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How to find entities by spring data with same values of different fields?

I look for short (one-method-long) solution of my problem.
I want to write query in spring data that does the same as:
select * from Users
where city = #EnteredData data OR homeTown = #EnteredData
I tried
List<User> findDistinctCity_findByHomeTown(String city);
I have no ideas how to find users whose city or homeTown is equal to given value in one method.

I will go with the below version. I better write my own queries instead of trying to use the Spring Data generated DSL one
#Query("select u from User u where u.city = :city or u.homeTown = :city") public List<User> getUserInfo(#Param("city") String city);

U can use method query, Add Or in query
List<User> findByCityOrHometowm(String name);
JpqL snippet will create as
where x.city= ?1 or x.hometown = ?2

What is the LIMIT clause alternative in JPQL?

I'm working with PostgreSQL query implementing in JPQL.
This is a sample native psql query which works fine,
SELECT * FROM students ORDER BY id DESC LIMIT 1;
The same query in JPQL doesnt work,
#Query("SELECT s FROM Students s ORDER BY s.id DESC LIMIT 1")
Students getLastStudentDetails();
seems like LIMIT clause doesn't work in JPQL.
According to JPA documentation we can use setMaxResults/setFirstResult, Can anyone tell me how can I use that in my above query?

You are using JPQL which doesn't support limiting results like this. When using native JPQL you should use setMaxResults to limit the results.
However you are using Spring Data JPA which basically makes it pretty easy to do. See here in the reference guide on how to limit results based on a query. In your case the following, find method would do exactly what you want.
findFirstByOrderById();
You could also use a Pageable argument with your query instead of a LIMIT clause.
#Query("SELECT s FROM Students s ORDER BY s.id DESC")
List<Students> getLastStudentDetails(Pageable pageable);
Then in your calling code do something like this (as explained here in the reference guide).
getLastStudentDetails(PageRequest.of(0,1));
Both should yield the same result, without needing to resort to plain SQL.

As stated in the comments, JPQL does not support the LIMIT keyword.
You can achieve that using the setMaxResults but if what you want is just a single item, then use the getSingleResult - it throws an exception if no item is found.
So, your query would be something like:
TypedQuery<Student> query = entityManager.createQuery("SELECT s FROM Students s ORDER BY s.id DESC", Student.class);
query.setMaxResults(1);
If you want to set a specific start offset, use query.setFirstResult(initPosition); too

Hello for fetching single row and using LIMIT in jpql we can tell the jpql if it's a native query.
( using - nativeQuery=true )
Below is the use
#Query("SELECT s FROM Students s ORDER BY s.id DESC LIMIT 1", nativeQuery=true)
Students getLastStudentDetails();

You can not use Limit in HQL because Limit is database vendor dependent so Hibernate doesn't allow it through HQL query.
A way you can implement is using a subquery:
#Query("FROM Students st WHERE st.id = (SELECT max(s.id) FROM Students s)")
Students getLastStudentDetails();

The correct way is to write your JPA interface method like this
public interface MyRepository extends PagingAndSortingRepository<EntityClass, KeyClass> {
List<EntityClass> findTop100ByOrderByLastModifiedDesc();
}
In the method name, "100" denotes how many rows you want which you would have otherwise put in the limit clause. also "LastModified" is the column which you want to sort by.
PagingAndSortingRepository or CrudRepository, both will work for this.
For the sake of completeness, OP's interface method would be
List<Students> findTop1ByIdDesc();

JPQL does not allow to add the limit keyword to the query generated by the HQL. You would get the following exception.
org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token:
LIMIT near line 1
But don't worry there is an alternative to use the limit keyword in the query generated by the HQL by using the following steps.
Sort.by(sortBy).descending() // fetch the records in descending order
pageSize = 1 // fetch the first record from the descending order result set.
Refer the following service class
Service:
#Autowired
StudentRepository repository;
public List<Student> getLastStudentDetails(Integer pageNo, Integer pageSize, String sortBy)
{
Integer pageNo = 0;
Integer pageSize = 1;
String sortBy = "id";
Pageable paging = PageRequest.of(pageNo, pageSize, Sort.by(sortBy).descending());
Slice<Student> pagedResult = repository.findLastStudent(paging);
return pagedResult.getContent();
}
Your repository interface should implement the PagingAndSortingRepository
Repository:
public interface StudentRepository extends JpaRepository<Student,Long>, PagingAndSortingRepository<Student,Long>{
#Query("select student from Student student")
Slice<Student> findLastStudent(Pageable paging);
}
This will add the limit keyword to you query which you can see in the console. Hope this helps.

Hardcode the pagination(new PageRequest(0, 1)) to achieve fetch only one record.
#QueryHints({ #QueryHint(name = "org.hibernate.cacheable", value = "true") })
#Query("select * from a_table order by a_table_column desc")
List<String> getStringValue(Pageable pageable);
you have to pass new PageRequest(0, 1)to fetch records and from the list fetch the first record.

Here a Top Ten Service (it's a useful example)
REPOSITORY
(In the Query, I parse the score entity to ScoreTo ( DTO class) by a constructor)
#Repository
public interface ScoreRepository extends JpaRepository<Scores, UUID> {
#Query("SELECT new com.example.parameters.model.to.ScoreTo(u.scoreId , u.level, u.userEmail, u.scoreLearningPoints, u.scoreExperiencePoints, u.scoreCommunityPoints, u.scoreTeamworkPoints, u.scoreCommunicationPoints, u.scoreTotalPoints) FROM Scores u "+
"order by u.scoreTotalPoints desc")
List<ScoreTo> findTopScore(Pageable pageable);
}
SERVICE
#Service
public class ScoreService {
#Autowired
private ScoreRepository scoreRepository;
public List<ScoreTo> getTopScores(){
return scoreRepository.findTopScore(PageRequest.of(0,10));
}
}

You can use something like this:
#Repository
public interface ICustomerMasterRepository extends CrudRepository<CustomerMaster, String>
{
#Query(value = "SELECT max(c.customer_id) FROM CustomerMaster c ")
public String getMaxId();
}

As your query is simple, you can use the solution of the accepted answer, naming your query findFirstByOrderById();
But if your query is more complicated, I also found this way without need to use a native query:
#Query("SELECT MAX(s) FROM Students s ORDER BY s.id DESC")
Students getLastStudentDetails();
Here a practical example where the named query method cannot be used.

