Develop Reference

One to many Spring Result Mapping

This is my first time posting problem hope someone can help me on this mapping the result into:
List<Clients>
So here's the sample result from database:
NameID | Name | Notes | Date
1 | Client1 | Get this on monday. | null
1 | Client1 | Get this on wednesday. | null
2 | Client2 | Meet on saturday. | null
So here's my mode (java class).
Name.java
private int NameId;
private String ClientName;
.. getter and setter
Notes.java
private int NotesId;
private String ClientNote;
.. getter and setter
Clients.java
private Name name;
private List<Notes> notes;
.. getter and setter
ClientResultSet.java
class ClientResultSet implements ResultSetExtractor<List<Clients>>{
#Override
public List<Clients> extractData(ResultSet rs) throws SQLException, DataAccessException {
Map<int, Clients> map = new HashMap<int, Clients>();
while(rs.next()) {
int NameId= rs.getInt("NameId");
Clients clients= map.get(contactId);
if (clients== null) {
// I'm losing my thoughts here. :'(
}
}
return null;
}
}
Result I want to achieve:
[
{
name:{
NameId: 1,
Name: "Client1"
},
notes[
{
NotesId: 1,
ClientNote: "Get this on monday",
Date: null
},
{
NoteId: 2,
ClientNote: "Get this on wednesday",
Date: null
}
]
},
{
name:{},
notes:[{},{}]
}
... and so on
]
I'm reading ResultSetExtractor but I don't know how to implement it. Thanks in advance and have a good day.

This is how your database should look like, you will create 4 tables, client, name, note, client_note. the client_note table should have the client id and the note id for one to many relationship. so if you want to get all notes for a client with an id of 1; your sql will look like this
SELECT note_id FROM client_note WHERE clientId = 1;
then you can use the note id's you get to query the note table for the actual note objects;
client
int id;
int nameId (foreignKey referencing name.id);
name
int id;
String ClientName;
note
int id;
String ClientNote;
client_note
int id;
int clientId (foreignKey referencing client.id);
int noteId (foreignKey referencing note.id);
Hope this helps?

I think everyone misunderstood my question and I don't blame everyone. I love that my question gets notice and I thank everyone who gave their suggestion to me.
I found the answer here if anyone have the same problem as me:
multiple one-to-many relations ResultSetExtractor
Again, thank you everyone and have a nice day.

Spring Data JPA Multi-FK Join

The problem I have is with the querying of data from a chain of 3 tables via the JPA entity framework in a Spring Boot v1.5.3 application. The models are defined as follows:
ValidationField {
#Id id,
name,
#OneToMany(
orphanRemoval = true,
fetch = FetchType.EAGER,
cascade = CascadeType.ALL,
mappedBy = "validationMessage")
Set<ValidationFieldMessage> messages
}
ValidationFieldMessage {
#Id id,
#ManyToOne(fetch = FetchType.LAZY)
ValidationField validationField
#ManyToOne(fetch = FetchType.EAGER)
#JoinColumn(name = "validation_message_id")
ValidationMessage validationMessage;
}
ValidationMessage {
#Id id,
text
}
The Spring Data repository method is defined as follows:
List<ValidationField> findByName(String name);
Data example:
Validation_Field
id | name
0 first_name
Validation_Field_Message
id | validation_message_id | validation_field_id
0 0 0
1 1 0
2 2 0
Validation_Message
id | text
0 "There should be no spaces"
1 "No special characters are allowed"
The result from the execution however yields the following:
field {
id:0,
name:first_name,
messages: {
[
id:0,
validationMessage: [
id:1,
text: "No special characters are allowed"
], [
id:1,
text: "No special characters are allowed"
]
]
}
}
The messages are duplicated instead of individually listed.
I've tried with #Query using joins as well, but without success.
Is there something I'm missing in the model definition?

