How can I count the number of query results without actually

1badpny · **Joined:** Fri Apr 15, 2005 4:53 pm **Posts:** 29

How can I count the number of query results without actually returning them?

"( (Integer) session.iterate("select count(*) from ....").next() ).intValue()"

thats from hibernate faq...but session doesnt have any iterate method :-s

thx

vgiguere · **Posted:** Sun May 22, 2005 11:02 am

You can use the Criteria API

Code:

session.createCriteria(MyObject.class).setProjection(Projections.count("fieldToCount")).uniqueResult();

1badpny · **Joined:** Fri Apr 15, 2005 4:53 pm **Posts:** 29

vgiguere wrote:

You can use the Criteria API

Code:

session.createCriteria(MyObject.class).setProjection(Projections.count("fieldToCount")).uniqueResult();

What object instance is returned from the following statement:

session.createSQLQuery ("select count(*) from users").list().get(0);

Its not letting me typecast it to Integer or a String.

Thanks

alesj · **Posted:** Tue May 24, 2005 1:48 am

Debug and you will see :-).

But you should be able to cast it to Integer or at least a Number.

1badpny · **Joined:** Fri Apr 15, 2005 4:53 pm **Posts:** 29

alesj wrote:

Debug and you will see :-).

But you should be able to cast it to Integer or at least a Number.

I've tried debugging but all it says is "java.lang.Object"

*confused*

alesj · **Posted:** Wed May 25, 2005 3:24 am

How could you run that sql query without setting entity or scalar?

>> org.hibernate.QueryException: addEntity() or addScalar() must be called on a sql query before executing the query. [select count(*) from users]

Your sql should probably look something like this:
SELECT {person}.NAME AS {person.name}, {person}.AGE AS {person.age}, {person}.SEX AS {person.sex}
FROM PERSON {person} WHERE {person}.NAME LIKE 'Hiber%'

vgiguere's solution should definitely work.

vgiguere · **Posted:** Wed May 25, 2005 3:16 pm

If you debug it, you will see that it is an Integer.

So, you can do:

Code:

Query query = session.createQuery("select count(*) from whatever");
Integer myAnswer = (Integer) query.uniqueResult();

You should use uniqueResult() if you are using an aggregation function. Otherwise, if you like using .list()just get the first result of the returned collection. It will still be an Integer.