JPA Inheritance: Query fails with unknown column - jpa

Executing the following JPA query while using EclipseLink v2.7.4:
SELECT pr FROM AbstractProduct pr WHERE pr.shelve.superMarket.id = :1 ORDER BY pr.sortOrder
Gives the following error:
Unknown column 't0.SORTORDER' in 'order clause'
Error Code: 1054
Call: SELECT t2.ID, t2.SORTORDER, t2.SHELVE_ID FROM APPLE t2, SHELVE t1 WHERE ((t1.SUPERMARKET_ID = ?) AND (t1.ID = t2.SHELVE_ID)) ORDER BY t0.SORTORDER
bind => [12]
The query is refering to t0 but nowhere in the generated query does it define which table t0 is.
These are the entities that I'm using:
#Entity
public class SuperMarket {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long id;
#OneToMany(mappedBy = "superMarket")
List<Shelve> shelves;
}
#Entity
public class Shelve {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long id;
protected SuperMarket superMarket;
#OneToMany(mappedBy = "shelve")
protected List<AbstractProduct> products;
}
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class AbstractProduct {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long id;
protected Shelve shelve;
protected long sortOrder;
}
#Entity
public class Apple extends AbstractProduct {
}
#Entity
public class Banana extends AbstractProduct {
}
Which results in the following queries:
CREATE TABLE SUPERMARKET (ID BIGINT NOT NULL, PRIMARY KEY (ID));
CREATE TABLE SHELVE (ID BIGINT NOT NULL, SUPERMARKET_ID BIGINT, PRIMARY KEY (ID));
CREATE TABLE APPLE (ID BIGINT NOT NULL, SORTORDER BIGINT, SHELVE_ID BIGINT, PRIMARY KEY (ID));
CREATE TABLE BANANA (ID BIGINT NOT NULL, SORTORDER BIGINT, SHELVE_ID BIGINT, PRIMARY KEY (ID));
CREATE TABLE ABSTRACTPRODUCT (SHELVE_ID BIGINT);
The last table ABSTRACTPRODUCT should not be created since it is an Abstract Java Entity and I'm using the table per class inheritance style. This seems to be a bug in eclipselink it is also discussed in this question: Understanding of TABLE_PER_CLASS with keys in eclipselink It is the combination of the inheritance with the OneToMany relations that seems to trigger the create table statement. Not sure if this bug is related to the query error that I mentioned at the start. I think not as this table doesn't even have the sort order field.
When I remove the ORDER BY clause the query will execute succesfully. When I change the query to only go one level up it will also execute succesfully:
SELECT pr FROM AbstractProduct pr WHERE pr.shelve.id = :1 ORDER BY pr.sortOrder
For a test I got rid of the inheritance and let the Shelve entity have a OneToMany relation to Apple directly, where Apple did not extend any other class, in that case the query is also executed succesfully. But I need the abstract class and inheritance.
Any idea why the generated query is wrong in this case?
As is noted in an answer below I could use a different inheritance strategy to solve this problem in a different way. I choose the table per class type because that allows me to use the abstract entity in queries and the concrete classes get a table with all the fields in it. I was hoping that this helps with performance when doing a lot of inserts and selects from the concrete classes because that would only involve a single db table.
Update
I think this is an error in EclipseLink I have created two bugreports:
https://bugs.eclipse.org/bugs/show_bug.cgi?id=549866 for the abstract class creating a table
https://bugs.eclipse.org/bugs/show_bug.cgi?id=549868 for the error in the query

I was able to reproduce your problem with the inheritance strategy TABLE_PER_CLASS. The query executed as expected once I changed it to InheritanceType.JOINED and recreated the schema of the database (in my case: PostgreSQL 10.9).
So the code should be changed to:
#Entity
#Inheritance(strategy = InheritanceType.JOINED)
public abstract class AbstractProduct {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long id;
#ManyToOne
protected Shelve shelve;
protected long sortOrder;
}
Note well:
I added the missing #ManyToOne annotation to the attribute shelve.
#ManyToOne should also be added to protected SuperMarket superMarket; in the entity Shelve.
The resulting DB schema looks as follows:
CREATE TABLE public.abstractproduct
(
id bigint NOT NULL,
dtype character varying(31) COLLATE pg_catalog."default",
sortorder bigint,
shelve_id bigint,
CONSTRAINT abstractproduct_pkey PRIMARY KEY (id),
-- FK definitions left out for brevity
)
And Apple, for instance, becomes:
CREATE TABLE public.apple
(
id bigint NOT NULL,
CONSTRAINT apple_pkey PRIMARY KEY (id),
-- FK definitions left out for brevity
)
Hope it helps.