JPA double relation with the same Entity

I have these Entities:
#Entity
public class Content extends AbstractEntity
{
#NotNull
#OneToOne(optional = false)
#JoinColumn(name = "CURRENT_CONTENT_REVISION_ID")
private ContentRevision current;
#OneToMany(mappedBy = "content", cascade = CascadeType.ALL, orphanRemoval = true)
private List<ContentRevision> revisionList = new ArrayList<>();
}
#Entity
public class ContentRevision extends AbstractEntity
{
#NotNull
#ManyToOne(optional = false)
#JoinColumn(name = "CONTENT_ID")
private Content content;
#Column(name = "TEXT_DATA")
private String textData;
#Temporal(TIMESTAMP)
#Column(name = "REG_DATE")
private Date registrationDate;
}
and this is the db mapping:
CONTENT
+-----------------------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------------------+--------------+------+-----+---------+----------------+
| ID | bigint(20) | NO | PRI | NULL | auto_increment |
| CURRENT_CONTENT_REVISION_ID | bigint(20) | NO | MUL | NULL | |
+-----------------------------+--------------+------+-----+---------+----------------+
CONTENT_REVISION
+-----------------------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------------------+--------------+------+-----+---------+----------------+
| ID | bigint(20) | NO | PRI | NULL | auto_increment |
| REG_DATE | datetime | YES | | NULL | |
| TEXT_DATA | longtext | YES | | NULL | |
| CONTENT_ID | bigint(20) | NO | MUL | NULL | |
+-----------------------------+--------------+------+-----+---------+----------------+
I have also these requirements:
Content.current is always a member of Content.revisionList (think about Content.current as a "pointer").
Users can add a new ContentRevision to an existing Content
Users can add a new Content with an initial ContentRevision (cascade persist)
Users can change Content.current (move the "pointer")
Users can modify Content.current.textData, but saves Content (cascade merge)
Users can delete ContentRevision
Users can delete Content (cascade remove to ContentRevision)
Now, my questions are:
Is this the best approach? Any best practice?
Is it safe to cascade merge when the same entity is referenced twice? (Content.current is also Content.revisionList[i])
Are Content.current and Content.revisionList[i] the same instance? (Content.current == Content.revisionList[i] ?)
Thanks
#jabu.10245 I'm very grateful for your effort. Thank you, really.
However, there's a problematic (missing) case from your tests: when you run it inside a container using CMT:
#RunWith(Arquillian.class)
public class ArquillianTest
{
#PersistenceContext
private EntityManager em;
#Resource
private UserTransaction utx;
#Deployment
public static WebArchive createDeployment()
{
// Create deploy file
WebArchive war = ShrinkWrap.create(WebArchive.class, "test.war");
war.addPackages(...);
war.addAsResource("persistence-arquillian.xml", "META-INF/persistence.xml");
war.addAsManifestResource(EmptyAsset.INSTANCE, "beans.xml");
// Show the deploy structure
System.out.println(war.toString(true));
return war;
}
#Test
public void testDetached()
{
// find a document
Document doc = em.find(Document.class, 1L);
System.out.println("doc: " + doc); // Document#1342067286
// get first content
Content content = doc.getContentList().stream().findFirst().get();
System.out.println("content: " + content); // Content#511063871
// get current revision
ContentRevision currentRevision = content.getCurrentRevision();
System.out.println("currentRevision: " + currentRevision); // ContentRevision#1777954561
// get last revision
ContentRevision lastRevision = content.getRevisionList().stream().reduce((prev, curr) -> curr).get();
System.out.println("lastRevision: " + lastRevision); // ContentRevision#430639650
// test equality
boolean equals = Objects.equals(currentRevision, lastRevision);
System.out.println("1. equals? " + equals); // true
// test identity
boolean same = currentRevision == lastRevision;
System.out.println("1. same? " + same); // false!!!!!!!!!!
// since they are not the same, the rest makes little sense...
// make it dirty
currentRevision.setTextData("CHANGED " + System.currentTimeMillis());
// perform merge in CMT transaction
utx.begin();
doc = em.merge(doc);
utx.commit(); // --> ERROR!!!
// get first content
content = doc.getContentList().stream().findFirst().get();
// get current revision
currentRevision = content.getCurrentRevision();
System.out.println("currentRevision: " + currentRevision);
// get last revision
lastRevision = content.getRevisionList().stream().reduce((prev, curr) -> curr).get();
System.out.println("lastRevision: " + lastRevision);
// test equality
equals = Objects.equals(currentRevision, lastRevision);
System.out.println("2. equals? " + equals);
// test identity
same = currentRevision == lastRevision;
System.out.println("2. same? " + same);
}
}
since they are not the same:
if I enable cascading on both properties, an Exception is thrown
java.lang.IllegalStateException:
Multiple representations of the same entity [it.shape.edea2.jpa.ContentRevision#1] are being merged.
Detached: [ContentRevision#430639650];
Detached: [ContentRevision#1777954561]
if I disable cascade on current, the change get lost.
the strange thing is that running this test outside the container results in successful execution.
Maybe it's lazy loading (hibernate.enable_lazy_load_no_trans=true), maybe something else, but it's definitely NOT SAFE.
I wonder if there's a way to get the same instance.

