In the django orm I can do something like the following:
people = Person.objects.filter(first_name='david')
for person in people:
print person.last_name
How would I do the equivalent in Java Hibernate's orm? So far, I've been able to do a single get, but not a filter clause:
Person p = session.get(Person.class, "david");
What would be the correct way to do this though?
you can use native SQL
session.beginTransaction();
Person p = getSingleResult(session.createNativeQuery("SELECT * FROM People where name = 'david'",Person.class));
session.getTransaction().commit();
and the function getSingleResult would be somthing like this :
public static <T> T getSingleResult(TypedQuery<T> query) {
query.setMaxResults(1);
List<T> list = query.getResultList();
if (list == null || list.isEmpty()) {
return null;
}
return list.get(0);
}
you can get a list like this :
List<Person> list = session
.createNativeQuery("SELECT * FROM People", Person.class)
.getResultList();
There are several approaches to do this so here goes:
Lazy way - possibly bad if you have a tons of data is to just load up
the whole list of persons, stream it and apply a filter to it to
filter out objects not matching the given first name.
Use a HQL query (Hibernate Query Language) to create a select query
https://docs.jboss.org/hibernate/orm/5.3/userguide/html_single/chapters/query/hql/HQL.html
Use Hibernate's Criteria API to achieve the above
https://docs.jboss.org/hibernate/orm/5.3/userguide/html_single/chapters/query/criteria/Criteria.html
Alternatively you can even use a native SQL query to do the above.
I know I can pass a list to named query in JPA, but how about NamedNativeQuery? I have tried many ways but still can't just pass the list to a NamedNativeQuery. Anyone know how to pass a list to the in clause in NamedNativeQuery? Thank you very much!
The NamedNativeQuery is as below:
#NamedNativeQuery(
name="User.findByUserIdList",
query="select u.user_id, u.dob, u.name, u.sex, u.address from user u "+
"where u.user_id in (?userIdList)"
)
and it is called like this:
List<Object[]> userList = em.createNamedQuery("User.findByUserIdList").setParameter("userIdList", list).getResultList();
However the result is not as I expected.
System.out.println(userList.size()); //output 1
Object[] user = userList.get(0);
System.out.println(user.length); //expected 5 but result is 3
System.out.println(user[0]); //output MDAVERSION which is not a user_id
System.out.println(user[1]); //output 5
System.out.println(user[2]); //output 7
The above accepted answer is not correct and led me off track for many days !!
JPA and Hibernate both accept collections in native query using Query.
You just need to do
String nativeQuery = "Select * from A where name in :names"; //use (:names) for older versions of hibernate
Query q = em.createNativeQuery(nativeQuery);
q.setParameter("names", l);
Also refer the answers here which suggest the same (I picked the above example from one of them)
Reference 1
Reference 2 which mentioned which cases paranthesis works which giving the list as a parameter
*note that these references are about jpql queries, nevertheless the usage of collections is working with native queries too.
A list is not a valid parameter for a native SQL query, as it cannot be bound in JDBC. You need to have a parameter for each argument in the list.
where u.user_id in (?id1, ?id2)
This is supported through JPQL, but not SQL, so you could use JPQL instead of a native query.
Some JPA providers may support this, so you may want to log a bug with your provider.
Depending on your database/provider/driver/etc., you can, in fact, pass a list in as a bound parameter to a JPA native query.
For example, with Postgres and EclipseLink, the following works (returning true), demonstrating multidimensional arrays and how to get an array of double precision. (Do SELECT pg_type.* FROM pg_catalog.pg_type for other types; probably the ones with _, but strip it off before using it.)
Array test = entityManager.unwrap(Connection.class).createArrayOf("float8", new Double[][] { { 1.0, 2.5 }, { 4.1, 5.0 } });
Object result = entityManager.createNativeQuery("SELECT ARRAY[[CAST(1.0 as double precision), 2.5],[4.1, 5.0]] = ?").setParameter(1, test).getSingleResult();
The cast is there so the literal array is of doubles rather than numeric.
More to the point of the question - I don't know how or if you can do named queries; I think it depends, maybe. But I think following would work for the Array stuff.
Array list = entityManager.unwrap(Connection.class).createArrayOf("int8", arrayOfUserIds);
List<Object[]> userList = entityManager.createNativeQuery("select u.* from user u "+
"where u.user_id = ANY(?)")
