I understand some might simply answer this question with "Why didn't you just Google it"... But I did, and the more I researched this the more confused I got. I'm trying to query my database with Hibernate, the query has a 'where' clause.
Now creating a database entry is easy enough, in the case where I have a 'User' class, I simply do this:
// Gets a new session
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
// Creates a new User object
User user = new User("John", "p#55w0rd*", "john#doe.com");
// Save and commit
session.save(user);
session.getTransaction().commit();
But what do I do when I what to for instance
select * from Users where id = '3';
My Google searches pointed to something called HQL, which makes me wonder why I couldn't of just used straight JDBC then. Also it doesn't seem very object oriented. And then there's something like
session.createCriteria(.......
But I'm not sure how to use this.. Any help? Thanks guys.
When you use Native Query (non HQL ) you need to tell hibernate explicitely to handle it like below :
In below query createSQLQuery is special function to handle native sql's
String sql = "SELECT * FROM EMPLOYEE WHERE id = :employee_id";
SQLQuery query = session.createSQLQuery(sql);
query.addEntity(User.class);
query.setParameter("employee_id", 3);
List<User> results = query.list();
This can be done using criteria as well for that following is good starting point:
Criteria criteria = sess.createCriteria( User.class);
List<User> users= criteria.list();
http://www.developerhelpway.com/framework/hibernate/criteria/index.php
First of all, you need a hibernate.cfg.xml which contains properties for hibernate. This is e.g url, username and password, the driver and dialect. This file is placed in a package called resources.
You have to choose between using Hibernate Annotations example
or using hbm.xml files example
This is how you tell hibernate what your database is like. It wil automatically create queries for you based on how you annotates or defines in e.g user.hbm.xml.
Create a HibernateUtil.java class which holds the session factory.
You can fetch data from the database with
Criteria crit = getSessionFactory().getCurrentSession().createCriteria(User.class);
Example using queries:
List<?> hibTuppleResultList = currentSession.createQuery(
"from Person p, Employment e "
+ "where e.orgno like ? and p.ssn = e.ssn and p"
+ ".bankno = ?")
.setString(0, orgNo).setString(1, bankNo).list();
for (Object aHibTuppleResultList : hibTuppleResultList)
{
Object[] tuple = (Object[]) aHibTuppleResultList;
Person person = (Person) tuple[0];
hibList.add(person);
}
In the end all I really wanted was to know that if you don't want to use HQL you get something called 'Criteria Queries', and that in my case I'd do something like this:
Criteria cr = session.createCriteria(User);
cr.add(Restrictions.eq("id", 3));
List results = cr.list();
Me: "Thanks!"
Me: "No problem :)"
PS - we can really delete this question.
Query q = session.createQuery("from User as u where u.id = :u.id");
q.setString("id", "3");
List result = q.list();
Query with Criteria:
Criteria cr = session.createCriteria(User.class);
List results = cr.list();
Restrictions with Criteria:
Criteria cr = session.createCriteria(User.class);
cr.add(Restrictions.eq("id", 3));
// You can add as many as Restrictions as per your requirement
List results = cr.list();
You could also use it like this
List results = session.createCriteria(User.class).add(Restrictions.eq("id", 3)).list();
Some example for Crieteria Rsetriction query
Criteria cr = session.createCriteria(Employee.class);
// To get records having salary more than 2000
cr.add(Restrictions.gt("salary", 2000));
// To get records having salary less than 2000
cr.add(Restrictions.lt("salary", 2000));
// To get records having fistName starting with zara cr.add(Restrictions.like("firstName", "zara%"));
// Case sensitive form of the above restriction.
cr.add(Restrictions.ilike("firstName", "zara%"));
// To get records having salary in between 1000 and 2000
cr.add(Restrictions.between("salary", 1000, 2000));
// To check if the given property is null
cr.add(Restrictions.isNull("salary"));
// To check if the given property is not null
cr.add(Restrictions.isNotNull("salary"));
// To check if the given property is empty
cr.add(Restrictions.isEmpty("salary"));
// To check if the given property is not empty
cr.add(Restrictions.isNotEmpty("salary"));
You can create AND or OR conditions using LogicalExpression restrictions as follows:
Criteria cr = session.createCriteria(Employee.class);
Criterion salary = Restrictions.gt("salary", 2000);
Criterion name = Restrictions.ilike("firstNname","zara%");
// To get records matching with OR condistions
LogicalExpression orExp = Restrictions.or(salary, name);
cr.add( orExp );
// To get records matching with AND condistions
LogicalExpression andExp = Restrictions.and(salary, name);
cr.add( andExp );
List results = cr.list();
I think this will help you
Related
query should be sorted by 'lastUsedTimestamp' in ASC. If these entities lastUsedTimestamp is null or expired or do not have the field, we just remove them from the collection with defined limit.