Is it safe to cascade merge when the same entity is referenced twice?
Yes. If you manage an instance of Content, then it's Content.revisionList and Content.current are managed as well. Changes in any of those will be persisted when flushing the entity manager. You don't have to call EntityManager.merge(...) manually, unless you're dealing with transient objects that need to be merged.
If you create a new ContentRevision, then call persist(...) instead of merge(...) with that new instance and make sure it has a managed reference to the parent Content, also add it to the content's list.
Are Content.current and Content.revisionList[i] the same instance?
Yes, should be. Test it to be sure.
Content.current is always a member of Content.revisionList (think about Content.current as a "pointer").
You could do that check in in SQL with a check constraint; or in Java, although you'd have to be sure the revisionList is fetched. By default it's lazy fetched, meaning Hibernate will run another query for this list if you access the getRevisionList() method. And for that you need a running transaction, otherwise you'll be getting a LazyInitializationException.
You could instead load the list eagerly, if that's what you want. Or you could define a entity graph to be able to support both strategies in different queries.
Users can modify Content.current.textData, but saves Content (cascade merge)
See my first paragraph above, Hibernate should save changes on any managed entity automatically.
Users can delete ContentRevision
if (content.getRevisionList().remove(revision))
entityManager.remove(revision);
if (revision.equals(content.getCurrentRevision())
content.setCurrentRevision(/* to something else */);
Users can delete Content (cascade remove to ContentRevision)
Here I'd prefer to ensure that in the database schema, for instance
FOREIGN KEY (content_id) REFERENCES content (id) ON DELETE CASCADE;
UPDATE
As requested, I wrote a test. See this gist for the implementations of Content and ContentRevision I used.
I had to make one important change though: Content.current cannot really be #NotNull, especially not the DB field, because if it were, then we couldn't persist a content and revision at the same time, since both have no ID yet. Hence the field must be allowed to be NULL initially.
As a workaround I added the following method to Content:
#Transient // ignored in JPA
#AssertTrue // javax.validation
public boolean isCurrentRevisionInList() {
return current != null && getRevisionList().contains(current);
}
Here the validator ensures that the there is always a non-null current revision and that it is contained in the revision list.
Now here are my tests.
This one proves that the references are the same (Question 3) and that it is enough to persist content where current and revisionList[0] is referencing the same instance (question 2):
#Test #InSequence(0)
public void shouldCreateContentAndRevision() throws Exception {
// create java objects, unmanaged:
Content content = Content.create("My first test");
assertNotNull("content should have current revision", content.getCurrent());
assertSame("content should be same as revision's parent", content, content.getCurrent().getContent());
assertEquals("content should have 1 revision", 1, content.getRevisionList().size());
assertSame("the list should contain same reference", content.getCurrent(), content.getRevisionList().get(0));
// persist the content, along with the revision:
transaction.begin();
entityManager.joinTransaction();
entityManager.persist(content);
transaction.commit();
// verify:
assertEquals("content should have ID 1", Long.valueOf(1), content.getId());
assertEquals("content should have one revision", 1, content.getRevisionList().size());
assertNotNull("content should have current revision", content.getCurrent());
assertEquals("revision should have ID 1", Long.valueOf(1), content.getCurrent().getId());
assertSame("current revision should be same reference", content.getCurrent(), content.getRevisionList().get(0));
}
The next ensures that it's still true after loading the entity:
#Test #InSequence(1)
public void shouldLoadContentAndRevision() throws Exception {
Content content = entityManager.find(Content.class, Long.valueOf(1));
assertNotNull("should have found content #1", content);
// same checks as before:
assertNotNull("content should have current revision", content.getCurrent());
assertSame("content should be same as revision's parent", content, content.getCurrent().getContent());
assertEquals("content should have 1 revision", 1, content.getRevisionList().size());
assertSame("the list should contain same reference", content.getCurrent(), content.getRevisionList().get(0));
}
And even when updating it:
#Test #InSequence(2)
public void shouldAddAnotherRevision() throws Exception {
transaction.begin();
entityManager.joinTransaction();
Content content = entityManager.find(Content.class, Long.valueOf(1));
ContentRevision revision = content.addRevision("My second revision");
entityManager.persist(revision);
content.setCurrent(revision);
transaction.commit();
// re-load and validate:
content = entityManager.find(Content.class, Long.valueOf(1));
// same checks as before:
assertNotNull("content should have current revision", content.getCurrent());
assertSame("content should be same as revision's parent", content, content.getCurrent().getContent());
assertEquals("content should have 2 revisions", 2, content.getRevisionList().size());
assertSame("the list should contain same reference", content.getCurrent(), content.getRevisionList().get(1));
}
SELECT * FROM content;
id | version | current_content_revision_id
----+---------+-----------------------------
1 | 2 | 2
UPDATE 2
It was hard to reproduce that situation on my machine, but I got it to work. Here is what I've done so far:
I changed all #OneToMany relations to use lazy fetching (the default) and rerun the following test case:
#Test #InSequence(3)
public void shouldChangeCurrentRevision() throws Exception {
transaction.begin();
entityManager.joinTransaction();
Document document = entityManager.find(Document.class, Long.valueOf(1));
assertNotNull(document);
assertEquals(1, document.getContentList().size());
Content content = document.getContentList().get(0);
assertNotNull(content);
ContentRevision revision = content.getCurrent();
assertNotNull(revision);
assertEquals(2, content.getRevisionList().size());
assertSame(revision, content.getRevisionList().get(1));
revision.setTextData("CHANGED");
document = entityManager.merge(document);
content = document.getContentList().get(0);
revision = content.getCurrent();
assertSame(revision, content.getRevisionList().get(1));
assertEquals("CHANGED", revision.getTextData());
transaction.commit();
}
The test passed with lazy fetching. Note that lazy fetching requires it to be executed within a transaction.
For some reason the content revision instance you're editing is not the same as the one in the one-to-many list. To reproduce that I've modified my test as follows:
#Test #InSequence(4)
public void shouldChangeCurrentRevision2() throws Exception {
transaction.begin();
Document document = entityManager.find(Document.class, Long.valueOf(1));
assertNotNull(document);
assertEquals(1, document.getContentList().size());
Content content = document.getContentList().get(0);
assertNotNull(content);
ContentRevision revision = content.getCurrent();
assertNotNull(revision);
assertEquals(2, content.getRevisionList().size());
assertSame(revision, content.getRevisionList().get(1));
transaction.commit();
// load another instance, different from the one in the list:
revision = entityManager.find(ContentRevision.class, revision.getId());
revision.setTextData("CHANGED2");
// start another TX, replace the "current revision" but not the one
// in the list:
transaction.begin();
document.getContentList().get(0).setCurrent(revision);
document = entityManager.merge(document); // here's your error!!!
transaction.commit();
content = document.getContentList().get(0);
revision = content.getCurrent();
assertSame(revision, content.getRevisionList().get(1));
assertEquals("CHANGED2", revision.getTextData());
}
And there, I got exactly your error. Then I modified the cascading setting on the #OneToMany mapping:
#OneToMany(mappedBy = "content", cascade = { PERSIST, REFRESH, REMOVE }, orphanRemoval = true)
private List<ContentRevision> revisionList;
And the error disappeared :-) ... because I removed CascadeType.MERGE.