.setParameter(1, list)
.getResultList();
I don't have the same schema as OP, so I haven't checked this exactly, but I think it should work - again, at least on Postgres & EclipseLink.
Also, the key was found in: http://tonaconsulting.com/postgres-and-multi-dimensions-arrays-in-jdbc/
Using hibernate, JPA 2.1 and deltaspike data I could pass a list as parameter in query that contains IN clause. my query is below.
#Query(value = "SELECT DISTINCT r.* FROM EVENT AS r JOIN EVENT AS t on r.COR_UUID = t.COR_UUID where " +
"r.eventType='Creation' and t.eventType = 'Reception' and r.EVENT_UUID in ?1", isNative = true)
public List<EventT> findDeliveredCreatedEvents(List<String> eventIds);
can be as simple as:
#Query(nativeQuery =true,value = "SELECT * FROM Employee as e WHERE e.employeeName IN (:names)")
List<Employee> findByEmployeeName(#Param("names") List<String> names);
currently I use JPA 2.1 with Hibernate
I also use IN condition with native query. Example of my query
SELECT ... WHERE table_name.id IN (?1)
I noticed that it's impossible to pass String like "id_1, id_2, id_3" because of limitations described by James
But when you use jpa 2.1 + hibernate it's possible to pass List of string values. For my case next code is valid:
List<String> idList = new ArrayList<>();
idList.add("344710");
idList.add("574477");
idList.add("508290");
query.setParameter(1, idList);
In my case ( EclipseLink , PostGreSQL ) this works :
ServerSession serverSession = this.entityManager.unwrap(ServerSession.class);
Accessor accessor = serverSession.getAccessor();
accessor.reestablishConnection(serverSession);
BigDecimal result;
try {
Array jiraIssues = accessor.getConnection().createArrayOf("numeric", mandayWorkLogQueryModel.getJiraIssues().toArray());
Query nativeQuery = this.entityManager.createNativeQuery(projectMandayWorkLogQueryProvider.provide(mandayWorkLogQueryModel));
nativeQuery.setParameter(1,mandayWorkLogQueryModel.getPsymbol());
nativeQuery.setParameter(2,jiraIssues);
nativeQuery.setParameter(3,mandayWorkLogQueryModel.getFrom());
nativeQuery.setParameter(4,mandayWorkLogQueryModel.getTo());
result = (BigDecimal) nativeQuery.getSingleResult();
} catch (Exception e) {
throw new DataAccessException(e);
}
return result;
Also in query cannot use IN(?) because you will get error like :
Caused by: org.postgresql.util.PSQLException: ERROR: operator does not exist: numeric = numeric[]
'IN(?)' must be swapped to '= ANY(?)'
My solution was based on Erhannis concept.
In jpa, it worked for me
#Query(nativeQuery =true,value = "SELECT * FROM Employee as e WHERE e.employeeName IN (:names)")
List<Employee> findByEmployeeName(#Param("names") List<String> names);
Tried in JPA2 with Hibernate as provider and it seems hibernate does support taking in a list for "IN" and it works. (At least for named queries and I believe it will be similar with named NATIVE queries)
What hibernate does internally is generate dynamic parameters, inside the IN same as the number of elements in the passed in list.
So in you example above
List<Object[]> userList = em.createNamedQuery("User.findByUserIdList").setParameter("userIdList", list).getResultList();
If list has 2 elements the query will look like
select u.user_id, u.dob, u.name, u.sex, u.address from user u "+
"where u.user_id in (?, ?)
and if it has 3 elements it looks like
select u.user_id, u.dob, u.name, u.sex, u.address from user u "+
"where u.user_id in (?, ?, ?)
you should do this:
String userIds ="1,2,3,4,5";
List<String> userIdList= Stream.of(userIds.split(",")).collect(Collectors.toList());
Then, passes like parameter inside your query, like this:
#NamedNativeQuery(name="User.findByUserIdList", query="select u.user_id, u.dob, u.name, u.sex, u.address from user u where u.user_id in (?userIdList)")
It's not possible with standard JPA. Hibernate offers the proprietary method setParameterList(), but it only works with Hibernate sessions and is not available in JPA's EntityManager.
I came up with the following workaround for Hibernate, which is not ideal but almost standard JPA code and has some nice properties to it.
