如何使用连接和基于行的限制(分页)来获得不同的休眠结果？

You can achieve the desired result by requesting a list of distinct ids instead of a list of distinct hydrated objects.

Simply add this to your criteria:

criteria.setProjection(Projections.distinct(Projections.property("id")));

Now you'll get the correct number of results according to your row-based limiting. The reason this works is because the projection will perform the distinctness check as part of the sql query, instead of what a ResultTransformer does which is to filter the results for distinctness after the sql query has been performed.

Worth noting is that instead of getting a list of objects, you will now get a list of ids, which you can use to hydrate objects from hibernate later.

I am using this one with my codes.

Simply add this to your criteria:

criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);

that code will be like the select distinct * from table of the native sql. Hope this one helps.

The solution:

criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);

works very well.

NullPointerException in some cases! Without criteria.setProjection(Projections.distinct(Projections.property("id"))) all query goes well! This solution is bad!

Another way is use SQLQuery. In my case following code works fine:

List result = getSession().createSQLQuery(
"SELECT distinct u.id as usrId, b.currentBillingAccountType as oldUser_type,"
+ " r.accountTypeWhenRegister as newUser_type, count(r.accountTypeWhenRegister) as numOfRegUsers"
+ " FROM recommendations r, users u, billing_accounts b WHERE "
+ " r.user_fk = u.id and"
+ " b.user_fk = u.id and"
+ " r.activated = true and"
+ " r.audit_CD > :monthAgo and"
+ " r.bonusExceeded is null and"
+ " group by u.id, r.accountTypeWhenRegister")
.addScalar("usrId", Hibernate.LONG)
.addScalar("oldUser_type", Hibernate.INTEGER)
.addScalar("newUser_type", Hibernate.INTEGER)
.addScalar("numOfRegUsers", Hibernate.BIG_INTEGER)
.setParameter("monthAgo", monthAgo)
.setMaxResults(20)
.list();

Distinction is done in data base! In opposite to:

criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);

where distinction is done in memory, after load entities!

A small improvement to @FishBoy's suggestion is to use the id projection, so you don't have to hard-code the identifier property name.

criteria.setProjection(Projections.distinct(Projections.id()));

A slight improvement building on FishBoy's suggestion.

It is possible to do this kind of query in one hit, rather than in two separate stages. i.e. the single query below will page distinct results correctly, and also return entities instead of just IDs.

Simply use a DetachedCriteria with an id projection as a subquery, and then add paging values on the main Criteria object.

It will look something like this:

DetachedCriteria idsOnlyCriteria = DetachedCriteria.forClass(MyClass.class);
//add other joins and query params here
idsOnlyCriteria.setProjection(Projections.distinct(Projections.id()));


Criteria criteria = getSession().createCriteria(myClass);
criteria.add(Subqueries.propertyIn("id", idsOnlyCriteria));
criteria.setFirstResult(0).setMaxResults(50);
return criteria.list();

I will now explain a different solution, where you can use the normal query and pagination method without having the problem of possibly duplicates or suppressed items.

This Solution has the advance that it is:

• faster than the PK id solution mentioned in this article
• preserves the Ordering and don’t use the 'in clause' on a possibly large Dataset of PK’s

The complete Article can be found on my blog

Hibernate gives the possibility to define the association fetching method not only at design time but also at runtime by a query execution. So we use this aproach in conjunction with a simple relfection stuff and can also automate the process of changing the query property fetching algorithm only for collection properties.

First we create a method which resolves all collection properties from the Entity Class:

public static List<String> resolveCollectionProperties(Class<?> type) {
List<String> ret = new ArrayList<String>();
try {
BeanInfo beanInfo = Introspector.getBeanInfo(type);
for (PropertyDescriptor pd : beanInfo.getPropertyDescriptors()) {
if (Collection.class.isAssignableFrom(pd.getPropertyType()))
ret.add(pd.getName());
}
} catch (IntrospectionException e) {
e.printStackTrace();
}
return ret;
}

After doing that you can use this little helper method do advise your criteria object to change the FetchMode to SELECT on that query.

Criteria criteria = …


//    … add your expression here  …


// set fetchmode for every Collection Property to SELECT
for (String property : ReflectUtil.resolveCollectionProperties(YourEntity.class)) {
criteria.setFetchMode(property, org.hibernate.FetchMode.SELECT);
}
criteria.setFirstResult(firstResult);
criteria.setMaxResults(maxResults);
criteria.list();

Doing that is different from define the FetchMode of your entities at design time. So you can use the normal join association fetching on paging algorithms in you UI, because this is most of the time not the critical part and it is more important to have your results as quick as possible.

Below is the way we can do Multiple projection to perform Distinct

    package org.hibernate.criterion;


import org.hibernate.Criteria;
import org.hibernate.Hibernate;
import org.hibernate.HibernateException;
import org.hibernate.type.Type;


/**
* A count for style :  count (distinct (a || b || c))
*/
public class MultipleCountProjection extends AggregateProjection {


private boolean distinct;


protected MultipleCountProjection(String prop) {
super("count", prop);
}


public String toString() {
if(distinct) {
return "distinct " + super.toString();
} else {
return super.toString();
}
}


public Type[] getTypes(Criteria criteria, CriteriaQuery criteriaQuery)
throws HibernateException {
return new Type[] { Hibernate.INTEGER };
}


public String toSqlString(Criteria criteria, int position, CriteriaQuery criteriaQuery)
throws HibernateException {
StringBuffer buf = new StringBuffer();
buf.append("count(");
if (distinct) buf.append("distinct ");
String[] properties = propertyName.split(";");
for (int i = 0; i < properties.length; i++) {
buf.append( criteriaQuery.getColumn(criteria, properties[i]) );
if(i != properties.length - 1)
buf.append(" || ");
}
buf.append(") as y");
buf.append(position);
buf.append('_');
return buf.toString();
}


public MultipleCountProjection setDistinct() {
distinct = true;
return this;
}


}

ExtraProjections.java

package org.hibernate.criterion;


public final class ExtraProjections
{
public static MultipleCountProjection countMultipleDistinct(String propertyNames) {
return new MultipleCountProjection(propertyNames).setDistinct();
}
}

Sample Usage:

String propertyNames = "titleName;titleDescr;titleVersion"


criteria countCriteria = ....


countCriteria.setProjection(ExtraProjections.countMultipleDistinct(propertyNames);

Referenced from https://forum.hibernate.org/viewtopic.php?t=964506

session = (Session) getEntityManager().getDelegate();
Criteria criteria = session.createCriteria(ComputedProdDaily.class);
ProjectionList projList = Projections.projectionList();
projList.add(Projections.property("user.id"), "userid");
projList.add(Projections.property("loanState"), "state");
criteria.setProjection(Projections.distinct(projList));
criteria.add(Restrictions.isNotNull("this.loanState"));
criteria.setResultTransformer(Transformers.aliasToBean(UserStateTransformer.class));

This helped me :D

if you want to use ORDER BY, just add:

criteria.setProjection(
Projections.distinct(
Projections.projectionList()
.add(Projections.id())
.add(Projections.property("the property that you want to ordered by"))
)
);