Related

Use postgres table sequence instead of sharing hibernate_sequence

When I do anything with a table, it always show the error:
Hibernate: select nextval ('hibernate_sequence')
2019-07-20 16:15:44.877 WARN 58376 --- [nio-9000-exec-1] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 0, SQLState: 42P01
2019-07-20 16:15:44.877 ERROR 58376 --- [nio-9000-exec-1] o.h.engine.jdbc.spi.SqlExceptionHelper : ERROR: relation "hibernate_sequence" does not exist
I DON'T want to use hibernate_sequence to share id sequence between tables, but want to define id seq for each table and use them respectively.
I use Spring Boot 2.1.6.RELEASE, Spring Data JPA (Hibernate 5.3.10.Final), and Postgres 11.2, and define the id field with BigSerial type and hope to use the id sequence of each table in respective entity class.
The demo repo is here: https://github.com/Redogame/share_hibernate_sequence
Create user table (use identity as table name because user is a Postgres reserved keyword).
By defining id with bigserial type, Postgres will create a identity_id_seq automatically, and I verified that identity_id_seq has been created successfully.
create table identity
(
id bigserial not null
constraint identity_pkey
primary key,
name varchar(255) not null
constraint identity_name_key
unique
constraint identity_name_check
check ((name)::text <> ''::text),
created_date timestamp not null,
created_by_id bigint not null
constraint identity_identity_id_fk
references identity,
last_modified_date timestamp not null,
last_modified_by_id bigint not null
constraint identity_identity_id_fk_2
references identity,
version bigint not null
);
Specify a sequence generator to use this id sequence:
#Table(name = "identity")
public class UserEntity extends Auditable<Long> {
#Id
#SequenceGenerator(name="identity_id_seq", sequenceName = "identity_id_seq", initialValue=1, allocationSize=1)
#GeneratedValue(strategy=GenerationType.SEQUENCE, generator="identity_id_seq")
private Long id;
But it doesn't work. I also tried to config spring.jpa.hibernate.use-new-id-generator-mappings and spring.jpa.properties.hibernate.id.new_generator_mappings, but still not work.
spring:
jpa:
hibernate:
use-new-id-generator-mappings: false
properties:
hibernate:
id:
new_generator_mappings: false
I expect not to use hibernate_sequence, that is: don't execute select nextval ('hibernate_sequence') before/after any SQL statement.
Try the below steps
CREATE SEQUENCE IF NOT EXISTS manual_seq;
Change create table script
create table identity
(
id integer NOT NULL DEFAULT nextval('manual_seq'::regclass),
name varchar(255) not null
constraint identity_name_key
unique
constraint identity_name_check
check ((name)::text <> ''::text),
created_date timestamp not null,
created_by_id bigint not null,
last_modified_date timestamp not null,
last_modified_by_id bigint not null,
version bigint not null,
CONSTRAINT manual_seq_pkey PRIMARY KEY (id)
);
I removed the foreign key constraint for testing purposes.
Update the entity mapping
#Entity
#Table(name = "identity")
#JsonIgnoreProperties(ignoreUnknown = true)
public class UserEntity extends Auditable<Long> {
#Id
#SequenceGenerator(name="manual-seq", sequenceName = "manual_seq",allocationSize = 1)
#GeneratedValue(generator="manual-seq")
private Long id;
#Basic
#Column(name = "name", nullable = false)
private String name;
#MappedSuperclass
#JsonIgnoreProperties({"new", "createdDate", "createdById", "lastModifiedDate", "lastModifiedById", "version"})
abstract class Auditable<PK extends Serializable>{
#NotAudited
#CreatedDate
#Temporal(TemporalType.TIMESTAMP)
private Date createdDate;
#NotAudited
#CreatedBy
private Long createdById;
#LastModifiedDate
#Temporal(TemporalType.TIMESTAMP)
private Date lastModifiedDate;
#LastModifiedBy
private Long lastModifiedById;
#NotAudited
#Version
private Long version;
Revert the spring.jpa.hibernate.use-new-id-generator-mappings
The issue was extending AbstractPersistable because of which database sequence was not getting used. Also, note I have removed the audit for testing purposes.
The same problem was happening to me. I explicitly set spring.jpa.properties.hibernate.id.new_generator_mappings=false but select nextval ('hibernate_sequence') was still run by Hibernate.
I found that when we use the #GeneratedValue annotation without setting a strategy, it defaults to AUTO, which means, Hibernate will try to generate the ID value using hibernate_sequence and then it wil fail because it doesn't exist in the database.
So, I made #GeneratedValue (strategy = GenerationType.IDENTITY) and tried again. In this case, the ID value was generated by my identity column in the database (the primary key that is automatically incremented) and not by hibernate_sequence.
create table users (id serial not null, name varchar(250), primary key (id));

How exactly work the #OneToMany JPA annotation in this example? Is it related to a table column or an entity class field?

