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Filter JSON column from MySQL table in spring boot

Below is my query to filter the json column in MySQL table ,
public interface MpaCustomizationRepository extends
JpaRepository<MpaCustomization, Serializable> {
#Query(nativeQuery = true, value = "select * from MPA_CUSTOMIZATION where json_contains(domain_based_properties, '{\"id\" : ?1}')")
MpaCustomization findByDomainBaseId(String domainId);
}
above method throws below SQL error,
but if I hard code the variable and run the query like below its works fine,
#Query(nativeQuery = true, value = "select * from MPA_CUSTOMIZATION where json_contains(domain_based_properties, '{\"id\" : 2}')")
MpaCustomization findByDomainBaseId(String domainId);
How to fix this issue?

Use $.id for values when using native query, as SQL accepts like it only.
Other way, you may use :id and bind parameter {\"id\":\"" + <value> + "\"} to the query.

Maybe the following will work.
#Query(nativeQuery = true, value = "select * from MPA_CUSTOMIZATION where json_contains(domain_based_properties, ?1 , '$.id'")
MpaCustomization findByDomainBaseId(String domainId);
Source

Named parameter not bound criteria native query [duplicate]

I want to execute a simple native query, but it does not work:
#Autowired
private EntityManager em;
Query q = em.createNativeQuery("SELECT count(*) FROM mytable where username = :username");
em.setProperty("username", "test");
(int) q.getSingleResult();
Why am I getting this exception?
org.hibernate.QueryException: Not all named parameters have been set: [username]

Named parameters are not supported by JPA in native queries, only for JPQL. You must use positional parameters.
Named parameters follow the rules for identifiers defined in Section 4.4.1. The use of named parameters applies to the Java Persistence query language, and is not defined for native queries. Only positional parameter binding may be portably used for native queries.
So, use this
Query q = em.createNativeQuery("SELECT count(*) FROM mytable where username = ?1");
q.setParameter(1, "test");
While JPA specification doesn't support named parameters in native queries, some JPA implementations (like Hibernate) may support it
Native SQL queries support positional as well as named parameters
However, this couples your application to specific JPA implementation, and thus makes it unportable.

After many tries I found that you should use createNativeQuery And you can send parameters using # replacement
In my example
String UPDATE_lOGIN_TABLE_QUERY = "UPDATE OMFX.USER_LOGIN SET LOGOUT_TIME = SYSDATE WHERE LOGIN_ID = #loginId AND USER_ID = #userId";
Query query = em.createNativeQuery(logQuery);
query.setParameter("userId", logDataDto.getUserId());
query.setParameter("loginId", logDataDto.getLoginId());
query.executeUpdate();

You are calling setProperty instead of setParameter. Change your code to
Query q = em.createNativeQuery("SELECT count(*) FROM mytable where username = :username");
em.setParameter("username", "test");
(int) q.getSingleResult();
and it should work.

I use EclipseLink. This JPA allows the following way for the native queries:
Query q = em.createNativeQuery("SELECT * FROM mytable where username = ?username");
q.setParameter("username", "test");
q.getResultList();

Use set Parameter from query.
Query q = (Query) em.createNativeQuery("SELECT count(*) FROM mytable where username = ?1");
q.setParameter(1, "test");

This was a bug fixed in version 4.3.11
https://hibernate.atlassian.net/browse/HHH-2851
EDIT:
Best way to execute a native query is still to use NamedParameterJdbcTemplate
It allows you need to retrieve a result that is not a managed entity ; you can use a RowMapper and even a Map of named parameters!
private NamedParameterJdbcTemplate namedParameterJdbcTemplate;
#Autowired
public void setDataSource(DataSource dataSource) {
this.namedParameterJdbcTemplate = new NamedParameterJdbcTemplate(dataSource);
}
final List<Long> resultList = namedParameterJdbcTemplate.query(query,
mapOfNamedParamters,
new RowMapper<Long>() {
#Override
public Long mapRow(ResultSet rs, int rowNum) throws SQLException {
return rs.getLong(1);
}
});

Query not giving solution in DAO class

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");

JPA and JSON operator native query

I'm trying to make this query work in JPA:
SELECT * FROM contrat WHERE contrat_json #> '{"nom" :"hever"}';
It works perfectly with postgresql but when I integrate it with JPA, I get the following error:
Parameter with that position [1] did not exist
My code:
#Transactional
#Query(nativeQuery = true,value = "select p from Contrat p where contrat_json #> '{\"nom\":\":nom\"}'")
public List<Contrat> findByNomRestrict(#Param("nom") String nom);
I think it does not recognize #> despite native query, do you have an idea?

Parameter holders are not understood inside literals: '...:nom...' will contain the characters :nom, not the bound values of nom.
For PostgreSQL 9.5 (and later), use:
SELECT * FROM contrat WHERE contrat_json #> jsonb_build_object('nom', :nom)
For 9.4:
SELECT * FROM contrat WHERE contrat_json #> CAST(json_build_object('nom', :nom) AS jsonb)
For 9.3 (and earlier), there is no JSON containment operator (neither the jsonb type).
http://rextester.com/AUHP11519

I had similar problem with my native query.
The jsonb field name is called data, and it's simple
{
"name" : "genderList",
"displayName" : "gender list"
}
I want to find by name with JpaRepository, and here is my Repository
#Repository
public interface LookupListRepository extends JpaRepository<LookupList, UUID>
{
#Query(value = "SELECT * FROM lookup_list WHERE data->>'name' = :name",
nativeQuery = true)
List<LookupList> findByName(#Param("name") String name);
}
You need nativeQuery = true.
With nativeQuery = true, this works as well.
SELECT * FROM lookup_list WHERE jsonb_extract_path_text(data, 'name') = :name
I see your #Transactional annotation, I assume you have the native query on top of application service method. Can you try moving all native query's in repository and use JpaRepository, and use the repository method in your application service?
Here is how my application service uses the repository.
public class LookupListServiceImpl implements LookupListService
{
#Autowired
LookupListRepository lookupListRepository;
#Override
#Transactional
public void changeLookupList(LookupListDto lookupListDto)
{
List<LookupList> lookupLists = lookupListRepository.findByName(lookupListDto.getName());
...
}
}
Reference for JPA repository
http://docs.spring.io/spring-data/jpa/docs/1.3.0.RELEASE/reference/html/jpa.repositories.html

With PostgreSQL and JSON you'll probably run into needing ? or other strange operators, so it's better you just use their function equivalents, instead. You can look them up in the psql console like this \doS+ #>.
Your query is not native, as the parameter says.
select p from Contrat p where...
Will only give you an error when it reaches the database.
Try something like
#Query(nativeQuery = true, value = "select * from Contrat where jsonb_contains(contrat_json, :nom )")
and just bind "{\"nom\":\"" + param + "\"}" as the parameter

Select top 1 result using JPA

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.