JPA CriteriaBuilder - How to use ?IN? comparison operator



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JPA CriteriaBuilder - How to use ?IN? comparison operator
0 Answer(s)      5 years and 9 months ago
Posted in : Hibernate

Can you please help me how to convert the following codes to using "in" operator of criteria builder? I need to filter by using list/array of usernames using "in".

***I also tried to search using JPA CriteriaBuilder - "in" method but cannot find good result. So I would really appreciate also if you can give me reference URLs for this topic. Thanks.***

Here is my codes:

    //usersList is a list of User that I need to put inside IN operator 

    CriteriaBuilder builder = getJpaTemplate().getEntityManagerFactory().getCriteriaBuilder();
    CriteriaQuery<ScheduleRequest> criteria = builder.createQuery(ScheduleRequest.class);

    Root<ScheduleRequest> scheduleRequest = criteria.from(ScheduleRequest.class);
    criteria =;

    List<Predicate> params = new ArrayList<Predicate>();

    List<ParameterExpression<String>> usersIdsParamList = new ArrayList<ParameterExpression<String>>();

    for (int i = 0; i < usersList.size(); i++) {
    ParameterExpression<String> usersIdsParam = builder.parameter(String.class);
    params.add(builder.equal(scheduleRequest.get("createdBy"), usersIdsParam) );

    criteria = criteria.where(params.toArray(new Predicate[0]));

    TypedQuery<ScheduleRequest> query = getJpaTemplate().getEntityManagerFactory().createEntityManager().createQuery(criteria);

    for (int i = 0; i < usersList.size(); i++) {
    query.setParameter(usersIdsParamList.get(i), usersList.get(i).getUsername());

    List<ScheduleRequest> scheduleRequestList = query.getResultList();

The internal Query String is converted to below, so I don't get the records created by the two users, because it is using "AND".

select generatedAlias0 from ScheduleRequest as generatedAlias0 where ( generatedAlias0.createdBy=:param0 ) and ( generatedAlias0.createdBy=:param1 ) order by generatedAlias0.trackingId asc

Please need your advise. Thank you in advance.

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