? You can index with term vectors, and at query time, look up the term vector for the document, construct a query using the terms, and submit the query.