I saw this example for native query which use countQuery:
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "SELECT * FROM USERS WHERE LASTNAME = ?1 ORDER BY ?#{#pageable}",
countQuery = "SELECT count(*) FROM USERS WHERE LASTNAME = ?1",
nativeQuery = true)
Page<User> findByLastname(String lastname, Pageable pageable);
}
And the second example:
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value="SELECT * FROM USERS WHERE LASTNAME = ?1", nativeQuery=true)
Page<User> findByLastname(String lastname, Pageable pageable);
}
The second example does not use countQuery, so what in the best practice? Can you explain ?
I tried both and both work well. What is the best practice ?
You can omit it if it is the same as your original query.
From the documentation :
countQuery
Defines a special count query that shall be used for pagination queries to lookup the total number of elements for a page. If none is configured we will derive the count query from the original
query or countProjection() query if any.
countQuery is useful when you need a different query to compute total items (this should almost never be the case though).
Note that countQuery can also be needed when you have a Pageable along with a non-native query, and need to fetch lazy loaded associations.
For example, let's assume your User entity has a lazy loaded registrations association :
This will not compile (query validation should fail with the below error message) :
#Query(value = "select u from Users u left join fetch u.registrations where lastName = ?1")
Page<User> findByLastname(String lastname, Pageable pageable);
org.hibernate.query.SemanticException: query specified join fetching,
but the owner of the fetched association was not present in the select
list
This will work :
#Query(value = "select u from Users u left join fetch u.registrations where lastName = ?1",
countQuery = "select u from Users u left join fetch u.registrations where lastName = ?1")
Page<User> findByLastname(String lastname, Pageable pageable);
(Note that you could have used #EntityGraph in that later case, this is just to demonstrate a use case of the countQuery where it is required).
I`m using Repository like this:
public interface UserRepository extends Repository<User, Long> {
List<User> findByEmailAddressAndLastname(String emailAddress, String lastname);
}
But what I need is execute my own sql statement.
select u.* from users u where exists ( select 1 from expires_users where users_id = u.id )
Please, past link references in answers.
I think you can do as pointed in the documentation of Spring Data JPA.
Here is an example:
public interface UserRepository extends JpaRepository<User, Long> {
#Query("select u.* from User u
where exists(select 1 from ExpiredUser e where e.id = u.id)")
Page<User> findExpiredUsers(Pageable pageable);
}
The Page and Pageable parts are for paging results, assuming that this query could return many more results than you would like to process at once. More information on paging results can be found here.
I have a ClassRoom class, in my Spring Boot application. Here is what it looks like :
public class ClassRoom {
#Id #GeneratedValue Long classRoomID;
ArrayList<User>adminList=new ArrayList<>();
}
And in my ClassRoomRepository class, I have :
public interface ClassRoomRepository extends JpaRepository<ClassRoom,Long> {
#Query("select ClassRoom from ClassRoom c where c.adminList = ?1")
ArrayList<ClassRoom> findByAdminList(ArrayList<User> adminList);
/*
#Query("select ClassRoom from ClassRoom c where c. = ?1")
ArrayList<ClassRoom> findByAdmin(User admin);
*/
}
I can query to select ClassRoom where ArrayList of ClassRoom gets passed parameter.
But I want to query to select ClassRoom where I pass only one User as parameter and returns ArrayList of ClassRoom.(Commented section-nothing done so far)
If it is possible in this interface, how can I do so?
Asuming you habe many-to-many association using join table, this is what you need,
#Query(value = "SELECT DISTINCT cr FROM ClassRoom cr JOIN cr.admins a WHERE a = ?1")
List<ClassRoom> findAllByAdmin(final Admin admin);
When mapping collection , Hibernate requires it to declare it using interface type such as List , Set , Map etc. But you are using ArrayList to declare the admin list , I am guessing the whole list will be stored to a single column with some binary data type in DB . No one would store the data in this funny way when using RDBMS unless you have strong reason to do it.
From what you describe , ClassRoom and User seem to be many to many. For demonstration convenience , I would map it as #ManyToMany:
#Entity
#Table
public class ClassRoom {
#ManyToMany
#JoinTable(name ="classroom_user",
joinColumns = #JoinColumn(name = "classroom_id"),
inverseJoinColumns = #JoinColumn(name = "user_id"))
private List<User> admins = new ArrayList();
}
#Entity
#Table
public class User{
#ManyToMany(mappedBy = "admins")
private List<ClassRoom> classRooms = Lists.newArrayList();
}
Then given an user admin ,to get all of his administrative class rooms :
#Query("select distinct cr from ClassRoom cr left join fetch cr.admins admin where admin = ?1 ")
public List<ClassRoom> findByAdmin(User admin);
Or :
#Query("select distinct cr from ClassRoom cr where ?1 member of cr.admins")
public List<ClassRoom> findByAdmin(User admin);
I have a question. when I use spring data jpa, I want to it return Map Collections, but it wrong. Then I search on the internet found a solution. Flowing.
#Transactional(readOnly = true)
public interface GoodsRepository extends JpaRepository<TbGoodsEntity, Integer> {
#Query(value = "select new map(t.id as id, t.goodsName as goodsName) from TbGoodsEntity t group by t.goodsName")
public List<Map<String, Object>> getGoodsNames();// it`s ok,
#Query(value = "select * from tb_goods t group by t.goodsName", nativeQuery = true)
public List<Map<String, Object>> getGoods();//it`s error
}
But I don't think to use new map method its best solution, I`d like to ask if there any other solutions. Thanks.
If use use "native" query then each row will be returned as "List" since spring doesn't have row transformer. So your output becomes List<List<Object>>.
If you try the below query then you should get List<Map<String,Object>>
#Query(value = "select t from tb_goods t group by t.goodsName")
public List<Map<String, Object>> getGoods();
Note: I am guessing your DB column name is "goodsName" so not commenting if query is correct or not.
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!