Develop Reference

liferay 7.2 service-builder error.NoClassDefFoundError: com/liferay/portal/kernel/dao/orm/ArgumentsResolver when deploying the service

I am working on a portal project using Liferay 7.2
and I have created a service by using a service-builder template, I followed the tutorial step by step.
but I am getting the following error when I deploy the service.
2021-07-28 20:24:59.092 ERROR [pipe-start 2245][RatingPersistenceImpl:93] bundle com.arabiait.mawareeth.service:1.0.0 (2245)[com.arabiait.mawareeth.service.persistence.impl.RatingPersistenceImpl(5523)] : The activate method has thrown an exception
java.lang.NoClassDefFoundError: com/liferay/portal/kernel/dao/orm/ArgumentsResolver
at com.arabiait.mawareeth.service.persistence.impl.RatingPersistenceImpl.activate(RatingPersistenceImpl.java:1997)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
=======================================================================
there is a nested static class that implements this ArgumentResolver interface already generated in the RatingPersistenceImpl.java as follows:
private static class RatingModelArgumentsResolver
implements ArgumentsResolver {
}
==========================================================================
the line that throws the exception is inside the RatingPersistenceImpl.java as follows:
_argumentsResolverServiceRegistration = _bundleContext.registerService(
ArgumentsResolver.class, new RatingModelArgumentsResolver(),
MapUtil.singletonDictionary(
"model.class.name", Rating.class.getName()));
service.xml
<?xml version="1.0"?>
<!DOCTYPE service-builder PUBLIC "-//Liferay//DTD Service Builder 7.3.0//EN" "http://www.liferay.com/dtd/liferay-service-builder_7_3_0.dtd">
<service-builder dependency-injector="ds" package-path="com.arabiait.mawareeth">
<namespace>MW</namespace>
<!--<entity data-source="sampleDataSource" local-service="true" name="Foo" remote-service="false" session-factory="sampleSessionFactory" table="foo" tx-manager="sampleTransactionManager uuid="true"">-->
<entity local-service="true" name="Rating" remote-service="true" uuid="true">
<!-- PK fields -->
<column name="ratingId" primary="true" type="long" />
<!-- Group instance -->
<column name="groupId" type="long" />
<!-- Audit fields -->
<column name="companyId" type="long" />
<column name="userId" type="long" />
<column name="userName" type="String" />
<column name="createDate" type="Date" />
<column name="modifiedDate" type="Date" />
<!-- Other fields -->
<!-- Order -->
<column name="name" type="String"></column>
<column name="comment" type="String"></column>
<column name="rate" type="int"></column>
<column name="pageName" type="String"></column>
<!-- References -->
<reference entity="AssetEntry" package-path="com.liferay.portlet.asset" />
<reference entity="AssetTag" package-path="com.liferay.portlet.asset" />
</entity>
</service-builder>
If someone can help me, I will be so grateful as I am overwhelmed by this issue for 3 days till now.

Hibernate Mapping - Join of two table to one flat class

I have two tables: patient_data and patient_diagnosis
Patient_data contains personal data of patient like: pid (pkey), gender, birth_date
patient_diagnosis contains the diagnosis data of the registered patients. It has fields like: record_id (pkey), pid (fkey to patient_data(pid)), diagnosis_date and other related fields.
Now, I want to join these two tables on pid and have all these fields in a single type of object.
Here is the mapping file:
<hibernate-mapping>
<class catalog="emr" name="in.Models.Emr" table="patient_diagnosis">
<id name="recordid" type="long">
<column name="record_id"/>
</id>
<property name="diagnosisDate" type="timestamp">
<column length="19" name="diagnosis_date" not-null="true"/>
</property>
<property name="snomedTermPrimary" type="long">
<column name="snomed_term_primary" not-null="true" />
</property>
<property name="snomedTermSecondary" type="string">
<column name="snomed_term_secondary" />
</property>
<property name="episodeNo" type="long">
<column name="episode_no" not-null="true" />
</property>
<property name="pid" type="long">
<column name="pid" not-null="true" />
</property>
<join table="patient_data">
<key column="pid"/>
<property name="gender" type="string">
<column name="gender" not-null="true"/>
</property>
<property name="birthDate" type="timestamp">
<column length="19" name="birth_date" not-null="true"/>
</property>
</join>
</class>
</hibernate-mapping>
But, the join applies on patient_diagnosis.record_id = patient_data.pid instead of patient_diagnosis.pid = patient_data.pid i.e. HQL applies on primary key of first table with mentioned column from second table.
Please provide the solution so that join can be applied on mentioned column from first with mentioned column from second table. Or is there another way out?
Please note that in case I didn't create classes for patient_data or patient_diagnosis. But, just Emr class having combination of fields of these tables is created.

