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Spring Data JPA Specifications - Distinct + order by column in join

I have some specifications that I am combining with "and":
Specification.where(predicate1).and(predicate2).and(predicate3);
One of them has distinct set :
query.distinct(true);
Another one makes an order by on a column that is in a join.
query.orderBy(builder.desc(bJoin.get("orderbyColumn")));
This fails with a SQLGrammarException stating that order by column should be in distinct.
So basically we have entity A, the main entity, and some nested entity B, we select from A but want to order by B, and in generated sql it only selects columns from A. The only way I found to make it work (= making it select from B as well) is to replace the join by a fetch :
Fetch < A, B > bFetch = root.fetch("joinCol", JoinType.INNER);
Join < A, B > bJoin = (Join < A, B > ) bFetch;
that worked for some time, was testing locally in H2, but then after some time started getting another error :
org.hibernate.QueryException: query specified join fetching, but the
owner of the fetched association was not present in the select list
I solved it somehow in my local pointing to H2 by requiring some columns to not be null, but in real server using PostgreSQL, it's not working at all, getting that error for all cases when a fetch is present.
My question is : what is the right way to use distinct along with orderby on a nested entity that is not fetched? Is my solution with fetch ok and it just needs to be fixed (and if so how?) or I should go for another option entirely?
For the actual query I am using this method :
findAll(Specification<>, Pageable)
Isn't there a way to have distinct wrapping the whole query with order by (some sort of subquery?) and bypassing all this nightmare? Have it generate a query like this:
select distinct colA1, colA2, coAl3 from (select colA1, colA2, coAl3
from A inner join B b on ........ order by b.colB1)
Do I need to convert my specification to predicate manually and do something else with it to try to solve my issues (some kind of hybrid approach)?
Any pieces of advice will be greatly appreciated.

I encountered same error but actually it was not error :)
findAll(Specification<>, Pageable) this method throws 2 different queries.
First one is count query where you have to be careful.
Second is the rows query where you actually did it.
You can check the query type with code below
if (query.getResultType() != Long.class && query.getResultType() != long.class){
root.fetch("entity1");
}

Specfiy columns which should be considered for distinct calculation

I'm using javax.persistence.criteria.CriteriaBuilder and javax.persistence.criteria.CriteriaQuery to select some entities.
I now want to select only the entities that are unique which should be specified by a certain column.
There is the method javax.persistence.criteria.CriteriaQuery#distinct which only returns unique entities.
I would rather need something like
CriteriaQuery<T> distinct(String... columnNames)
Do you know how I can bake such a distinct in my JPA CriteriaQuery?
It seems to be possible with Hibernate.

The following statement has no sense:
I now want to select only the entities that are unique which should be
specified by a certain column.
The result sets are filtered by 'distinct' if they are 'exactly the same'.
The entities are not the same if only some fields are the same.
You can make distinct clause on resultset in the following manner:
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery query = builder.createQuery();
Root<Friend> c = query.from(Friend.class);
query.multiselect(c.get(Friend_.firstName),c.get(Friend_.lastName)).distinct(true);
then you will get unique combination of firstName and lastName from Friend entities.
So for instance... 'Give me all unique combinations from my Friends where the firstName and lastName of the friend is unique.' But it doesn't mean give me unique friends.

distinct takes a boolean as parameter. You can use multiselect to select more than one column like in this example:
CriteriaQuery<Country> q = cb.createQuery();
Root<Country> c = q.from(Country.class);
q.multiselect(c.get("currency"), c.get("countryCode")).distinct(true);

JPA CriteriaQuery with LEAST and GREATEST functions

I am have a problem where i need to join two tables using the LEAST and GREATEST functions, but using JPA CriteriaQuery. Here is the SQL that i am trying to duplicate...
select * from TABLE_A a
inner join TABLE_X x on
(
a.COL_1 = least(x.COL_Y, x.COL_Z)
and
a.COL_2 = greatest(x.COL_Y, x.COL_Z)
);
I have looked at CriteriaBuilder.least(..) and greatest(..), but am having a difficult time trying to understand how to create the Expression<T> to pass to either function.

