Related
I want to assign the result of a query to a DTO object. The DTO looks like this:
#Getter
#Setter
#NoArgsConstructor
public class Metric {
private int share;
private int shareholder;
public Metric(int share, int shareholder) {
this.share = share;
this.shareholder = shareholder;
}
}
And the query looks like the following:
#RepositoryRestResource(collectionResourceRel = "shareholders", path = "shareholders")
public interface ShareholderRepository extends PagingAndSortingRepository<Shareholder, Integer> {
#Query(value = "SELECT new com.company.shareholders.sh.Metric(SUM(s.no_of_shares),COUNT(*)) FROM shareholders s WHERE s.attend=true")
Metric getMetrics();
}
However, this didn't work, as I got the following exception:
Caused by:org.hibernate.QueryException: could not resolve property: no_of_shares of:com.company.shareholders.sh.Shareholder[SELECT new com.company.shareholders.sh.Metric(SUM(s.no_of_shares),COUNT(*)) FROM com.company.shareholders.sh.Shareholder s WHERE s.attend=true]
In my project I've used projections to this like shown below:
#Repository
public interface PeopleRepository extends JpaRepository<People, Long> {
#Query(value = "SELECT p.name AS name, COUNT(dp.people_id) AS count " +
"FROM people p INNER JOIN dream_people dp " +
"ON p.id = dp.people_id " +
"WHERE p.user_id = :userId " +
"GROUP BY dp.people_id " +
"ORDER BY p.name", nativeQuery = true)
List<PeopleDTO> findByPeopleAndCountByUserId(#Param("userId") Long userId);
#Query(value = "SELECT p.name AS name, COUNT(dp.people_id) AS count " +
"FROM people p INNER JOIN dream_people dp " +
"ON p.id = dp.people_id " +
"WHERE p.user_id = :userId " +
"GROUP BY dp.people_id " +
"ORDER BY p.name", nativeQuery = true)
Page<PeopleDTO> findByPeopleAndCountByUserId(#Param("userId") Long userId, Pageable pageable);
}
The interface to which the result is projected:
public interface PeopleDTO {
String getName();
Long getCount();
}
The fields from the projected interface must match the fields in this entity. Otherwise field mapping might break.
Also if you use SELECT table.column notation always define aliases matching names from entity as shown in example.
In your case change #Query like shown below:
#Query(value = "SELECT new " +
"SUM(s.no_of_shares) AS sum,COUNT(*) AS count FROM " +
"shareholders s WHERE s.attend=true", nativeQuery = true)
MetricDTO getMetrics();
And create interface MetricDTO like shown below:
public interface MetricDTO {
Integer getSum();
Long getCount();
}
Also make sure the return type of getSum() and getCount() is correct this may vary based not database.
First, you can have a look at the Spring Data JPA documentation, you can find some help at this section : Class-based Projections (DTOs).
There is also a paragraph titled Avoid boilerplate code for projection DTOs, where they advise you to use Lombok's #Value annotation, to produce an immutable DTO. This is similar to Lombok's #Data annotation, but immutable.
If you apply it to your example, the source will look like :
#Value
public class MetricDto {
private int share;
private int shareholder;
}
Then, as your query is a NativeQuery, specifiy it in your Spring Data Repository.
You can find help in the documentation : Native Queries.
You will need something like :
#Query(value = "SELECT new
com.company.shareholders.sh.MetricDto(SUM(s.no_of_shares),COUNT(*)) FROM
shareholders s WHERE s.attend=true", nativeQuery = true)
MetricDto getMetrics();
Query query = sessionFactory.getCurrentSession()
.createNativeQuery(stringQuery).unwrap(org.hibernate.query.Query.class);
((NativeQueryImpl) query).setResultTransformer(new AliasToBeanResultTransformer(DtoClass.class));
You are writing a mixed query of native and jpql; no_of_shares is your column name in the database, but jpa is expecting you to provide not native syntax so try to replace no_of_shares with the corresponding field in your entity class. Or just add nativeQuery = true to make jpa understand it's a native query.
