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JPA Query using " where in" and setParameter

Wally Hartshorn
Ranch Hand

Joined: Jan 30, 2003
Posts: 77
I'm using the JPA Query Language for the first time and have run into a problem. I'm trying to do a "select where in (list of values)" query, but can't figure out the syntax to avoid hard-coding the list of values.

This works:

Replacing the hard-coded values with a parameter and an array gives me a class cast exception:

I'm using JDK 1.5, Hibernate 3, and JPA.

Can anyone give me some pointers on how to do this? (Or a better way to accomplish it?)


Wally Hartshorn
Debu Panda
Ranch Hand

Joined: Jan 21, 2007
Posts: 100
JPA supports IN clause either with a static list or dynamic list with a sub query.

You can probably work around by dynamically building the query:

"select t from Thing t where t.status in ("+ statusCodes +")" )

Author: EJB 3 In Action (
Dejan Mratinkovic
Ranch Hand

Joined: Nov 20, 2008
Posts: 65
Any updates on this? It's almost two years since original post. Is it possible that this is only way to create dynamic "where in"?

Can it be achieved in more elegant way?
Shailen Karambe

Joined: Nov 05, 2008
Posts: 2
You can check this link JPA doc

and query help

let me know whether your problem got resolved..
Konstantin Yankovski

Joined: Apr 25, 2009
Posts: 1
I do following and it works fine:

Hibernate generates following SQL for this JPA query:

'storeIds' parameter must be Collection, it doesn't work with an array!
Mandar Kaduskar

Joined: May 01, 2008
Posts: 3
absolutely there is a solution two ways you can achieve this, as mentioned by Debu Panda you can create a comma separated String and dynamically add it to a query
String statesCode= "'A', 'E'";
select t from Thing t where t.status in (statesCode)")


List<String> inList = new ArrayList<String>();
select t from Thing t where t.status in (:statesCode)")
query.setParameter("statesCode", inList );

Hope this helps.
Mandar Kaduskar

Joined: May 01, 2008
Posts: 3
Also dont forget that if your IN parameter has numeric data type like Long, smallInt, BigInt etc then you will need to create the list of the appropriate data type instead of Strings. Hope makes sense.
Amit Ghorpade

Joined: Jun 06, 2007
Posts: 2764

"app developer " please check your private messages for an important administrative matter. You can check them by clicking the My Private Messages link above.

|Asking Good Questions|
lilili bbb

Joined: Sep 07, 2011
Posts: 1
try to change setParameter(1,something);
to setLong(1,something);
(there are setString etc..)
that helps me....
good luck!
Yasin Turk

Joined: Feb 23, 2008
Posts: 4
You should use setParameterList() method for "in" type parameters
I agree. Here's the link:
subject: JPA Query using " where in" and setParameter