How to avoid hibernate default order by id?

I make this query:
String query = FROM Account acc WHERE acc.id = ? OR acc.id = ? or acc.id = ?...
I have array of ids:
long[] accountIds= {327913,327652,327910,330511,330643};
Then I make
getHibernateTemplate().find(query, accountIds);
I see that the list of accounts I get back from this query is:
327652,327910,327913,330511,330643, obviously , ordered by id.
Any chance I get it back in the order I wrote the ids?
Will appreciate all the help

You may want to use Criteria and its addOrder.
Something like this:
DetachedCriteria cr = DetachedCriteria.forClass(entityClass);
//Add expressions or restrictions to your citeria
//And add your ordering
cr.addOrder(Order.asc("yourID"));
List<T> ls = getHibernateTemplate().findByCriteria(cr);
return ls;

You can't do it on query level.
You can sort after loading from db, something like this
long[] accountIds= {327913,327652,327910,330511,330643};
List<Account> afterHql = getHibernateTemplate().find(query, accountIds);
List<Account> sortedList = new ArrayList<Acount>();
for (long id : accountIds)
{
for (Account account : afterHql)
{
if (account.getId() == id)
{
sortedList.add(account);
}
}
}

It is not possible to fetch results by providing any entity in OR Query or Restrictions.in(). As by deafult when you fire this kind of query it will search for the results id wise. So it will give you results id wise only. You can change the order by using Criteria either in asc or desc. And if you want to have results as per you enter id, then second answer is the only option.

You can only order by column values returned by the query, in a sequence defined by the data type . Wouldn't it be better to pre-order the IDs you supply, and order the query result by ID, so they come out in the same order?

Using HibernateTemplate.find(...) with a String[]

I'm having troubles with HQL again :(
My desired SQL is this:
select employee.idemployee as id, employee.age as age,
employee.birthday as birthday, employee.firstName as firstName,
employee.gender as gender, employee.lastName as lastName from employee
employee
inner join employee_skillgroups skillgroup1 on
employee.idemployee=skillgroup1.idemployee
inner join employee_skillgroups skillgroup2 on
employee.idemployee=skillgroup.idemployee
where skillgroup1.idskillgroup = 'Sprachen'
and skillgroup.idskillgroup = 'SoftSkills'
But i just can't get HQL to generate me this ...
"Sprachen" and "SoftSkills" are two string coming out of a String[] I'm giving the method as a parameter. The method currently looks like this:
public List<Employee> findEmployeeWithTwoSkillGroups(final String[] skillGroups) {
return template.find("from Employee e join e.skillGroups as s where s in ?", Arrays.asList(skillGroups).toString
().substring(1, Arrays.asList(skillGroups).toString().length()-1));
}
I "cast" the array to a list, execute toString() on it (so i get "[Sprachen, SoftSkills]") and cut off the first and the last char (so i get "Sprachen, SoftSkills").
I guess the problem is that HQL generates "[...].idskillgroup in ('Sprachen, SoftSkills')", like it treats the two strings like ONE string ...
And i just can't get it to work like i want it to :/
Can someone please help me and give me a hint what to try/do next? :-)
Greetz
gilaras
Simmilar question:
HQL to get elements that possess all items in a set

The proper HQL statement should go more or less like this:
from Employee e join e.skillGroups as s where s in ( 'Foo', 'Bar', 'Baz' )
so you are missing the ' ' for each word.
EDIT: Now that I got what you want to achieve here :-)
I would recommend you doing this:
Query query = session.createQuery(
"from Employee e join e.skillGroups as s where s in (:skillGroups)");
// skillGroups must be a Collection type
query.setParameterList("skillGroups", skillGroups);
List list = query.list();
In case you need your result to be the AND of all elements of your String[] array, you can do the following:
Query query = session.createQuery(
"from Employee e join e.skillGroups as s where s in (:skillGroups) group by e having count(s) = :skillGroupsLength");
query.setParameterList("skillGroups", skillGroups);
query.setParameter("skillGroupsLength", skillGroups.length);
List list = query.list();

It's a prepared statement, not string concatenation. It handles ('Sprachen, SoftSkills') like a single string because it is a single string. You want to insert a collection in your query. I'm not sure how to do this in hibernateTemplate, but hibernate itself supports adding a collection. see the hibernate documentation.

HQL Object set from set

I got a Set<UserDTO> collection in a not Hibernate object, and I got a User domain entity in Hibernate.
UserDTO contains less information about user (only id and name)
How can I select full Hibernate User Set/List from the DTO object?
Like this?
Set<UserDTO> setDTO = .....
String hql = "FROM User WHERE id IN (:userDTO )";
Query query = entityManager.createQuery(hql);
query.setParameter("userDTO", setDTO);
return query.getResultList();
Thanks

Almost. But you'd have to extract the IDs in a separate collection first:
Set<Long> ids = new HashSet<Long>(setDTO.size());
for (UserDTO dto : setDTO) {
ids.add(dto.getId());
}
Then proceed with the query, and pass the ids set as param.

Don't forget that you need to use Query#setParameterList() instead of Query#setParameter.