I have a controller that communicates with my service to get the id of the logged in user. I first get the username using principal object
#ResponseBody
#RequestMapping(value = "/lid",method = RequestMethod.GET)
public Integer lid(Principal p) {
String name = p.getName();
Integer gotid = personService.getDbId(name);
return gotid;
}
then call the service passing the username to the dao
public Integer getDbId(String name){
Session currentSession = sessionFactory.getCurrentSession();
#SuppressWarnings("rawtypes")
Query theQuery = currentSession.createQuery("select id from app_user where sso_id=:name");
//set multiple parameters from the user
//theQuery.setParameter("city", city);
theQuery.setParameter("sso_id", name);
//Get the int or the big int
//Get results into list object : List list = query.list();
int theid = (int)theQuery.getSingleResult();
//uniqueResult
return theid;
//org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: from near line 1, column 8 [select from app_user where sso_id=:name]
}
However, i get this error
java.lang.IllegalArgumentException:
org.hibernate.hql.internal.ast.QuerySyntaxException: app_user is not
mapped [select id from app_user where sso_id=:name]
How can i fix it to give me the one result that i want?.
You are trying to execute pure SQL but hibernate expects HQL. So it tries to find an Entity with app_user name but fails - there is no such class.
Try instead
Query theQuery = currentSession.createSQLQuery("...");
Try with this
Query theQuery = currentSession.createQuery("select id from AppUser where ssoId=:name");
AppUser might be your class name and ssoId is field name.
use your Domain class name and domain field names whatever it is because query requires HQL.
I'm using Spring Data JPA, and when I use #Query to to define a query WITHOUT Pageable, it works:
public interface UrnMappingRepository extends JpaRepository<UrnMapping, Long> {
#Query(value = "select * from internal_uddi where urn like %?1% or contact like %?1%",
nativeQuery = true)
List<UrnMapping> fullTextSearch(String text);
}
But if I add the second param Pageable, the #Query will NOT work, and Spring will parse the method's name, then throw the exception No property full found. Is this a bug?
public interface UrnMappingRepository extends JpaRepository<UrnMapping, Long> {
#Query(value = "select * from internal_uddi where urn like %?1% or contact like %?1%",
nativeQuery = true)
Page<UrnMapping> fullTextSearch(String text, Pageable pageable);
}
You can use pagination with a native query. It is documented here: Spring Data JPA - Reference Documentation
"You can however use native queries for pagination by specifying the count query yourself:
Example 59. Declare native count queries for pagination at the query method using #Query"
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "SELECT * FROM USERS WHERE LASTNAME = ?1",
countQuery = "SELECT count(*) FROM USERS WHERE LASTNAME = ?1",
nativeQuery = true)
Page<User> findByLastname(String lastname, Pageable pageable);
}
A similar question was asked on the Spring forums, where it was pointed out that 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. This means that where you wrote:
select * from internal_uddi where urn like
You should instead have:
select * from internal_uddi iu where iu.urn like ...
Considering that the UrnMapping class is mapped to the internal_uddi table, I would suggest this:
#Repository
public interface UrnMappingRepository extends JpaRepository<UrnMapping, Long> {
#Query(value = "select iu from UrnMapping iu where iu.urn like %:text% or iu.contact like %:text%")
Page<UrnMapping> fullTextSearch(#Param("text") String text, Pageable pageable);
}
Please note that you might have to turn off native queries with dynamic requests.
With #Query , we can use pagination as well where you need to pass object of Pageable class at end of JPA method
For example:
Pageable pageableRequest = new PageRequest(page, size, Sort.Direction.DESC, rollNo);
Where,
page = index of page (index start from zero)
size = No. of records
Sort.Direction = Sorting as per rollNo
rollNo = Field in User class
UserRepository repo
repo.findByFirstname("John", pageableRequest);
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "SELECT * FROM USER WHERE FIRSTNAME = :firstname)
Page<User> findByLastname(#Param("firstname") String firstname, Pageable pageable);
}
Please reference :Spring Data JPA #Query, if you are using Spring Data JPA version 2.0.4 and later. Sample like below:
#Query(value = "SELECT u FROM User u ORDER BY id")
Page<User> findAllUsersWithPagination(Pageable pageable);
Declare native count queries for pagination at the query method by using #Query
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "SELECT * FROM USERS WHERE LASTNAME = ?1",
countQuery = "SELECT count(*) FROM USERS WHERE LASTNAME = ?1",
nativeQuery = true)
Page<User> findByLastname(String lastname, Pageable pageable);
}
Hope this helps
https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#jpa.query-methods
Rewrite your query to:
select iu from internal_uddi iu where iu.urn....
