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Got some troubles with JPQL (JPA) query SELECT DISTINCT ORDERED BY

as in title i got some problems with query in JPQL
that's the query I'm using:
#Query(value = "SELECT DISTINCT o.idWhom FROM Message o WHERE o.idWho = ?1 ORDER BY o.date DESC")
List<Users> allCorrespondents(Users user);
Class message:
public class Message {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
#NotBlank
private String content;
#Temporal(TemporalType.TIMESTAMP)
private Date date = new Date();
boolean read = false;
#ManyToOne
#JoinColumn(name = "id_who")
private Users idWho;
#ManyToOne
#JoinColumn(name = "id_whom")
private Users idWhom;
}
Error I've got:
2022-06-17 13:02:22.435 ERROR 2304 --- [nio-8080-exec-2] c.e.h.S.CustomAuthorizationFilter :
Error logging in: Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessResourceUsageException: could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet

Although I am not able to prove it through issue tickets, I have done enough investigation to believe that this issue occurs to hibernate version previous from 5. (Example Hibernate 4!) with database being used Oracle.
What the issue seems to be is the following:
When you have written the JPQL query SELECT DISTINCT o.idWhom FROM Message o you would expect that the entities would be filtered to be distinct after the execution of the SQL query, where the mapper would filter out the distinct entities.
For some reason however in those versions I mention above what actually happens is that the distinct is passed in the SQL query and what actually executes is the following:
SELECT DISTINCT idh.id_column FROM O_TABLE o JOIN IDWHOM_TABLE idh on (o.id_foreign_key_column = idh.id_column
As you can see this select query does not return in result set all columns that are needed from mapper to build the entity but only returns the id column. Therefore the mapper is not able to extract from ResultSet all columns needed to build the entity and you get the error that you get could not extract ResultSet.
A workaround I have found, is that it is able to work when distinct is placed on some primitive field and not on an entity, so you can use this to your advantage to bypass this issue with the following approach:
This supposes that the entity Users has as id a field named id.
#Query(value = "SELECT us FROM Users us where us.id in (SELECT DISTINCT o.idWhom.id FROM Message o WHERE o.idWho = ?1)")
List<Users> allCorrespondents(Users user);
I have excluded the ordering which you have in your original query since even from logical point of view seems wrong. A User entity can have multiple messages. How could you order the returned users just from created date of messages? This field is not able to order Users themselves. It could be used to order messages only.

Unable to use OffsetDateTime as a query parameter using JPA Data/JPQL

I have an Entity CustomEntity with a data member updateDate of type OffsetDateTime.
I have defined a Repository for this Entity which has a simple method to retrieve list of records matching updateDate as
List<CustomEntity> findByUpdateDate(OffsetDateTime updateDate);
Now, when this method is called from Controller/Service bean, I can see no matching record is retrieved; however, when I execute the generated SQL in the DB, I can see matching rows available.
I can retrieve the records based on other data members of the entity; its just an issue with OffsetDateTime and LocalDateTime
I got to understand that java.time package support was not in JPA 2.1; however I am using JPA 2.3.1. Do I need to use Converters (as suggested for JPA 2.1?
Any help is much appreciable.
EDIT :-
Below is the code for Entity
#Entity
#Table(name = "SAMPLE_TABLE")
public class CustomEntity {
#Id
#GeneratedValue
private Long id;
#Column
private OffsetDateTime updateDate;
//Getters & Setters
}
I am using Microsoft SQL Server and the generated SQL query (hibernate generated) looks something like below
select sample0_.id as id1_10, sample0_.updateDate as update2_10 from sample_table sample0_ where sample0_.updateDate=?
binding parameter [1] as [TIMESTAMP] - [2021-07-27T17:22:34.597Z]

How to query with Spring JPA on jsonb columns?

I'm using spring JPA with PostgreSQL database. I have an Entity as follow:
#Entity
#TypeDef(name="json_binary", typeClass = com.vladmihalcea.hibernate.type.json.JsonBinaryType.class)
public class LiveTokens {
#Id
#GeneratedValue()
private Long id;
private String username;
#Type(type="json_binary")
#Column(columnDefinition = "jsonb")
private Token token
}
and Token:
public class Token {
private Long expireTime;
private String accessToken;
}
For saving object to column with Hibernate, I use Hibernate Types project.
Now I want to get All LiveTokens that expired. I can't do it with Spring JPA. How do I query to posgresql jsonb column with Spring data JPA?