I have them written like below but it is giving null
Criteria fieldsCriteria1 = Criteria.where("lastAccessTimestamp").lte(date);
Criteria fieldsCriteria2 = Criteria.where("lastAccessTimestamp").exists(false);
Query query2 = new Query();
query2.limit(3);
query2.with(Sort.by(Sort.Direction.ASC,"lastAccessTimestamp")); // to set in ASC order
query2.addCriteria(fieldsCriteria1); // to set the expired time
If i have only these above criteria added it works fine, the problem occurs when i add the below criteria
query2.addCriteria(fieldsCriteria2); // to get if the lastAccessTimestamp field is empty
I am new to Mongo Db, also I am not sure which is the best way to fulfill the above query.
I had figured out the solution, the below query worked for me.
var fieldsCriteria = new Criteria()
.orOperator(Criteria.where(LAST_ACCESS_TIMESTAMP).lt(date)
,Criteria.where(LAST_ACCESS_TIMESTAMP).exists(false)
);
var query = new Query();
query.limit(limit);
query.with(Sort.by(Sort.Direction.ASC, LAST_ACCESS_TIMESTAMP));
query.addCriteria(fieldsCriteria);
List<Document> list=mongoTemplate.find(query,Document.class,collectionName);
I'm trying to delete all the records from a MySQL table (46 records).
The code I have tried. Any suitable answer?
Session hs = connection.NewHibernateUtil.getSessionFactory().openSession();
Criteria cr = hs.createCriteria(Bookmark.class);
Bookmark b;
List<Bookmark> li = cr.list();
for (Bookmark s : li) {
b = new Bookmark();
b.setId(s.getId());
Transaction tr = hs.beginTransaction();
hs.delete(b);
tr.commit();
hs.flush();
hs.close();
}
Error
org.hibernate.NonUniqueObjectException: a different object with the same identifier value was already associated with the session: [mypojos.Bookmark#7]
You cant delete objects like that. You would first have to get the object from db and then you can delete using hs.delete(b); this is usually used when you have to cascade changes to associated objects.
Best approach in this case is to use HQL query something like this.
String stringQuery = "DELETE FROM tablename";
Query query = session.createQuery(stringQuery);
query.executeUpdate();
I've a database with many thousands of tables that have been (and continue to be) created with a naming strategy - one table per calendar day:
data_2010_01_01
data_2010_01_02
...
data_2020_01_01
All tables contain sensor data from the same system in the same shape. So a single entity (lets call it SensorRecord) will absolutely map to all tables.
I'd imagined something like this would work:
#Query(nativeQuery = true, value = "SELECT * FROM \"?1\"")
Collection<SensorRecord> findSensorDataForDate(String tableName);
But it does not, and reading around the topic seems to suggest I am on the wrong path. Most posts on dynamic naming seem to state explicitly that you need one entity per table, but generating thousands of duplicate entities also seems wrong.
How can I use JPA (JPQL?) to work with this data where the table name follows a naming convention and can be changed as part of the query?
Parameters are only allowed in the where clause.
You can create custom repository method returns collection of SensorRecord dto. No need to map so many entities. You should get List<Object []> as query result and manually create dto objects.
#Autowired
EntityManager entityManager;
public List<SensorRecord> findSensorDataForDate(LocalDate date) {
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy_MM_dd");
String tableName = "data_" + date.format(formatter);
Query query = entityManager.createNativeQuery(
"select t.first_column, t.second_column from " + tableName + " t");
List<Object[]> queryResults = query.getResultList();
List<SensorRecord> sensorRecords = new ArrayList<>();
for (Object[] row : queryResults) {
SensorRecord record = new SensorRecord();
record.setFirstParameter((Integer) row[0]);
record.setSecondParameter((String) row[1]);
sensorRecords.add(record);
}
return sensorRecords;
}
Could it be just syntax error?
This has worked for me:
#Query(value = "select * from job where job.locked = 1 and job.user = ?1", nativeQuery = true)
public List<JobDAO> getJobsForUser(#Param("user") String user);
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");
I have a method like below
public List<String> getSimilarResourceNames(String resourceName){
String searchString = "%"+resourceName+"%";
Session session = getSession();
Criteria criteria = session.createCriteria(Resource.class);
criteria.add(Restrictions.like("name", searchString));
return criteria.list()
}
This will return me the entire resource from the DB, but what i need is just the name of the resource. How can I accomplish that ?
Use Projection, you can find examples in Hibernate documentation.
Criteria criteria = session.createCriteria(Resource.class);
criteria.setProjection(Property.forName("name"))
criteria.add(Restrictions.like("name", searchString));
By using Projection you will get other fields (Which you did not got by Projection) in your Pojo setted to default values. In HQL you can get specified column values as follow:
Query query = session.createQuery("select u.fullname from Users u");
List<Object[]> rows = query.list();
List<String> fullnames = new ArrayList<String>();
for (Object[] row: rows) {
fullnames.add(row[0]);
}
I hope this will help you.