Can we use two In clause and both will take list of integer in a JPA method.
Ex: findByGroupIdAndUserId(List groupIdList, List userIdList)
Multiple column with IN clause in not yet supported by Spring data.
You can use #Query annotation for custom query as below:
#Query( "select o from MyObject o where groupId in :gids and userId in :uids" )
List findByIds(#Param("gids") List groupIdList, #Param("uids") List userIdList);
Related
I have researched this for a few days but can't seem to find the right information.
Here is what I need, I have a Database, with multiple tables, I need to join a few tables together to make a sort of "search" API. I have to implement the ability to dynamically search fields (from various tables in the query), sortable, with pagination.
I have found that I cannot combine the #Query annotation with Specification API, and I looked into using the specification API to do the joins I needed but, the problem is the root must be one table/repository.
For example:
If I have a users table that has to join on addresses, phone_numbers, and preferences
the base repository will be UserResposiory and it will return the User entity model, but I need it to return a custom DTO
AccountUserDTO which contains fields from the User, Address, PhoneNumber, and Preference entities.
Would anyone know if this is possible at all??
I am at wits end here and I really want to build this the correct way.
Cheers!
You may do this way:
Build hql query as an string, depend on how the filter condition is requested, you can build the corresponding query, eg:
if (hasParam(searchName)) {
queryString = queryString + " myEntity.name = :queryName"
}
Query query = session.createQuery(queryString);
and the parameter providing
if (hasParam(searchName)) {
query.setParameter("queryName", searchName);
}
...
and execute it.
To create a customized object, the easiest way is treating the object as an array of field:
Query query = session.createQuery("select m.f1, m.f2, m.f3 from myTable m");
List managers = query.list();
Object[] manager = (Object[]) managers.get(0); //first row
System.out.println(manager[0]) //f1
System.out.println(manager[1]) //f2
System.out.println(manager[2]) //f3
There is also some other solution to select, such as
String query = "select new mypackage.myclass(m.f1, m.f2, m.f3) from myTable m";
-> And when execute the above query, it will return a list of object.
Or to be simpler, make your own view in db and map it to one entity.
I have a method -> findByfileNameLike(fileName,1, pageable) and its declaration in a repository that extends JPA Repository is
#Query(value = QUERY)
#Transactional(propagation = Propagation.REQUIRED, readOnly = true)
#LogExecutionTime
Page<BatchDTO> findByfileNameLike(String
fileName,#Param("departmentId")Integer departmentId, Pageable pageable)
Query is Select new DTO(bdm.id.batch.status) from Table bdm where bdm.id.departmentId =:departmentId and bdm.id.batch.status <> 7";
I want to filter the query by the column fileName.I have read to give the method name as given according to the doc of spring data jpa.But its not working.
Where and how will i give the fileName to be filtered?Should it be first parameter in the method?
Once you specify a query using the annotation #Query, Spring data jpa will not automatically create a query for you based on the method name and it will rely on the query provided by using the annotation.
The method findByfileNameLike will not make any difference here as a query is provided explicitly. Hope that answers your question
I want to write a Query for getting specific columns.using
entitymanager.createQuery("SELECT u.name FROM Department u").getResultList();
however it returns object is not an instance of declaring class.
What is the correct way to get specific column from table in jpa.as a provider i am using Hibernate
It returns list of objects (it would be hard for JPA to guess what types are you returning). But luckily you can give it a hint with TypedQuery like this
TypedQuery<String> query = em.createQuery("SELECT u.name FROM Department u",
String.class);
List<String> departmentNames = query.getResultList();
I want to learn if we could write a query that has a condition like
List<Entity> findbyField1EqualsField2();
This method should not take any parameter . It should fetch entities which has a field1 equals field2. It is just a simple sql :
select * from entity where field1=field2.
But I could not find any solution yet. Thanks.
Create an operation with a query as next:
#Query("select t from entity t where t.field1 like t.field2")
List<T> findByField1LikeField2();
I don't think findByField1LikeField2()works.... you would need to do it passing a param findByField1Like(String param) and for this maybe you would need to load the entity before to get the value of field2.
Is it possible to add an OrderBy clause in to JPA Named query at runtime?
Named queries are processed by the persistence provider when the EntityManagerFactory is created. You can't change/modify/append anything about a named query dynamically at runtime.
If you are using JPA 2.0 and you need a way to do high-performance dynamic queries at runtime, you should look into the Criteria API.
From Java EE 5 Documentation : "Is used to specify a named query in the Java Persistence query language, which is a static query expressed in metadata. Query names are scoped to the persistence unit".
As it says, it is static & you can't change the query at runtime.
Rather use custom query or if ordering element is fixed then you can use annotation;
Field:
#OrderBy(value="nickname")
List<Person> friends;
Method:
#OrderBy("nickname ASC")
public List<Person> getFriends() {...};
Even if you can't add order-by clause to your named query, you can provide a parametric orderBy. The following is a perfectly valid query:
SELECT u FROM User u ORDER BY :orderBy
And you are going to change ordering with something like:
query.setParameter("orderBy", "id");