The simplest way to compare two columns and get the least/greatest value is to use the CASE statement.
In JPQL, the query would look like
select a from EntityA a join a.entityXList x
where a.numValueA=CASE WHEN x.numValueY <= x.numValueZ THEN x.numValueY ELSE x.numValueZ END
and a.numValueB=CASE WHEN x.numValueY >= x.numValueZ THEN x.numValueY ELSE x.numValueZ END
You can code the equivalent using CriteriaBuilder.selectCase() but I've never been a big fan of CriteriaBuilder. If requirements forces you to use CriteriaBuilder then please let me know and I can try to code the equivalent.
CriteriaBuilder least/greatest is meant to get the min/max value of all the entries in one column. Let's say you want to get the Entity that had the alphabetically greatest String name. The code would look like
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery query = cb.createQuery(EntityX.class);
Root<EntityX> root = query.from(EntityX.class);
Subquery<String> maxSubQuery = query.subquery(String.class);
Root<EntityX> fromEntityX = maxSubQuery.from(EntityX.class);
maxSubQuery.select(cb.greatest(fromEntityX.get(EntityX_.nameX)));
query.where(cb.equal(root.get(EntityX_.nameX), maxSubQuery));
I created a sample Spring Data JPA app that demonstrates these JPA examples at
https://github.com/juttayaya/stackoverflow/tree/master/JpaQueryTest

It turns out that CriteriaBuilder does support calling LEAST and GREATEST as non-aggregate functions, and can be accessed by using the CriteriaBuilder.function(..), as shown here:
Predicate greatestPred = cb.equal(pathA.get(TableA_.col2),
cb.function("greatest", String.class,
pathX.get(TableX_.colY), pathX.get(TableX_.colZ)));

Need help creating JPA criteria query

I'm building my first Java EE web application using Glassfish and JSF. I'm fairly new to the criteria query and I have a query I need to perform but the javaee6 tutorial seems a little thin on examples. Anyway, I'm having a hard time creating the query.
Goal: I want to pull the company with the most documents stored.
Companies have a OneToMany relationship with Documents.
Documents has a ManyToOne relationship with several tables, the "usertype" column distinguishes them.
MySQL query:
SELECT USERID, COUNT(USERID) AS CNT
FROM DOCUMENTS
WHERE USERTYPE="COMPANY"
GROUP BY USERID
ORDER BY CNT DESC
Thanks
--update--
Based on user feedback, here is what I have so far:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Documents> cqry = cb.createQuery(Documents.class);
//Intersting Stuff
Root<Documents> root = cqry.from(Documents.class);
Expression userid = root.get("userID");
Expression usertype = root.get("userType");
Expression count = cb.count(userid);
cqry.multiselect(userid, count);
Predicate userType = cb.equal(usertype, "COMPANY");
cqry.where(userType);
cqry.groupBy(userid);
cqry.orderBy(cb.desc(count));
//more boilerplate
Query qry = em.createQuery(cqry);
List<Documents> results = qry.getResultList();
The error I get is:
Exception Description: Partial object queries are not allowed to maintain the cache or be edited. You must use dontMaintainCache().
Typical error, means nothing to me!

Your query doesn't return a complete entity object as you're only selecting two fields of the given table (this is why you're getting an error that says yadayadapartialyadayada).
Your solution is almost right, here's what you need to change to make it work—making it partial.
Instead of a plain CriteriaQuery<...> you have to create a tuple CriteriaQuery<..> by calling CriteriaBuilder.createTupleQuery(). (Basically, you can call CriteriaBuilder.createQuery(...) and pass Tuple.class to it as an argument. Tuple is a sort of wildcard entity class.)
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Tuple> cq= cb.createTupleQuery();
Root<Documents> root = cq.from(Documents.class);
Expression<Integer> userId = root.get("USERID");
Expression<String> userType = root.get("USERTYPE");
Expression<Long> count = cb.count(userId);
cq.multiselect(userId.alias("USERID"), count.alias("CNT"));
cq.where(cb.equal(userType, "COMPANY");
cq.groupBy(userId);
cq.orderBy(cb.desc(count));
TypedQuery<Tuple> tq = em.createQuery(cq);
for (Tuple t : tq.getResultsList()) {
System.out.println(t.get("USERID"));
System.out.println(t.get("CNT"));
}
(Accessing fields of a Tuple gave me an error if I didn't use aliases for them (in multiselect(...)). This is why I've used aliases, but this can be tackled more cleanly by using JPA 2's Metamodel API, which is described in the specification quite thoroughly. )
The documentation for CriteriaQuery.multiselect(...) describes the behaviour of queries using Tuple objects more deeply.

