I need to bring from DB only one single result. How can I do that with JPA?
Select top 1 * from table
I tried
"select t from table t"
query.setMaxResults(1);
query.getSingleResult();
but didn't work. Any other ideas?
Try like this
String sql = "SELECT t FROM table t";
Query query = em.createQuery(sql);
query.setFirstResult(firstPosition);
query.setMaxResults(numberOfRecords);
List result = query.getResultList();
It should work
UPDATE*
You can also try like this
query.setMaxResults(1).getResultList();
To use getSingleResult on a TypedQuery you can use
query.setFirstResult(0);
query.setMaxResults(1);
result = query.getSingleResult();
2021 att: you can use TOP or FIRST in Spring Data JPA
example:
User findFirstByOrderByLastnameAsc();
User findTopByOrderByAgeDesc();
Page<User> queryFirst10ByLastname(String lastname, Pageable pageable);
Slice<User> findTop3ByLastname(String lastname, Pageable pageable);
List<User> findFirst10ByLastname(String lastname, Sort sort);
List<User> findTop10ByLastname(String lastname, Pageable pageable);
doc: https://docs.spring.io/spring-data/jpa/docs/2.5.4/reference/html/#repositories.limit-query-result
The easiest way is by using #Query with NativeQuery option like below:
#Query(value="SELECT 1 * FROM table ORDER BY anyField DESC LIMIT 1", nativeQuery = true)
Use a native SQL query by specifying a #NamedNativeQuery annotation on the entity class, or by using the EntityManager.createNativeQuery method. You will need to specify the type of the ResultSet using an appropriate class, or use a ResultSet mapping.
Related
I have situation in which I have to compose a JPQL query in some method, then pass this query to spring data query method to be used as the query in #Query annotation.
#Query(value = ":DyanamicQuery")
List<PrizeInsuranceConfiguration> filterConfigurPrizeInsurance(String DyanamicQuery);
This is not possible. As it throws compile error. Can you give me an alternat way to achieve this?
#Query(value = ":DyanamicQuery")
List<PrizeInsuranceConfiguration> filterConfigurPrizeInsurance(String DyanamicQuery);
expected to get Query that I made will go and sit at value = ":DyanamicQuery" I also tried giving #Query("#{#entityName}").
You can create a JPQL query as a string, but you cannot pass it as a parameter to the value attribute of the #Query annotation.
You can use below example -
String dyanamicQuery= "SELECT * FROM table_name WHERE condition= :condition";
#Query(value = dyanamicQuery)
List<Entity> findByField(#Param("condition") String condition);
Or you can use below example also -
#Query(value = "SELECT * FROM table_name WHERE condition= :condition")
List filterConfigurPrizeInsuranceByField(#Param("condition") String condition);
I'd like to select distinct records based on a column where not null and return a page. The following is throwing java.sql.SQLSyntaxErrorException: ORA-01791: not a SELECTed expression.
(If I can get this to work I'd like to select distinct by a few additional columns.)
#Query(value="SELECT DISTINCT pm.batchId from PrintManifest pm",
// + "where pm.batchId is not null"
countQuery = "select count(DISTINCT pm.batchId) from PrintManifest pm",
nativeQuery = true
)
Page<PolicyPrintManifest> findDistinctBatchId(Pageable pageable);
I also tried with no luck:
Page<PolicyPrintManifest> findDistinctByBatchIdExists(Pageable pageable);
What's the right way to do this query?
Is PrintManifest name of the table or the entity. Remember that in native query, you have to use table name not entity name
when i run my query in database visualizer its working perfectly, but i think there are some issues in syntax when i convert it in my DAO class method.
I want to get whole data against the name provided
In Visualizer:
SELECT first_name,last_name,nic,phone,email FROM x_hr_user where (first_name = 'Irum');
Now in Dao
public List<XHrUser> findXHrUserByNameInTable()
{
String name ="Irum";
Query query = em.createQuery("SELECT xHrNewUserObj.firstName,xHrNewUserObj.lastName, xHrNewUserObj.nic, xHrNewUserObj.phone, xHrNewUserObj.emil FROM XHrUser xHrNewUserObj where (xHrNewUserObj.firstName) = (name)");
List<XHrUser> list = query.getResultList();
return list;
}
Instead of showing single row, it displays whole data Table
Thank you
Your current query is not valid JPQL. It appears that you intended to insert the raw name string into your query, which could be done via a native query, but certainly is not desirable. Instead, use a named parameter in your JPQL query and then bind name to it.
String name = "Irum";
Query query = em.createQuery("SELECT x FROM XHrUser WHERE x.firstName = :name")
.setParameter("name", name);
List<XhrUser> list = query.getResultList();
You have to write query as below. where : is used for variable
Query query = em.createQuery("SELECT xHrNewUserObj.firstName,xHrNewUserObj.lastName, xHrNewUserObj.nic, xHrNewUserObj.phone, xHrNewUserObj.emil FROM XHrUser xHrNewUserObj where (xHrNewUserObj.firstName) = :name");
I know that I can write so:
Query query = session.createSQLQuery(
"select s.stock_code from stock s where s.stock_code = :stockCode")
.setParameter("stockCode", "7277");
List result = query.list();
How I must do if I use list values
select count(*) from skill where skill.id in (1,2,4)
I want replace hardcode values.
Maybe:
Query query = session.createSQLQuery("select count(*) from skill where skill.id in :ids")
.setParameter("ids", Arrays.asList(1,2,4));
Query interface have setParameterList(List<any>) function to set the value in IN Clause in HQL. But in HQL IN Clause have a limit to set the element. If the limit is exceed, memory overflow exception occur.
Have you tried something like this?
Query query = session.createSQLQuery(
"select s.stock_code from stock s where s.stock_code in (:stockCodes)")
.setParameter("stockCodes", "1,2,4");
Does it work for you?
I want to execute a query using hibernate where the requirment is like
select * from user where regionname=''
that is select all the users from user where region name is some data
How to write this in hibernate
The below code is giving result appropraitely
Criteria crit= HibernateUtil.getSession().createCriteria(User.class);
crit.add(Restrictions.eq("regionName", regionName));
Well as you alaready said you can either use the Criteria API or create a HQL query:
// Criteria
List<User> users = HibernateUtil.getSession().createCriteria(User.class);
crit.add(Restrictions.eq("regionName", regionName)).list();
// HQL
String query = "SELECT FROM User WHERE regionName = :region";
List<User> users = HibernateUtil.getSession().createQuery(query).setString("region", regionName).list();
String hql = "SELECT u FROM User u WHERE regionName=:regionName";
Query q = session.createQuery(hql);
q.setParameter("regionName", regionName);
List result = q.list();