Hql Query to Bind a Data Member - java

Query qry = session.createQuery("From RegistrationBean where ? = ?");
qry.setString(0,searchCriteria);
qry.setString(1,searchField);
searchList =(ArrayList<RegistrationBean>) qry.list();
RegistrationBean Entity class has userName, address, age fields..
I want to search a user by search criteria such as userName, address etc. using the above single query...
But the query is returning me zero results even though the user exist..
what's the problem?

Both parameters are set to 0 position, the second is not set. The position parameter should be set sequentially.
qry.setParameter(0,searchCriteria);
qry.setParameter(1,searchField);
But the field name should pass in the following way
String queryString = "from RegistrationBean as model where model." + propertyName + "= ?";
Query queryObject = getSession().createQuery(queryString);
queryObject.setParameter(0, value);
return queryObject.list();

Try in the following way:
Query qry = session.createQuery("select * from RegistrationBean where :searchCrit = :searchValue");
qry.setString(":searchCrit",searchCriteria);
qry.setString(":searchValue",searchField);
searchList =(ArrayList<RegistrationBean>) qry.list();
It is more appropriate to use it like this instead of setting indexes (your problem is that you set both to 0), because if you change something in your query you might need to change your setters afterwards.

Related

Dyamic table name for JPQL / Hibernate query

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);

How to update an entire object using jdbcTemplate without specifying column names?

I was looking at the reference here. We can do this -
String orderSql = "select * from order where id = ?";
jdbcTemplate.query(orderSql, new BeanPropertyRowMapper<>(Order.class), orderId);
This reads from the database and can directly deserialise into the Order object. This is well and good.
What I want to do is -
String updateSql = "update order ? where id = ?";
jdbcTemplate.save(updateSql, new BeanPropertyRowMapper<>(Order.class), order, orderId);
Is there a way of doing this?
No, it's not possible. JdbcTemplate comes with the ability to map an object using query as you've said without having to define columns/attributes, but there is no equivalent usage for update. You will have to pass in your SQL and the relevant parameters. As someone has mentioned if you really want to do this you could consider using hibernate.
Use this query for update :-
String updateSql = "update <table_name> set order= ? where id = ?";
jdbcTemplate.update(updateSql, new Object[]{order, orderId});
It will return affected row update count.
No ! You can only do this !
jdbcTemplate.update("update ordertable set order = ? where id = ?", order, orderId);

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 native query control entity(part 3)

Retrieve company table
GetEntity.java
String table = "company";
String q = "select * from " +table;
Query query = em.createNativeQuery(q, Company.class);
List<Company> list = query.getResultList();
...
Retrieve staff table
GetEntity.java
String table = "staff";
String q = "select * from " +table;
Query query = em.createNativeQuery(q, Staff.class);
List<Staff> list = query.getResultList();
...
My questions is how do I control the ? from the following:
em.createNativeQuery(q, ?);
List<?> list = q.getResultList();
Any ideas or suggestion?
Another option is to pass the Class entityClass as an argument to your find method and then you can try and derive table name from entityClass by using reflection and use the entityClass as the type argument to your createNativeQuery method.
Hope this helps!

How to query a database using Hibernate?

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

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