I have a PostgreSQL table where channel_ids field is BIGINT[].
In console I can search like so:
select * FROM user where 10000=ANY(channel_ids).
But I can't figure out how to build up the Specification for search in JpaRepository.
public interface UserRepository extends JpaRepository<User, Long> {
Page<User> findAll(Specification<User> spec, Pageable pageable);
}
Is it even supported?
Related
I have employee class
#Entity
class Employee {
#Id
Integer id;
String name;
Integer age;
}
and another class EmployeeInfo:
class EmployeeInfo {
Integer id;
String name;
}
Now, I need to build a service to get a paginated list of EmployeeInfo by using findAll(Pageable pageable)
from the repository
#Repository
public interface EmployeeRepository extends JpaRepository<Employee, Long> {
what is the best way to do that? I want to avoid getting the page from findAll and create a new object of EmployeeInfo then adding it to a list in a loop
You can utilize spring-data-jpa projections.
There are many ways to use them (open/closed projections, class or interface based, etc.), but since you already have a EmployeeInfo class, it can be achieved by defining a new method in your repository:
#Repository
public interface EmployeeRepository extends JpaRepository<Employee, Long> {
Page<EmployeeInfo> getAllBy(Pageable pageable);
}
Note that your projection DTO class properties must exactly match properties in the aggregate root (entity class).
Also reference documentation suggest to define .equals() and .hashcode() methods.
Other methods can be found in the official documentation.
Normally I use annotiations:#Query("SELECT c FROM Country c") with JpaRepositoryor predefined methods like findAll
but in my case I want to generate dynamic query.
String baseQuery =SELECT c FROM Country c`
if(age!=null)
baseQuery+="WHERE c.age=20"
I need to perform same query from code level like this:
Query q1 = em.createQuery("SELECT c FROM Country c");
but I dont use EntityManager in spring boot
How can I generate query from code level?
If you would like to create dynamic queries from code you can take advantage of Spring's JdbcTemplate. Using spring boot it is as simple as injecting JdbcOperations bean to your repository class (assuming you have provided spring-boot-starter-jdbc module to your project).
But remember! This solution uses SQL, not JPQL. That's why you have to use proper tables and columns names in queries and properly map result to objects (i.e. using RowMapper)
This simple example worked fine for me (with different entity, but in same manner - I've adapted it to your example):
#Repository
public class CountryRepository {
#Autowired
private JdbcOperations jdbcOperations;
private static String BASIC_QUERY = "SELECT * FROM COUNTRY";
public List<Country> selectCoutry(Long age){
String query = BASIC_QUERY;
if (age != null){
query += " WHERE AGE = ";
query += age.toString();
}
//let's pretend that Country has constructor Conutry(String name, int age)
return jdbcOperations.query(query, (rs, rowNum) ->
{ return new Country(rs.getString("NAME"), rs.getInt("AGE");}
);
};
}
Then in service or whatever you inject CountryRepository and call method.
Since you're using Spring Boot, you can use Spring Data to create queries in your repository:
#Repository
public interface CountryRepository extends JpaRepository<Country, Long> {
}
Not a 100% on syntax, but should be something similar.
Now you can autowire this class:
#Autowired
public CountryRepository countryRepo;
And all basic methods are already available to you like:
countryRepo.findOne(id);
countryRepo.find();
If you want to make more advanced queries, you can use Spring Data e.g.:
#Repository
public interface CountryRepository extends JpaRepository<Country, Long> {
public Country findByNameAndContinent(String name, String continent);
}
This is just an example (a stupid one) of course and assumes your Country class has the field names 'name' and 'continent' and both are strings. More is available here:
http://docs.spring.io/spring-data/jpa/docs/current/reference/html/
Section 5.3 more specifically.
PS: Make sure your Country class has the #Entity annotation
My UserRepository:
public interface UserRepository extends CrudRepository<User, Integer> {
List<User> findAll(List<Integer> ids);
}
Error:
Caused by:
org.springframework.data.mapping.PropertyReferenceException: No
property findAll found for type User
Refer - http://docs.spring.io/spring-data/commons/docs/current/api/org/springframework/data/repository/CrudRepository.html?is-external=true#findAll-java.lang.Iterable-
Can some one tell me how to get list of User objects based on List of Id's.
This is working
#Query(" select new User(id,x,y,z) from User b where b.id in ?1 ")
List<User> findById(List<Integer> id);
Firstly, I would rename the repository to UserRepository, because having 2 User classes is confusing.
findAll(), by definition, is meant to get all the models with no criteria. You should add a method named
findByIdIn(Collection<Integer> ids)
Use List<User> findAll(Iterable<Integer> ids) or List<User> findByIdIn(List<Integer> ids)
I have a JpaRepository:
public interface UserRepository extends JpaRepository<User, Long> {
List<User> getByFirstName(String firstName);
}
But how to search in several values? I need something like this:
List<User> getByFirstNames(List<String> firstNames);
You need to change the signature of the method to:
List<User> getByFirstNameIn(List<String> firstNames);
Take a look at all the supported method of Spring Data JPA on their reference.
The following code is working,
Code in Repository,
List<Shop> findByNameIn(List<String> names);
Code in controller,
List<String> names=new ArrayList<String>();
names.add("gunaa");
names.add("pranav");
List<Shop> sl=shopService.findByNameIn(names);
for(Shop s:sl)
System.out.println(s.getName());
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.