description: http://forum.spring.io/forum/spring-projects/data/126415-is-it-possible-to-use-query-and-pageable?p=611398#post611398
I found it works different among different jpa versions, for debug, you'd better add this configurations to show generated sql, it will save your time a lot !
spring.jpa.show-sql=true
spring.jpa.properties.hibernate.format_sql=true
for spring boot 2.1.6.RELEASE, it works good!
Sort sort = new Sort(Sort.Direction.DESC, "column_name");
int pageNumber = 3, pageSize = 5;
Pageable pageable = PageRequest.of(pageNumber - 1, pageSize, sort);
#Query(value = "select * from integrity_score_view " +
"where (?1 is null or data_hour >= ?1 ) " +
"and (?2 is null or data_hour <= ?2 ) " +
"and (?3 is null or ?3 = '' or park_no = ?3 ) " +
"group by park_name, data_hour ",
countQuery = "select count(*) from integrity_score_view " +
"where (?1 is null or data_hour >= ?1 ) " +
"and (?2 is null or data_hour <= ?2 ) " +
"and (?3 is null or ?3 = '' or park_no = ?3 ) " +
"group by park_name, data_hour",
nativeQuery = true
)
Page<IntegrityScoreView> queryParkView(Date from, Date to, String parkNo, Pageable pageable);
you DO NOT write order by and limit, it generates the right sql
I had the same issue - without Pageable method works fine.
When added as method parameter - doesn't work.
After playing with DB console and native query support came up to decision that method works like it should. However, only for upper case letters.
Logic of my application was that all names of entity starts from upper case letters.
Playing a little bit with it. And discover that IgnoreCase at method name do the "magic" and here is working solution:
public interface EmployeeRepository
extends PagingAndSortingRepository<Employee, Integer> {
Page<Employee> findAllByNameIgnoreCaseStartsWith(String name, Pageable pageable);
}
Where entity looks like:
#Data
#Entity
#Table(name = "tblEmployees")
public class Employee {
#Id
#Column(name = "empID")
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#NotEmpty
#Size(min = 2, max = 20)
#Column(name = "empName", length = 25)
private String name;
#Column(name = "empActive")
private Boolean active;
#ManyToOne
#JoinColumn(name = "emp_dpID")
private Department department;
}
When using nativeQuery that is having (nativeQuery = true), you may do the pagination yourself in the query by adding (LIMIT :sizeValue OFFSET :page)
Note:
Your page value passed to this method should be offset * size
Example
#Query(value = "SELECT * FROM person " +
"LIMIT ?1 OFFSET ?2", nativeQuery = true)
Optional<List<TDriverJob>> findPersons(int size, int page);
I tried all above solution and non worked , finally I removed the Sorting from Pagination and it worked
the following tutorial helped me
-> https://www.baeldung.com/spring-data-jpa-query
At this point 4.3. Spring Data JPA Versions Prior to 2.0.4
VERY IMPORTANT to add \ n-- #pageable \ n
Without this I was wrong
Also the pagination setting must be without ordering
PageRequest paginaConf = new PageRequest ((param1 - 1)
, param2);
Finally to convert the Page <Object []>
Page <Object []> list = myQueryofRepo ();
List <XXXModel> lstReturn = myConversor (list.getContent ());
Page <XXXModel> ret = new PageImpl <XXXModel> (lstReturn, pageConf, param2);
This bugged me for a while but I managed with a very smooth solution.
The challenge is JPA did not automatically detect the count query so I resolved to use the countName which according JPA docs Returns the name of the javax.persistence.NamedQuery to be used to execute count queries when pagination is used. Will default to the named query name configured suffixed by .count.
