I am new to JPA and Spring Boot and trying to write a custom query which returns only one column from the table when the API is triggered. But I get an error doing so. It would be helpful if someone could guide me to a correct way of writing this custom query.
// Controller class
#Autowired
private UserTestInstanceDao usertestdao;
List<String> usertestinst = usertestdao.tempQuery();
// DAO class
public interface UserTestInstanceDao extends CrudRepository<UserTestInstance, Long> {
#Query("SELECT ti.test_name FROM test_instance ti")
public List<String> tempQuery();
}
I think that your query should look like this (if you follow conventions):
#Query("SELECT ti.testName FROM UserTestInstance ti")
For this query, you UserTestInstance should look like this:
public class UserTestInstance {
private String testName;
<getters and setters>
}
This is because you're using JPQL, and you should be querying your objects and their declared variables. It's Spring's Data job to translate to your database specific db query.
Spring Data documentation for reference.
It's ok but you have two options :
You must put your name class in the from clause #Query("SELECT ti.testName FROM UserTestInstance ti")
Use a native query whith the real name of Database #Query(value="SELECT ti.test_name FROM test_instance ti",nativeQuery=true)
I used JpaRepository for this purpose.
Controller class
#Autowired
private UserTestInstanceDao usertestdao;
//in method
List<String> usertestinst = usertestdao.tempQuery();
DAO class:
public interface UserTestInstanceDao extends JpaRepository<UserTestInstance, Long> {
#Query(value="SELECT test_name FROM ti", nativeQuery = true)
public List<String> tempQuery();
}
And these are in my application.properties:
# DATASOURCE
spring.datasource.url=jdbc:mysql://localhost/test_instance
spring.datasource.username=root
spring.datasource.password=password
spring.datasource.driver-class-name=com.mysql.jdbc.Driver
# JPA
spring.jpa.properties.hibernate.globally_quoted_identifiers=true
spring.jpa.hibernate.ddl-auto=update
spring.jpa.show-sql=true
spring.jpa.hibernate.naming-strategy=org.hibernate.cfg.ImprovedNamingStrategy
spring.data.jpa.repositories.enabled=true
spring.jpa.database-platform=org.hibernate.dialect.MySQL5InnoDBDialect
Related
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 .
I have a native query in Spring Data JPA:
#Repository
public interface BookRepository extends JpaRepository<Book, Integer>{
#Query(value = "select * from DEV.Book where find_in_set(market)", nativeQuery = true)
public List<Book> findBooksByMarcket(#Param("market") String market);
}
Now I want to change the DEV.Book based on my connection SIT/PROD dynamically, based on the connection I tried to pass a string into native query which didn't work. What is the best way to do it?
You should set Schema in your connection and leave the query without schema.
You should find some way to remove the namespace from the query.
You could for example have three databases without any namespace. If you configure the database connection via properties you can easily switch by supplying different startup properties or property files.
You can use it in this way:
#Override
public List<Book> findBooksByMarcket(String market) {
TypedQuery query = em.createNativeQuery("select * from DEV.Book where find_in_set(?)", Book.class);
query.setParameter(1, market);
return query.getResultList();
}
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 Document is
#QueryEntity #Data #Document(collection = "MyCol") public class MyCol {
#Id private String _id;
private String version;
I want to get all distinct version stored in the db.
My attempts:
public interface MyColDao extends MongoRepository<MyCol, String>, QueryDslPredicateExecutor<MyCol> {
#Query("{ distinct : 'MyCol', key : 'version'}")
List<String> findDistinctVersion();
}
Or just findDistinctVersion without the query annotation.
Most of the examples of github have a By-field like
List<Person> findDistinctPeopleByLastnameOrFirstname(String lastname, String firstname);
I don't need a By field.
Another example I found here.
#Query("{ distinct : 'channel', key : 'game'}")
public JSONArray listDistinctGames();
This doesn't seem to work for me.
I can't seem to find queryDSL/Morphia's documentation to do this.
public interface MyColDao extends MongoRepository<MyCol, String>, QueryDslPredicateExecutor<MyCol> {
#Query("{'yourdbfieldname':?0}")
List<String> findDistinctVersion(String version);
}
here version replaces your your db field name
more you can see here
This spring documentation provide the details, how to form a expression when you are want to fetch distinct values.
Link
I had a similar problem, but I couldn't work out how to do it within the MongoRepository (as far as I can tell, it's not currently possible) so ended up using MongoTemplate instead.
I believe the following would meet your requirement.
#AutoWired
MongoTemplate mongoTemplate
public List<String> getVersions(){
return mongoTemplate.findDistinct("version", MyCol.class, String.class);
}
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.