JPA CriteriaBuilder - How to use “IN” comparison operator

问题

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 = criteria.select(scheduleRequest);

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) );
usersIdsParamList.add(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

回答1:

If I understand well, you want to Join ScheduleRequest with User and apply the in clause to the userName property of the entity User.

I'd need to work a bit on this schema. But you can try with this trick, that is much more readable than the code you posted, and avoids the Join part (because it handles the Join logic outside the Criteria Query).

List<String> myList = new ArrayList<String> ();
for (User u : usersList) {
    myList.add(u.getUsername());
}
Expression<String> exp = scheduleRequest.get("createdBy");
Predicate predicate = exp.in(myList);
criteria.where(predicate);

In order to write more type-safe code you could also use Metamodel by replacing this line:

Expression<String> exp = scheduleRequest.get("createdBy");

with this:

Expression<String> exp = scheduleRequest.get(ScheduleRequest_.createdBy);

If it works, then you may try to add the Join logic into the Criteria Query. But right now I can't test it, so I prefer to see if somebody else wants to try.

回答2:

Not a perfect answer though may be code snippets might help.

public <T> List<T> findListWhereInCondition(Class<T> clazz,
            String conditionColumnName, Serializable... conditionColumnValues) {
        QueryBuilder<T> queryBuilder = new QueryBuilder<T>(clazz);
        addWhereInClause(queryBuilder, conditionColumnName,
                conditionColumnValues);
        queryBuilder.select();
        return queryBuilder.getResultList();

    }


private <T> void addWhereInClause(QueryBuilder<T> queryBuilder,
            String conditionColumnName, Serializable... conditionColumnValues) {

        Path<Object> path = queryBuilder.root.get(conditionColumnName);
        In<Object> in = queryBuilder.criteriaBuilder.in(path);
        for (Serializable conditionColumnValue : conditionColumnValues) {
            in.value(conditionColumnValue);
        }
        queryBuilder.criteriaQuery.where(in);

    }

来源：https://stackoverflow.com/questions/9321916/jpa-criteriabuilder-how-to-use-in-comparison-operator

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hibernate

jpa

orm

jpa-2.0

criteria-api