So I created a named query
#NamedNativeQuery(
name = "[queryName].count",
query = [your count query],
resultSetMapping = "[query resultSetMapping name]"
)
}
As indicated, the count query should be suffixed with .count
Count query returns Long so add the resultSetMapping as
#SqlResultSetMapping(
name="[query resultSetMapping name]",
columns={#ColumnResult(name="count", type = Long.class)})
Then in your repository, use the count query as indicated below
#Query(countName ="[queryName].count" , nativeQuery = true)
Page<Object> [mainQuery](...params)
Hope this helps!
What is the simplest way of declaring a Spring data JPA query that uses properties of an input parameter as query parameters?
For example, suppose I have an entity class:
public class Person {
#Id
private long id;
#Column
private String forename;
#Column
private String surname;
}
and another class:
public class Name {
private String forename;
private String surname;
[constructor and getters]
}
... then I would like to write a Spring data repository as follows:
public interface PersonRepository extends CrudRepository<Person, Long> {
#Query("select p from Person p where p.forename = ?1.forename and p.surname = ?1.surname")
findByName(Name name);
}
... but Spring data / JPA doesn't like me specifying property names on the ?1 parameter.
What is the neatest alternative?
This link will help you: Spring Data JPA M1 with SpEL expressions supported. The similar example would be:
#Query("select u from User u where u.firstname = :#{#customer.firstname}")
List<User> findUsersByCustomersFirstname(#Param("customer") Customer customer);
https://spring.io/blog/2014/07/15/spel-support-in-spring-data-jpa-query-definitions
Define the query method with signatures as follows.
#Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(#Param("surname") String lastname,
#Param("forename") String firstname);
}
For further details, check the Spring Data JPA reference
What you want is not possible. You have to create two parameters, and bind them separately:
select p from Person p where p.forename = :forename and p.surname = :surname
...
query.setParameter("forename", name.getForename());
query.setParameter("surname", name.getSurname());
You could also solve it with an interface default method:
#Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(#Param("surname") String lastname,
#Param("forename") String firstname);
default User findByName(Name name) {
return findByForenameAndSurname(name.getLastname(), name.getFirstname());
}
Of course you'd still have the actual repository function publicly visible...
You can try something like this:
public interface PersonRepository extends CrudRepository<Person, Long> {
#Query("select p from Person AS p"
+ " ,Name AS n"
+ " where p.forename = n.forename "
+ " and p.surname = n.surname"
+ " and n = :name")
Set<Person>findByName(#Param("name") Name name);
}
if we are using JpaRepository then it will internally created the queries.
Sample
findByLastnameAndFirstname(String lastname,String firstname)
findByLastnameOrFirstname(String lastname,String firstname)
findByStartDateBetween(Date date1,Date2)
findById(int id)
Note
if suppose we need complex queries then we need to write manual queries like
#Query("SELECT salesOrder FROM SalesOrder salesOrder WHERE salesOrder.clientId=:clientId AND salesOrder.driver_username=:driver_username AND salesOrder.date>=:fdate AND salesOrder.date<=:tdate ")
#Transactional(readOnly=true)
List<SalesOrder> findAllSalesByDriver(#Param("clientId")Integer clientId, #Param("driver_username")String driver_username, #Param("fdate") Date fDate, #Param("tdate") Date tdate);
The simplicity of Spring Data JPA is that it tries to interpret from the name of the function in repository without specifying any additional #Query or #Param annotations.
If you are supplying the complete name, try to break it down as firstname and lastname and then use something like this -
HotelEntity findByName(String name);
My HotelEntity does contain the field name so JPA tries to interpret on its own to infer the name of the field I am trying to query on and create a subsequent query internally.
Some more evidence from JPA documentation -
Further details - here
Are you working with a #Service too? Because if you are, then you can #Autowired your PersonRepository to the #Service and then in the service just invoke the Name class and use the form that #CuriosMind... proposed:
#Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(#Param("surname") String lastname,
#Param("forename") String firstname);
}
and when invoking the method from the repository in the service, you can then pass those parameters.
for using this, you can create a Repository for example this one:
Member findByEmail(String email);
List<Member> findByDate(Date date);
// custom query example and return a member
#Query("select m from Member m where m.username = :username and m.password=:password")
Member findByUsernameAndPassword(#Param("username") String username, #Param("password") String password);
#Autowired
private EntityManager entityManager;
#RequestMapping("/authors/{fname}/{lname}")
public List actionAutherMulti(#PathVariable("fname") String fname, #PathVariable("lname") String lname) {
return entityManager.createQuery("select A from Auther A WHERE A.firstName = ?1 AND A.lastName=?2")
.setParameter(1, fname)
.setParameter(2, lname)
.getResultList();
}
What would the equivalent of Oracle's DECODE() function be in the Hibernate Criteria API?