Try giving foreign key
<id name="pid" type="java.lang.Long">
<column name="pid" />
<generator class="foreign">
<param name="property">patient_data</param>
</generator>
</id>
I am not sure but maybe this should work.
And
<one-to-one name="patient_data" class="in.Models.Emr"
cascade="save-update"></one-to-one>
Similarly in Join class
<one-to-one name="patient_diagnosis" class="in.Models.Emr"
cascade="save-update"></one-to-one>
I hope this helps you.

Hibernate - load children

I have the following Hibernate Mappings:
<class name="Database.Content" table="..." schema="" catalog="...">
<id name="id">
<column name="id" sql-type="int" not-null="true"/>
</id>
<property name="week">
<column name="week" sql-type="int"/>
</property>
<property name="day">
<column name="day" sql-type="int"/>
</property>
<property name="hour">
<column name="hour" sql-type="int"/>
</property>
<property name="type">
<column name="type" sql-type="int" not-null="true"/>
</property>
<many-to-one name="group" class="Database.Group">
<column name="group"/>
</many-to-one>
<many-to-one name="table" class="Database.Table">
<column name="table" not-null="true"/>
</many-to-one>
<list name="entries" inverse="true" table="...">
<key>
<column name="content" not-null="true"/>
</key>
<list-index column="id"/>
<one-to-many not-found="ignore" class="Database.Entry"/>
</list>
</class>
<class name="Database.Entry" table="..." schema="" catalog="...">
<id name="id">
<column name="id" sql-type="int" not-null="true"/>
</id>
<property name="teacher">
<column name="teacher" sql-type="int" not-null="true"/>
</property>
<property name="course">
<column name="course" sql-type="int" not-null="true"/>
</property>
<property name="room">
<column name="room" sql-type="int" not-null="true"/>
</property>
<property name="p">
<column name="p" sql-type="int" not-null="true"/>
</property>
<many-to-one name="content" class="Database.Content" fetch="join" lazy="false">
<column name="content" not-null="true"/>
</many-to-one>
</class>
Now I am trying to query all contents with the corresponding entries:
List<Content> contents = session.createQuery("from Content c WHERE c.day IN :days ").setParameterList("days", days).list();
The query returns the correct response. However, when I do the following:
contents.get(0).getEntries()
there is a bunch of null values. What is the correct way to eager load all corresponding entries for each content?
I have about 20,000 content records, and most of the records have only one entry.
If I set lazy="false" for the list of entries, I get Java heap space error.
I ended up fetching entries and joining contents:
List<Entry> entries = session.createQuery("select e from Entry e Join e.content c WHERE c.day IN :days ").setParameterList("days", days).list();
I also changed lazy to proxy in:
<many-to-one name="content" class="Database.Content" fetch="join" lazy="proxy">
<column name="content" not-null="true"/>
</many-to-one>

Add lazy=false attribute:
<list name="entries" inverse="true" table="up_timetable_entries" lazy="false">
Hope helped you!

Try calling:
contents.get(0).getEntries().size();
To force hibernate to load the children.

Making join with Criteria

I'm terribly new to Hibernate. I've googled for two hours but I still can't figure out, how to make JOIN without using HQL, only by criteria. I have tables Clients(cID, name) and Visits(vID, vcID, date). The relation is one to many (one client can visit multiple times). I would also like to do it without setFetchMode. Just Criteria. Do I have to change the mappping xml?
UPDATE:
this is part of my mapping xml:
<class name="Client" table="Clients">
<id name="cID" column="cID"><generator class="native"/></id>
<property name="name" length="10" not-null="true"/>
</class>
<class name="Visit" table="Visits">
<id name="vID" column="vID"><generator class="native"/></id>
<property name="vcID" length="10" not-null="true"/>
<property name="date" length="25" not-null="true"/>
</class>

Having a class Client with a list-attribute "visits" that's mapping to your Visit-Entity:
Criteria criteria = session.createCriteria(Client.class);
criteria.addCriteria("visits");
This would create an inner join between your client-table and your visits-table.
Update:
Here you'll find some good examples: http://docs.jboss.org/hibernate/orm/3.3/reference/en/html/querycriteria.html#querycriteria-associations
Mapping Example
I hardly ever use hibernate mapping xml, however it should read similiar to:
<class name="Client" table="Clients">
<id name="cID" column="cID"><generator class="native"/></id>
<property name="name" length="10" not-null="true"/>
<bag name="visits">
<key column="vcId"/>
<one-to-many class="Visit"/>
</bag>
</class>
Tell Hibernate that there is a property "visits" which represents a one-to-many relationship.

