set table name with hibernate name parameters

Question

I need to set a table name dynamically so that I use query.setText(tname,abc)

e.g: select a.name from :tname where a.id = '2'

I used setText() because when I use setString() it says "tname is a invalid parameter" because I assume that Hibernate adds '' when setting string parameters.

But even setText() does not help and gives the same exception.

How can I set the table name dynamically?

Reply to PSR:

So you mean replace table name as a java string replacement. But then we can not take support of sql injections prevention etc from hibernate right? Also How we bind parameters in hibernate in a situation where like statement,

Eg: name like "%:name%"

This also gives me Illegal argument exception: Parameter does not exist as a named parameter when i try to bind it using query.setString(name,"def");

Andrew Lazarus · Accepted Answer · 2013-06-09 05:03:15Z

4

Hibernate will not do this for you, because it works with PreparedStatements, and you can't prepare a statement where the table being queried isn't known yet.

I don't see why you would be exposing table names to end users, so preventing SQL injection doing a regular string substitution should be easy. You use some sort of business logic to determine the correct table from a list that only you know. The table name isn't coming from user input at all.

Depending on your choice of RDBMS, you may find a discriminator column, or table inheritance with partitioning to be a better way of handling a situation where identical queries are made against different tables.

answered Jun 9, 2013 at 5:03

Andrew Lazarus

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PSR · Accepted Answer · 2013-06-09 02:18:36Z

1

It is not possible to set table name dynamically.You can set dynamically column names.it is not possible to set table name

try like this

select a.name from '+table name+'where a.id = '2'

answered Jun 9, 2013 at 2:18

PSR

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1 Comment

Harshana Over a year ago

please find the reply in the question

user1993667 · Accepted Answer · 2022-04-07 12:00:24Z

In my opinion, There are 2 ways to resolve this issue:

1- If you are using Spring and Hibernate together, you could use SpEL and it would be like #{#entityName} as it is described here

@Entity
public class User {

  @Id
  @GeneratedValue
  Long id;

  String lastname;
}

public interface UserRepository extends JpaRepository<User,Long> {

  @Query("select u from #{#entityName} u where u.lastname = ?1")
  List<User> findByLastname(String lastname);
}

2-You could use CriteriaBuilder like

CriteriaQuery<YourEntity> cr = cb.createQuery(YourEntity.class);
Root<YourEntity> root = cr.from(YourEntity.class);
cr.select(root);

I copied the source codes from the provided links and they are described there much better

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