Best Practice to Update Spring Repository - java

I have Spring Repo class which look like this :
#Repository
public interface EmployeeRepository
extends CrudRepository<Employee, Long> {
#Query("Select s from Employee s where s.Id = ?1")
Optional<Employee> findEmployeeByID(String Id);
#Transactional
#Modifying
#Query(" How to DO that")
UpdateEmployeeDetails()
}
How to efficiently update the Employee details, as employee tables contains more than 15 fields i can't write a function for each of these fields, how we can generate a query which will have common logic to update
For Example :
If we want to update name only so UpdateEmployeeDetails() will execute this query :
UPDATE Employee SET firstName=UPPER(firstName) WHERE id in :ids
If we want to update name and address :
UPDATE Employee SET firstName=UPPER(firstName), Address = "Address" WHERE id in :ids
How we can do this efficiently as there are may fields so we want to generate query dynamically

Related

Spring JPA count where group by

I want to transform a sql query in JPA.
SELECT status, count(*)
FROM abc
WHERE type='XXX'
GROUP BY status
I need something in a JPARepository with sql.
#Repository
public interface ABCRepository extends JpaRepository<abc, Long> {
long countByStatusAndType(final A type, final B status);
}
Is it Possible?
Firstly, create class containing parameters status and count for handling results, then create method in repository with query
#Query("SELECT status, count(*) as count FROM abc WHERE type=:type GROUP BY status")
List<CustomResultClass> countByStatus(String type);

Mapping #Query with multiple column selection to an Java object in a #Repository - is it possible out of the box?

Is it possible to map the results of a Hibernate #Query like this (in a #Repository interface extending #JpaRepository):
#Query("select u.id, u.email, u.status from user u")
public SimpleUserDTO getAllUsersSimpleData();
directly to a Java object like this:
public class SimpleUserDTO {
private Long id;
private String email;
private String status;
}
What I know, is that doing something like this:
Query query = session.createQuery("select u.id, u.email, u.status from user u");
the result can be extracted to a List<Object[]>:
List<Object[]> users = (List<Object[]>) query.list();
But is it possible to map it directly to a List<SimpleUserDTO> without writing additional method that will map the values to SimpleUserDTO?
You have two options.
Option 1: Constructor Expression:
#Query("select new <insert_package_here>.SimpleUserDTO(u.id, u.email, u.status) from user u")
public List<SimpleUserDTO> getAllUsersSimpleData();
Option 2: Use Interface projection
Turn your DTO into an interface and you can use it without #Query annotation
public List<SimpleUserDTO> getAllUsersSimpleData();
Please find out more about projections and Spring Data JPA in the documentation:
https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#projections

Is Spring's QueryByExampleExecutor usable with Projections

Is there a way to return Iterable< IUser > where IUser is a projection of User entity.
Example<User> userExample = Example.of(user, userMatcher);
Iterable<User> foundUsers = userRepository.findAll(userExample, Sort.by("createdAt").descending());
Instead of sticking with spring data auto generated query and response type.
You can create your own HQL query by using #Query annotation so that you can except your custom return type.
Example :
#Query(" select user from User as user where user.userName like %:username% order by createdAt desc")
public List<User> findAllByUserName(String username);

how to handle JSON_VAL() result set in CrudRepository hibernate

Could some one help me how to read result set from JSON_VAL() in db2. I am having the below db2 query which am executing from CrudRepository named query from hibernate.
SELECT SYSTOOLS.BSON2JSON(data) FROM TABLE WHERE JSON_VAL(data,'column','s:25')=:value
Could some one help me how can I read the value.
My Hibernate repository class is
#Repository
public interface MyRepository extends CrudRepository<MyClass, Integer> {
#Query(value = "SELECT SYSTOOLS.BSON2JSON(data) FROM TABLE WHERE JSON_VAL(data,'column','s:25')=:value", nativeQuery = true)
com.ibm.db2.jcc.DB2Clob findById(#Param("idd") String id);
}
thanks in advance .

How to use #NamedQuery in spring a CrudRepository #Query?

