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How to group by max value and get full entities from MongoDB in Spring Boot?

I am new to MongoDB and have trouble. I would be very grateful if someone could help me with it.
I have the entity like:
class FileRecord {
private ObjectId id;
private String fileId;
private EnumStatus status;
private Long userId;
private Integer year;
etc.
> }
A file can have many records but only one in the year.
I need to get all the last documents by criteria.
I don't understand how to make it right in Spring, but in the mongo query too if to be honest)
I understand how to make criteria for it:
var match = match(Criteria.where("userId").in(userIds).andOperator(Criteria.where("status").in(EnumStatus.availableStatues())));
And group by max:
var group = group("fileId", "year").max("year").as("max_year");
But how to make it to query in one that will return for me Colletion<FileRecord> after aggregation?
If I try it make it this way:
var aggregation = newAggregation(filter, group);
AggregationResults<FileRecord> aggregateResult =
mongoOperations.aggregate(aggregation, FileRecord.class, FileRecord.class);
Collection<FileRecord> records = aggregateResult.getMappedResults()
I get an exception:
readObjectId can only be called when CurrentBSONType is OBJECT_ID, not when CurrentBSONType is DOCUMENT.;
If someone has an idea I'm interested
Thank you in advance

I found the answer)
var group = group("fileId", "year").max("year").as("max_year").first("_id").as("enityId");
I needed to create the projection and after again make request to the DB and get the data that I needed by id. I don't have many of them, so the solution worked for me. But, as I understand, you can add a lot "first", and add the fields you need. And get all by one request

how to make values of column in entity class as lower case while storing using Annotation

I want an entity class through which I can store data to DB, for
specific columns I want in case-sensitive, is there any annotation to
be used above these specific columns to store as requested? please let
me if the question is not clear,Thanks in advance.
#Column(name = "EMAIL",unique=true) //i want store this email id as lower case,
private String email; //never mind in what case user enters

You can do this using #ColumnTransformer
#Column(name = "email")
#ColumnTransformer(write = "LOWER(email)")
private String email;

I try this and it works:
#Column
#ColumnTransformer(write = "LOWER(?)")
private String email;
expression of #ColumnTransformer.write would replace the original "?" placeholder in persisting sql statement. So must provide a compliant-sql expression with "?":
insert into member (email) values (?)
---->
insert into member (email) values (LOWER(?))
detail usage: https://docs.jboss.org/hibernate/orm/5.4/javadocs/org/hibernate/annotations/ColumnTransformer.html

Filter GAE Key Field with Substring

I have a kind named Posts and the key is postId.
The postId format is : YYMMDDXXX. Where XXX is 3 digits sequence number.
For example : 150703001, 150704001, 150704002
How do I get the sequence number from my entities in datastore? I want to convert this SQL Select nvl(max(substring(postId, 7, 3)), 0) from posts where substring(postId, 1, 6) = '150704' to Objectivy filter.
Please help, thanks a lot!

You will have to re-design your Kinds/Entities into a different structure to accomplish this. The datastore can not search by substring.
A possible approach could be this layout :
#Entity
class PostGroup {
#Id
String id; // this is will hold the formatted date YYMMDD
}
#Entity
class Post {
#Id
Long id; // auto generated ID
#Index
Long sequence; // the sequence number
#Parent
Key<PostGroup> group;
}
Now you can set the PostGroup has the ancestor for your Objectify query, sort by Post.sequence desc limit 1 to find the latest post.
String yymmdd = "150704";
Key<PostGroup> postGroupKey = Key.create(PostGroup.class, yymmdd);
Query<Post> query = ofy().type(Post.class).ancestor(postGroupKey).order("-sequence");
Post latestPost = query.first().now();
Note: you don't have to actually persist the PostGroups to use them as ancestors.

Excluding a bean's field in Hibernate criteria result

Here is how I'm getting my User beans from Database.
session.createCriteria(User.class).list();
That returns all of User records from database. Interesting part is that, I don't want to get the password field from DB. Just want to exclude that field while retrieving.
Options I have
1) using projections on other fields. That needs to be more code to add in to projections list. So dropped from that idea.
2) With Sql I need to write a manually query, which kills the theme of Hibernate.
Any possibility to exclude a column value of Bean ?

you can try:
Example example = Example.create(cat)
.excludeZeroes() //exclude zero valued properties
.excludeProperty("color") //exclude the property named "color"
.ignoreCase() //perform case insensitive string comparisons
.enableLike(); //use like for string comparisons
List results = session.createCriteria(Cat.class)
.add(example)
.list();
Reference :-
http://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/querycriteria.html#querycriteria-examples
http://docs.jboss.org/hibernate/orm/3.6/javadocs/org/hibernate/criterion/Example.html

Let's assume that following is your POJO:
User.java
private long id;
private String fName;
private String lName;
// getter - setter of all fields
Now suppose you want to get only id & fName fields and not lName.
Apart from the two approaches you've described, there's also a third way to achieve this, which uses HQL.
Query q1 = session.createQuery("Select new User(id, fName) from User");
List<User> users = q1.list();
Modify User.java to have another constructor as follows:
public User(long id, String fName)
{
this.id = id;
this.fName = fName;
}
In short, whatever fields you want to retrieve, you can list them in the constructor as well as in the query.
Hope this helps.

Hibernate: Criteria API: Query entities which do not contain a specified element in a CollectionOfElements?

Let's say I have the following two classes; User and Location. I want to create a DetachedCriteria to query the user table, and return all users who do not have a location with the name "xyz".
#Entity
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
Long id;
#CollectionOfElements
Set<Location> locations;
}
#Entity
public class Location{
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
Long id;
#Column
String name;
}
The following code will return all users who DO have a location with name "xyz" set:
DetachedCriteria dc = DetachedCriteria.forClass(User.class);
dc.createCriteria("locations")
dc.add(Restrictions.eq("name", "xyz"));
If I change to Restrictions.ne(), that doesn't work, because it will only return users who actually have locations set. Also, if there are a bunch of locations set for a user, it will duplicate that user over and over.
Any ideas?

Using the entities User and Location as described in your original post:
1) Check if the associated locations is empty.
Junction or = Restrictions.disjunction();
or.add(Restrictions.isEmpty("locations"));
2) Create an associated criteria for locations using a LEFT_JOIN and an alias to be used in the "ne" restriction. The LEFT_JOIN is required so we still get User records back even if the locations relationship is empty.
userCriteria.createCriteria("locations", "loc", Criteria.LEFT_JOIN);
or.add(Restrictions.ne("loc.name", "xyz"));
3) Add the disjunction to the the original User criteria.
userCriteria.add(or);

You'll probably need to use some combination of Restrictions.and(), Restrictions.not(), Restrictions.or() and Restrictions.in() to get your inverse-logic to work right. Check http://docs.jboss.org/hibernate/core/3.5/api/org/hibernate/criterion/Restrictions.html - it can definitely be done!
To ensure you only get at-most-one User, use a Distinct projection (http://docs.jboss.org/hibernate/core/3.5/api/org/hibernate/criterion/Distinct.html) via the Projections.distinct() factory method (http://docs.jboss.org/hibernate/core/3.5/api/org/hibernate/criterion/Projections.html)- i.e.:
dc.setProjection(Projections.distinct(Projections.property("id")));
Something like that should do what you need.