For starters you can keep the named native query nicely separated in a orm.xml file:
<named-native-query name="Item.FIND_BY_COLORS" result-class="com.example.Item">
<query>
SELECT i.*
FROM item i
WHERE i.color IN ('blue',':colors')
AND i.shape = :shape
</query>
</named-native-query>
The placeholder is wrapped in single quotes, so it's a valid native JPA query. It runs without setting a parameter list and would still return correct results when other matching color parameters are set around it.
Set the parameter list in your DAO or repository class:
#SuppressWarnings("unchecked")
public List<Item> findByColors(List<String> colors) {
String sql = getQueryString(Item.FIND_BY_COLORS, Item.class);
sql = setParameterList(sql, "colors", colors);
return entityManager
.createNativeQuery(sql, Item.class)
.setParameter("shape", 'BOX')
.getResultList();
}
No manual construction of query strings. You can set any other parameter as you normally would.
Helper methods:
String setParameterList(String sql, String name, Collection<String> values) {
return sql.replaceFirst(":" + name, String.join("','", values));
}
String getQueryString(String queryName, Class<?> resultClass) {
return entityManager
.createNamedQuery(queryName, resultClass)
.unwrap(org.hibernate.query.Query.class) // Provider specific
.getQueryString();
}
So basically we're reading a query string from orm.xml, manually set a parameter list and then create the native JPA query. Unfortunately, createNativeQuery().getResultList() returns an untyped query and untyped list even though we passed a result class to it. Hence the #SuppressWarnings("unchecked").
Downside: Unwrapping a query without executing it may be more complicated or impossible for JPA providers other than Hibernate. For example, the following might work for EclipseLink (untested, taken from Can I get the SQL string from a JPA query object?):
Session session = em.unwrap(JpaEntityManager.class).getActiveSession();
DatabaseQuery databaseQuery = query.unwrap(EJBQueryImpl.class).getDatabaseQuery();
databaseQuery.prepareCall(session, new DatabaseRecord());
Record r = databaseQuery.getTranslationRow();
String bound = databaseQuery.getTranslatedSQLString(session, r);
String sqlString = databaseQuery.getSQLString();
An alternative might be to store the query in a text file and add code to read it from there.
You can pass in a list as a parameter, but:
if you create a #NamedNativeQuery and use .createNamedQuery(), you don't use named param, you used ?1(positional parameter). It starts with 1, not 0.
if you use .createNativeQuery(String), you can use named param.
You can try this :userIdList instead of (?userIdList)
#NamedNativeQuery(
name="User.findByUserIdList",
query="select u.user_id, u.dob, u.name, u.sex, u.address from user u "+
"where u.user_id in :userIdList"
)
I have using the Native Query in JPA Repository and my query is as:
select u.*, max(a.version_no) as versionNo from users u left join andother_table a on u.id=a.user_id where u.title='abc' group by id;
from my Query i get the "versionNo" which is not mapped to mu user modal.I have put this also in our user modal like as
#Transient
private String versionNo;
with is getter/setter.but in view i will get versionNo is null.
Please help me.
It is null because you annotated it as #Transient, so it is not populated with DB values. You can execute the query without mapping the results directly to your class, and populate the class manually (the query will return a list of Object[]). Other than this, I'm not aware of any alternatives (because of #Transient).
List<Object[]> results = session.createNativeQuery("select max(a.version_no) as versionNo, u.* from users u left join andother_table a on u.id=a.user_id where u.title='abc' group by id").list();
List<MyClass> myClasses = new ArrayList<MyClass>();
for (Object[] result : results) {
MyClass mc = new MyClass();
...
mc.setVersionNo(result[0]);
mc.setSomethingElse(result[1])
...
myClasses.add(mc);
}
Each entry in results list is an object array representing a row returned by native query. Columns are ordered as you select them, so if you put versionNo in the first place in SELECT clause, it will be available with result[0].
I'm creating a complex query with multiple tables and need to list the result. Usually, I'm using the EntityManager and map the result to the JPA-Representation:
UserEntity user = em.find(UserEntity.class, "5");
Then I can access all values as the user UserEntity class defines it. But how can I access the field-values returned from a native, multiple-table query? What I get is a List of Objects. That's fine so far, but what "is" that Object? Array? Map? Collection? ...