I want to make use of a #NamedQuery inside a JpaRepository. But it does not work:
public interface MyEntityRepository extends JpaRepository<MyEntity, Long> {
#Query(name = MyEntity.FIND_ALL_CUSTOM)
List<MyEntity> findAllCustom(Pageable pageable);
}
#Entity
#NamedQuery(
name = MyEntity.FIND_ALL_CUSTOM, query = "select * from MyEntity me where me.age >= 18"
)
public class MyEntity {
public static final String FIND_ALL_CUSTOM = "findAllCustom";
}
Result:
org.springframework.data.mapping.PropertyReferenceException: No property findAllCustom found for type MyEntity!
at org.springframework.data.mapping.PropertyPath.<init>(PropertyPath.java:75)
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:327)
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:307)
at org.springframework.data.mapping.PropertyPath.from(PropertyPath.java:270)
at org.springframework.data.mapping.PropertyPath.from(PropertyPath.java:241)
at org.springframework.data.repository.query.parser.Part.<init>(Part.java:76)
at org.springframework.data.repository.query.parser.PartTree$OrPart.<init>(PartTree.java:235)
at org.springframework.data.repository.query.parser.PartTree$Predicate.buildTree(PartTree.java:373)
at org.springframework.data.repository.query.parser.PartTree$Predicate.<init>(PartTree.java:353)
at org.springframework.data.repository.query.parser.PartTree.<init>(PartTree.java:84)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery.<init>(PartTreeJpaQuery.java:61)
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$CreateQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:94)
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$CreateIfNotFoundQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:205)
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$AbstractQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:72)
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.<init>(RepositoryFactorySupport.java:369)
at org.springframework.data.repository.core.support.RepositoryFactorySupport.getRepository(RepositoryFactorySupport.java:192)
at org.springframework.data.repository.core.support.RepositoryFactoryBeanSupport.initAndReturn(RepositoryFactoryBeanSupport.java:239)
at org.springframework.data.repository.core.support.RepositoryFactoryBeanSupport.afterPropertiesSet(RepositoryFactoryBeanSupport.java:225)
at org.springframework.data.jpa.repository.support.JpaRepositoryFactoryBean.afterPropertiesSet(JpaRepositoryFactoryBean.java:92)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1633)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1570)
... 28 more
Update:
public interface MyEntityRepository extends JpaRepository<MyEntity, Long> {
List<MyEntity> findAllCustom(Pageable pageable);
}
#Entity
#NamedQuery(
name = "MyEntity.findAllCustom", query = "select * from MyEntity me where me.age >= 18"
)
public class MyEntity {
}
Still same exception:
PropertyReferenceException: No property findAllCustom found for type MyEntity!
Take a look at the documentation of Spring Data JPA - Using JPA NamedQueries.
I advise you follow the conventions set in the documentation (starting with the simple name of the configured domain class, followed by the method name separated by a dot). Cut the underscore and name the query like
#NamedQuery(name = "MyEntity.findAllCustom", query="...")
or even better add a suggestive name like findByAge or sth.
To allow execution of these named queries all you need to do is to specify MyEntityRepository as follows:
public interface MyEntityRepository extends JpaRepository <MyEntity, Long> {
List<MyEntity> findAllCustom();
}
I implemented it with the JpaRepository as the documentation exemplifies. But you could try with a simple CrudRepository and see if that works.
I think the problem was you where using #Query and the Queries annotated to the query method will take precedence over queries defined using #NamedQuery. Read the docs for the #Query usage, i think you where also using it wrong.
Update
To use the Pageable, according to this answer
to apply pagination, a second subquery must be derived. Because the
subquery is referring to the same fields, you need to ensure that your
query uses aliases for the entities/tables it refers to
that means you would rewrite your query like
query ="select * from MyEntity me where me.age >= 18".
The example was used for #Query, but that is also a named query so it should apply to your case as well. The only difference is that with #Query you actually bind them directly rather than annotating them to the domain class.
Update 2
I tried in my own app.
First off you should have the query using the alias instead of * (i.e me).
Secondly the string you use FIND_ALL_CUSTOM is not following the convention which is "MyEntity.findAllCustom".
Solution
Copy paste this:
public interface MyEntityRepository extends JpaRepository<MyEntity, Long> {
List<MyEntity> findAllCustom(Pageable pageable);
List<MyEntity> findAllCustom();
}
#Entity
#NamedQuery(
name = MyEntity.FIND_ALL_CUSTOM, query = "select me from MyEntity me where me.age >= 18"
)
public class MyEntity {
public static final String FIND_ALL_CUSTOM = "MyEntity.findAllCustom";
}
Both will work. For the one with the pageable method argument call it as myEntityRepository.allCustom(new PageRequest(0,20)). Ofc, you know that myEntityRepository is injected.

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