SQL JSON functions
If you want to call a SQL function that processes the JSON column, then you just need to make sure you register the function with Hibernate prior to using it.
JSON operators
For JSON operators, like the PostgreSQL ones (e.g., ->>), the only option you have is to use a native SQL query. JPQL or HQL don't support database-specific JSON operators.

Using EclipseLink and spring data jpa, if your data in db is something like: {"expireTime":102020230201, "accessToken":"SOMETHING" }, my first question is why to use long numbers for your dates instead of timestamps (ex '2019-09-14 12:05:00'). If you use timestamps there are also options to manage timezones (either from postgresql or from you source code).
Regarding your issue you may use the FUNC JPQL keyword of EclipseLink (Hibernate may have something similar) in order to run a database specific function. In the example below I use FUNC('jsonb_extract_path_text', lt.token, 'expireTime') to get the values of the json for token.expireTime.
PostgreSql method jsonb_extract_path_text returns text, thus you cannot do a less that condition, so I cast the output of the function using JPQL CAST keyword with (CAST -data- TO -type-).
#Repository
public interface MyRepository extends JpaRepository<LiveTokens, Integer> {
#Query(value = "SELECT lt FROM LiveTokens lt WHERE CAST(FUNC('jsonb_extract_path_text', lt.token, 'expireTime') AS LongType) < :expirirationThreshold")
List<LiveTokens> findByExpireTime(#Param("expirirationThreshold") Long expirirationThreshold);
}
Again, this is not tested.

This is how a native query using postgres JSON query operators would look like, incoorporating your example:
#Query(value="SELECT t.* FROM LiveTokens t WHERE CAST(t.token ->> 'expireTime' AS LONG) < now()", native=true)
Assuming your real tablename is LiveTokens, native queries do no longer use the JPA translations, and the tablename has to match the one in the DB. (You may also need to specify it's schema.)

Try it:
Service class:
Long currentTime = new Date().getTime();
Repository:
#Query("SELECT lt FROM LiveTokens lt WHERE lt.token.expireTime <= :currentTime")
List<LiveTokens> findExpiredLiveTokens(#Param("currentTime") long currentTime)

JPA: compare LocalDateTime attribute to builder.currentTimestamp()

Given:
#Entity
public class Paramter() {
#Id
private Long id;
private LocalDateTime startDate;
// getters & setters
}
// Extract from repository/dao method that grabs parameters:
...
final CriteriaBuilder builder = entityManager.getCriteriaBuilder();
final CriteriaQuery<Parameter> query = builder.createQuery(Parameter.class);
final Root<Parameter> param = query.from(Parameter);
// The line below is generates following compilation error due to mismatch
// of the types of LocalDateTime and Timestamp parameters:
// java: no suitable method found for greaterThanOrEqualTo(javax.persistence.criteria.Path<java.time.LocalDateTime>,javax.persistence.criteria.Expression<java.sql.Timestamp>)
query.where(builder.greaterThanOrEqualTo(param.get(Parameter_.startDate), builder.currentTimestamp())); ...
How to properly use java 8 date-time api (more specifically LocalDateTime property) with jpa criteria builder to achive: "select * from parameter where start_date >= sysdate"?
Please consider:
The solution can be Oracle specific but should use JPA criteria builder
I do not want to create a select query to grab the sysdate and than use it in another select query to grab the valid parameters. It should be done in one query like in the expected query example above.
The time from the server is not reliable - database time should be used
Using Hibernate 5.2.16 which seems to include JPA 2.1. Changing to JPA 2.2 doesn't seem to make a difference.

Hibernate #Formula annotation causing sql exception errors when hibernate inserts table reference inappropriately

I'm trying to use a Hibernate #Formula to import some calculated information into one of my Entity classes. This is the code:
#Formula("(select max(site_visits.`date`) from site_visits where site_visits.siteId = id)")
#Temporal(TemporalType.TIMESTAMP)
private Date lastSiteVisit;
I've tried the query manually and it works fine. The problem is that when I request the object, the SQL hibernate generates gets mangled: it insists on inserting its own table reference into the middle of my column reference expressions, so I get the following SQL error:
Unknown column 'site_visits.site0_.date' in 'field list'
Any idea how to fix this?

Use Alias
#Formula("(select max(sv.date) from site_visits sv where sv.siteId = id)")
Also don't use ` in queries