How to join on foreign key in querydsl

I am new to querydsl and I'm trying to use querydsl in pure java (no hibernate, JPA or anything).
We have a database where the tables are linked through minimum 3 columns
I followed the doc here and ended up with my schema duly created.
Here are my pseudo tables :
Item
Corporation (pk) mmcono
Item number (pk) mmitno
Environnement (pk) mmenv
Item description mmitds
Item_warehouse
Corporation (fk for Item) mbcono
Item number (fk for Item) mbitno
Environnement (fk for Item) mbenv
Warehouse number mbwhlo
Other properties (not important)
Inside the Item_wharehouse class, I manually added the foreignKey (because it's not defined in the actual db schema)
public final com.querydsl.sql.ForeignKey<QItemWharehouse > _ItemWharehouseFk = createInvForeignKey(Arrays.asList(mbitno, mbcono, mbenv), Arrays.asList("mmitno", "mmcono", "mbenv"));
I'm working on the following code in my main class:
SQLTemplates templates = SQLServer2012Templates.builder().printSchema().build();
Configuration configuration = new Configuration(templates);
QItem mm = new QItem ("mm");
QItemWarehouse mb = new QItemWarehouse("mb");
JtdsDataSource ds = getDataSource();
SQLQueryFactory queryFactory = new SQLQueryFactory(configuration, ds);
String toto = queryFactory.select(mm.mmitno, mm.mmitds)
.from(mm)
.join( ???????????? )
.where(mb.mbwhlo.eq("122"))
.fetch()
As per doc here I should be able to do something like this : AbstractSQLQuery.innerJoin(ForeignKey<E> key, RelationalPath<E> entity)
What I want in the end is to allow joining table without having to specify manually all the columns required for the join condition.
As stated before, my model starts with minimum 3 columns in the pk, and it's not uncommon to have 6 or 7 cols in the on clause! It's a lot of typing and very error prone, because you can easily miss one and get duplicate results.
I would like something like .join(mb._ItemWharehouseFk, ???) and let querydsl handle little details like generating the on clause for me.
My trouble is that I can't find the second parameter of type RelationalPath<E> entity for the join method.
I am doing something wrong ? What do I miss ? Is it even possible to accomplish what I want ?