If you are using Hibernate, this should work:
ProjectionList pl = Projections.projectionList()
.add(Projections.groupProperty("userid"))
.add(Projections.property("userid"))
.add(Projections.count("userid"));
Criteria criteria = session.createCriteria(Document.class)
.add(Restrictions.eq("usertype",usertype))
.setProjection(pl)
.addOrder(Order.desc("cnt"));
Hope it helps!

Take a look into this easy tutorial. It uses JPA2 and Criteria
http://www.jumpingbean.co.za/blogs/jpa2-criteria-api
Regards!

You need to add a constructor to Documents with only userid and count because you will need it on:
cqry.multiselect(userid, count);

Referring to an earlier aliased field in a criteria query

In this query:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Tuple> q = cb.createTupleQuery();
// FROM GamePlayedEvent gpe
Root<GamePlayedEvent> gpe = q.from(GamePlayedEvent.class);
// SELECT gameId, COUNT(*) AS count, AVG(duration)
// AS avDur, AVG(rewardCurrency) AS avCur, AVG(rewardXP) avXp
q.select(cb.tuple(
gpe.<String>get("gameId"),
cb.count(gpe).alias("count"),
cb.avg(gpe.<Double>get("duration")).alias("avDur"),
cb.avg(gpe.<Integer>get("rewardCurrency")).alias("avCur"),
cb.avg(gpe.<Integer>get("rewardXp")).alias("avXp")
));
// WHERE loginTime BETWEEN ...
q.where(cb.between(gpe.<Date>get("time"), fromTime, toTime));
// GROUP BY gameId
q.groupBy(gpe.<String>get("gameId"));
// ORDER BY count DESC
q.orderBy(cb.desc(???));
How can I add the ORDER BY count DESC, referring to the "count" defined in the SELECT clause?

What if you just captured the count expression, and used it directly?
Expression event_count = cb.count(gpe);
q.select(cb.tuple(
gpe.<String>get("gameId"),
event_count,
...
));
q.orderBy(cb.desc(event_count));

I came across the same problem today but none of the suggested solutions worked for me because I needed to reuse the expression not only in the order by clause but also in the group by clause.
One obvious solution would be to create a view on the database level but this is a bit clumsy, creates an unnecessary subquery and even not possible if the db user isn't granted enough privileges. A better option which I ended up implementing is to write something like this
q.select(cb.tuple(
gpe.<String>get("gameId"),
cb.count(gpe),
...
)).groupBy(cb.literal(2)).orderBy(cb.literal(2));
The first downside of this approach is that the code is errorprone. The other drawback is that the generated sql query contains ordinal position notation, which works on some databases (like postgresql or mariadb) but doesn't work on others (like sql server). In my case, however, I found this to be the best option.
Tested on jpa 2.1 with hibernate 5.2.3 as a provider and postgresql 9.6.

Even though the Pro JPA 2 book describes that the alias method can be used to generate a sql query alias (on page 251) I have had no success with making it work with neither EclipseLink or Hibernate. For your question I would say that your orderBy line should read:
q.orderBy(cb.desc(cb.count(gpe));
if it was supported by the different vendors.
As far as my research goes it seams that the alias method is only used for naming elements in the tuble used in the select (so only for projection).
I have one question though. Why would you want to use the JPA Criteria API for this query. It (the query) seams to be static in nature so why not use JPQL where you can define your query aliases directly.

Have you tried setting up a projection with an alias?
criteria.setProjection(Projections.projectionList()
.add(Projections.count("item.id"), "countItems"));
criteria.addOrder(Order.desc("countItems"));

For a sum aggregation field I have the following code which worked for me:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<T> cq = cb.createQuery(entity);
Root<T> root = cq.from(entity);
cq.orderBy(cb.desc(cb.sum(root.get(orderByString))));
// orderByString is string entity field that is being aggregated and which we want to put in orderby clause as well.