//simpleExample
Query query = em.createNativeQuery("SELECT u.name,s.something FROM user u, someTable s WHERE s.user_id = u.id");
List list = query.getResultList();
//do sth. with the list, for example access "something" for every result row.
I guess the answer is quite simple, but most examples out there just show the usage when directly casting to a targetClass.
PS: In the example I could use the class-mappings of course. But in my case someTable is not managed by JPA, and therefore I don't have the entity nor do I have a class-representation of it, and since I'm joining like 20 tables, I don't want to create all the classes just to access the values.
General rule is the following:
If select contains single expression and it's an entity, then result is that entity
If select contains single expression and it's a primitive, then result is that primitive
If select contains multiple expressions, then result is Object[] containing the corresponding primitives/entities
So, in your case list is a List<Object[]>.
Since JPA 2.0 a TypedQuery can be used:
TypedQuery<SimpleEntity> q =
em.createQuery("select t from SimpleEntity t", SimpleEntity.class);
List<SimpleEntity> listOfSimpleEntities = q.getResultList();
for (SimpleEntity entity : listOfSimpleEntities) {
// do something useful with entity;
}
If you need a more convenient way to access the results, it's possible to transform the result of an arbitrarily complex SQL query to a Java class with minimal hassle:
Query query = em.createNativeQuery("select 42 as age, 'Bob' as name from dual",
MyTest.class);
MyTest myTest = (MyTest) query.getResultList().get(0);
assertEquals("Bob", myTest.name);
The class needs to be declared an #Entity, which means you must ensure it has an unique #Id.
#Entity
class MyTest {
#Id String name;
int age;
}
The above query returns the list of Object[]. So if you want to get the u.name and s.something from the list then you need to iterate and cast that values for the corresponding classes.
I had the same problem and a simple solution that I found was:
List<Object[]> results = query.getResultList();
for (Object[] result: results) {
SomeClass something = (SomeClass)result[1];
something.doSomething;
}
I know this is defenitly not the most elegant solution nor is it best practice but it works, at least for me.
Here is the sample on what worked for me. I think that put method is needed in entity class to map sql columns to java class attributes.
//simpleExample
Query query = em.createNativeQuery(
"SELECT u.name,s.something FROM user u, someTable s WHERE s.user_id = u.id",
NameSomething.class);
List list = (List<NameSomething.class>) query.getResultList();
Entity class:
#Entity
public class NameSomething {
#Id
private String name;
private String something;
// getters/setters
/**
* Generic put method to map JPA native Query to this object.
*
* #param column
* #param value
*/
public void put(Object column, Object value) {
if (((String) column).equals("name")) {
setName(String) value);
} else if (((String) column).equals("something")) {
setSomething((String) value);
}
}
}
What if you create a bean with all required properties and cast the result using Java 8+ streams?
Like this:
public class Something {
private String name;
private String something;
// getters and setters
}
And then:
import javax.persistence.Query;
...
Query query = em.createNativeQuery("SELECT u.name,s.something FROM user u, someTable s WHERE s.user_id = u.id", Something.class);
List<?> list = query.getResultList();
return list
.stream()
.map(item -> item instanceof Something ? (Something) item : null)
.collect(Collectors.toList());
That way, you don't need to return List<Object[]> nor hide the warning with #SuppressWarnings("unchecked")
Ps.:
1 - I know that this post is very old. But... I'm here in 2021, so others will be coming here too =)
2 - This is wrong or bad practice? Let me know :D
You can also update your hibernate to a version greater than 5.4.30.final
I got a Set<UserDTO> collection in a not Hibernate object, and I got a User domain entity in Hibernate.
UserDTO contains less information about user (only id and name)
How can I select full Hibernate User Set/List from the DTO object?
Like this?
Set<UserDTO> setDTO = .....
String hql = "FROM User WHERE id IN (:userDTO )";
Query query = entityManager.createQuery(hql);
query.setParameter("userDTO", setDTO);
return query.getResultList();
Thanks
Almost. But you'd have to extract the IDs in a separate collection first:
Set<Long> ids = new HashSet<Long>(setDTO.size());
for (UserDTO dto : setDTO) {
ids.add(dto.getId());
}
Then proceed with the query, and pass the ids set as param.
Don't forget that you need to use Query#setParameterList() instead of Query#setParameter.