You need to update you mapping:
<class name="Client" table="Clients">
<id name="cID" column="cID"><generator class="native"/></id>
<property name="name" length="10" not-null="true"/>
<!-- Declare Set<Visit> visits in the Client class-->
<set name="visits" lazy="false" cascade="all">
<key column="vcID"/>
<one-to-many class="your.package.Visit"/>
</set>
</class>
<class name="Visit" table="Visits">
<id name="vID" column="vID"><generator class="native"/></id>
<!-- and add "Client client" property to your Visit class -->
<many-to-one name="client" column="vcID" lazy="false"/>
<property name="date" length="25" not-null="true"/>
</class>
Then:
Criteria criteria = session.createCriteria(Visit.class).addCriteria("client")
.add(Restriction.eq(...));
or
Criteria criteria = session.createCriteria(Client.class).addCriteria("visits")
.add(Restriction.eq(...));
And Hibernate will join them automatically.

Hibernate without primary keys generated by db?

I'm building a data warehouse and want to use InfiniDB as the storage engine. However, it doesn't allow primary keys or foreign key constraints (or any constraints for that matter).
Hibernate complains "The database returned no natively generated identity value" when I perform an insert.
Each table is relational, and contains a unique integer column that was previously used as the primary key - I want to keep that, but just not have the constraint in the db that the column is the primary key.
I'm assuming the problem is that Hibernate expects the db to return a generated key. Is it possible to override this behaviour so I can set the primary key field's value myself and keep Hibernate happy?
-- edit --
Two of the mappings are as follows:
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated Jun 1, 2010 2:49:51 PM by Hibernate Tools 3.2.1.GA -->
<hibernate-mapping>
<class name="com.example.project.Visitor" table="visitor" catalog="orwell">
<id name="id" type="java.lang.Long">
<column name="id" />
<generator class="identity" />
</id>
<property name="firstSeen" type="timestamp">
<column name="first_seen" length="19" />
</property>
<property name="lastSeen" type="timestamp">
<column name="last_seen" length="19" />
</property>
<property name="sessionId" type="string">
<column name="session_id" length="26" unique="true" />
</property>
<property name="userId" type="java.lang.Long">
<column name="user_id" />
</property>
<set name="visits" inverse="true">
<key>
<column name="visitor_id" />
</key>
<one-to-many class="com.example.project.Visit" />
</set>
</class>
</hibernate-mapping>
and:
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated Jun 1, 2010 2:49:51 PM by Hibernate Tools 3.2.1.GA -->
<hibernate-mapping>
<class name="com.example.project.Visit" table="visit" catalog="orwell">
<id name="id" type="java.lang.Long">
<column name="id" />
<generator class="identity" />
</id>
<many-to-one name="visitor" class="com.example.project.Visitor" fetch="join" cascade="all">
<column name="visitor_id" />
</many-to-one>
<property name="visitId" type="string">
<column name="visit_id" length="20" unique="true" />
</property>
<property name="startTime" type="timestamp">
<column name="start_time" length="19" />
</property>
<property name="endTime" type="timestamp">
<column name="end_time" length="19" />
</property>
<property name="userAgent" type="string">
<column name="user_agent" length="65535" />
</property>
<set name="pageViews" inverse="true">
<key>
<column name="visit_id" />
</key>
<one-to-many class="com.example.project.PageView" />
</set>
</class>
</hibernate-mapping>

As you noted in a comment, there're many id generators you can use. E.g., you many find 'increment' convenient. Complete overview
http://docs.jboss.org/hibernate/core/3.3/reference/en/html/mapping.html#mapping-declaration-id
And hibernate won't care about your db restrictions (both primary keys and foreign keys). In fact, hibernate has no way of knowing about certain db restriction until it's violated. And if restriction doesn't exist, it can never be violated :)

You can remove the <generator class="identity" /> element and set the id manually before saving the object.