Oups I found the problem : I had it all in the wrong order.
The foreign key was located in the itemWarehouse class.
it should have been named this way :
public final com.querydsl.sql.ForeignKey<QItem> _ItemFk = createInvForeignKey(Arrays.asList(mbitno, mbcono, mbenv), Arrays.asList("mmitno", "mmcono", "mbenv"));
that means that you just have to reverse the order in the statement this way :
SQLTemplates templates = SQLServer2012Templates.builder().printSchema().build();
Configuration configuration = new Configuration(templates);
QItem mm = new QItem ("mm");
QItemWarehouse mb = new QItemWarehouse("mb");
JtdsDataSource ds = getDataSource();
SQLQueryFactory queryFactory = new SQLQueryFactory(configuration, ds);
List<Tuple> toto = queryFactory.select(mm.mmitno, mm.mmitds)
.from(mb)
.join(mb._ItemFk, mm )
.where(mb.mbwhlo.eq("122"))
.fetch()
And you get your nice on clause generated. It's just a question of how you construct your relation.
#Enigma, I sincerely hope it will help you for your Friday afternoon. I wouldn't want your boss to be disappointed with you :-)

Hibernate not deleting/updating one to many

I am trying to learn how to work with hibernate, and until now i thought i was doing ok...
The problem is, i have a one to many relationship that i can't update/delete.
My DB is pretty basic, i have a ClientsBasic that has a one to many relationship with IndirectClients (which simply has a ClientsBasic ID and a URL, both keys because you can have for the same ID lots of URLs)
ClientBasic:
#OneToMany(fetch = FetchType.EAGER, mappedBy = "clientsBasic", cascade=CascadeType.ALL)
public List<IndirectClients> getIndirectClients() {
return this.indirectClients;
}
public void setIndirectClients(List<IndirectClients> indirectClients) {
// this.indirectClients = indirectClients;
this.indirectClients.clear();
this.indirectClients.addAll(indirectClients);
}
ClientDao:
public ClientsBasic save(ClientsBasic client) throws HibernateException {
Transaction tx = null;
tx = session.beginTransaction();
session.saveOrUpdate(client);
tx.commit();
log.info("Client saved with id: " + client.getClientId());
return client;
}
Now if i try to delete ClientsBasic, it will delete both ClientsBasic and all related indirectClients, so its working as expected, but if i simply try to update/delete and entry in indirectClients it doesn't work.
Example:
I create a new Client
ClientsBasic cb = new ClientsBasic("company_1", 1234, "company_1#email.com");
cbDao.save(cb);
And then a new Indirect Client
List<IndirectClients> indirectClientsSet= new ArrayList<IndirectClients>();
indirectClientsSet.add(new IndirectClients(new IndirectClientsId(cb.getClientId(), "www.url.test_1.com"), cb));
cb.setIndirectClients(indirectClientsSet);
cbDao.save(cb);
Now if i try to change the url like this
ClientsBasic cb = cbDao.findClientById(1);
List<IndirectClients> indC = cb.getIndirectClients();
indC.get(0).getId().setUrl("TEST");
cb.setIndirectClients(indC);
cbDao.save(cb);
no changes are made in the DB.
Can someone please help me?
Thank you.

If your IndirectClients is defined as an Entity it has its own life cycle, meaning you have to persist/delete instances separately from their ClientBasic parent.
If you want a scenario where all children are managed through their parent relation, consider using ElementCollection.
See also JPA: When to choose Multivalued Association vs. Element Collection Mapping