An SQL example of what I need to do:
SELECT DECODE(FIRST_NAME, NULL, LAST_NAME, FIRST_NAME) as NAME ORDER BY NAME;
Which returns LAST_NAME to NAME in the event that FIRST_NAME is NULL.
I would prefer to use the Criteria API but could use HQL if there's no other way.
Check out org.hibernate.criterion.Projections.sqlProjection(...).
Similar to this answer.
For the example you give, you could use COALESCE().
How to simulate NVL in HQL
You can use sqlRestriction to call the native decode function.
session.createCriteria(Table.class).add(Restrictions.sqlRestriction("decode({alias}.firstName,null,
{alias}.lastName,
{alias}.firstName)"))
With HQL, the Oracle dialect already has coalesce and nvl functions, or if you really need decode, you could subclass the dialect and add it as a custom function. I don't know if Hibernate supports a variable length number of arguments like decode does, but worst-case, you could create decode1, decode2, etc to support different numbers of arguments.
Or, if you aren't using the column in a where or group by, you could just bring both attributes back and do the check in Java.
Ended up adding a formula for it:
<property name="name" formula="coalesce(first_name, last_name)"/>
I'm concerned about cross-database problems and possibly efficiency problems with this approach so I'm willing to change the accepted answer.
You can Use Hibernate #Type attribute,Based on your requirement you can customize the annotation and apply on top of the fied. like :
public class PhoneNumberType implements UserType {
#Override
public int[] sqlTypes() {
return new int[]{Types.INTEGER, Types.INTEGER, Types.INTEGER};
}
#Override
public Class returnedClass() {
return PhoneNumber.class;
}
// other methods
}
First, the null SafeGet method:
#Override
public Object nullSafeGet(ResultSet rs, String[] names,
SharedSessionContractImplementor session, Object owner) throws HibernateException,
SQLException {
int countryCode = rs.getInt(names[0]);
if (rs.wasNull())
return null;
int cityCode = rs.getInt(names[1]);
int number = rs.getInt(names[2]);
PhoneNumber employeeNumber = new PhoneNumber(countryCode, cityCode, number);
return employeeNumber;
}
Next, the null SafeSet method:
#Override
public void nullSafeSet(PreparedStatement st, Object value,
int index, SharedSessionContractImplementor session)
throws HibernateException, SQLException {
if (Objects.isNull(value)) {
st.setNull(index, Types.INTEGER);
} else {
PhoneNumber employeeNumber = (PhoneNumber) value;
st.setInt(index,employeeNumber.getCountryCode());
st.setInt(index+1,employeeNumber.getCityCode());
st.setInt(index+2,employeeNumber.getNumber());
}
}
Finally, we can declare our custom PhoneNumberType in our OfficeEmployee entity class:
#Entity
#Table(name = "OfficeEmployee")
public class OfficeEmployee {
#Columns(columns = { #Column(name = "country_code"),
#Column(name = "city_code"), #Column(name = "number") })
#Type(type = "com.baeldung.hibernate.customtypes.PhoneNumberType")
private PhoneNumber employeeNumber;
// other fields and methods
}
This might solve your problem, This will work for all database. if you want more info refer :: https://www.baeldung.com/hibernate-custom-types
If you can use HQL the you can replace DECODE with CASE.
You can update your query from,
SELECT DECODE(FIRST_NAME, NULL, LAST_NAME, FIRST_NAME) as NAME ORDER BY NAME;
to,
SELECT CASE WHEN FIRST_NAME = NULL then LAST_NAME ELSE FIRST_NAME END as NAME ORDER BY NAME;