I have 2 DB tables named respectivelly T_ACCOUNT and T_ACCOUNT_BENEFICIARY.
These tables have the following structure:
create table T_ACCOUNT (ID integer identity primary key, NUMBER varchar(9), NAME varchar(50) not null, CREDIT_CARD varchar(16), unique(NUMBER));
create table T_ACCOUNT_BENEFICIARY (ID integer identity primary key, ACCOUNT_ID integer, NAME varchar(50), ALLOCATION_PERCENTAGE decimal(5,2) not null, SAVINGS decimal(8,2) not null, unique(ACCOUNT_ID, NAME));
And the T_ACCOUNT table is bound to the T_ACCOUNT_BENEFICIARY table with a one to many relationship, this is the graphical representation:
So this is the first class named Account that map the T_ACCOUNT table:
#Entity
#Table(name="T_ACCOUNT")
public class Account {
#Id
#GeneratedValue
#Column(name="id")
private Long entityId;
#Column(name="NUMBER")
private String number;
#Column(name="NAME")
private String name;
#OneToMany
#JoinColumn(name="ACCOUNT_ID")
private Set<Beneficiary> beneficiaries = new HashSet<Beneficiary>();
#Column(name="CREDIT_CARD")
private String creditCardNumber;
// GETTERS & SETTERS
}
And this is the Beneficiary class that map the T_ACCOUNT_BENEFICIARY table:
/**
* A single beneficiary allocated to an account. Each beneficiary has a name (e.g. Annabelle) and a savings balance
* tracking how much money has been saved for he or she to date (e.g. $1000).
*/
#Entity
#Table(name="T_ACCOUNT_BENEFICIARY")
public class Beneficiary {
#Id
#GeneratedValue
#Column(name="ID")
private Long entityId;
#Column(name="NAME")
private String name;
#Embedded
#AttributeOverride(name="value",column=#Column(name="ALLOCATION_PERCENTAGE"))
private Percentage allocationPercentage;
#Embedded
#AttributeOverride(name="value",column=#Column(name="SAVINGS"))
private MonetaryAmount savings = MonetaryAmount.zero();
As you can see into the Account I have the beneficiaries field that implement the one to may relationship
#OneToMany
#JoinColumn(name="ACCOUNT_ID")
private Set<Beneficiary> beneficiaries = new HashSet<Beneficiary>();
I know that, on the DB, this relationship is implemented by the ACCOUNT_ID field of the T_ACCOUNT_BENEFICIARY table (so multiple row of the T_ACCOUNT_BENEFICIARY table can have the same value of the ACCOUNT_ID field and this means that a single row of the T_ACCOUNT table can be associated to more than one rows of T_ACCOUNT_BENEFICIARY table).
As you can see in the previous sippet there is the #JoinColumn(name="ACCOUNT_ID") annotation.
My doubt is generated by the fact that I have an ACCOUNT_ID column on my T_ACCOUNT_BENEFICIARY table, infact:
create table T_ACCOUNT_BENEFICIARY (ID integer identity primary key, ACCOUNT_ID integer, NAME varchar(50), ALLOCATION_PERCENTAGE decimal(5,2) not null, SAVINGS decimal(8,2) not null, unique(ACCOUNT_ID, NAME));
but this column seems to not be mapped on the Beneficiary that map this T_ACCOUNT_BENEFICIARY table.
So my doubts is: the #JoinColumn(name="ACCOUNT_ID") is working at relational level performing the join operation on the ACCOUNT_ID column of the table mapped by the Beneficiary entity (T_ACCOUNT_BENEFICIARY) or am I missing something? How exactly is performed this join?
If my interpretation is right can I work at entity level and say to join the beneficiaries field of my Account entity class to a new accountId field inserted into my Beneficiary entity class and mapping the ACCOUNT_ID column of the T_ACCOUNT_BENEFICIARY table?
Tnx
It seems is a Unidirectional OneToMany relationship
In JPA 2.0 a #JoinColumn can be used on a OneToMany to define the foreign key
I'm not sure if I understand your question. But what you have done with your #JoinColumn annotation is correct and Hibernate will execute appropriate SQL statements when you have multiple beneficiaries for your account. For example executing multiple INSERTS if you have two Beneficiaries for an Account. And yes using the #JoinColumn annotation is at the hibernate level. If you want to access an Account from a Beneficiary entity you would need to define a Bidirectional relationship in the Beneficiary class like below.
#Entity
#Table("T_ACCOUNT_BENEFICIARY")
public class Beneficiary {
#ManyToOne(mappedBy = "beneficiaries")
Account account;
...
}

JPQL Query working in testing, not in production

I have two Entities related by a ManyToMany and I want to select them via a named Query. This works in my test (with a H2 DB set up) and throws exceptions at runtime (with postgresql set up). Other than the H2 and PG I am hard pressed to find differences between test and production.
The Entities and the Query look like so (abbreviated):
#Entity(name = "Enrichment")
#Table(name = "mh_Enrichment")
NamedQueries({
#NamedQuery(name = "findByLink",
query = "SELECT e FROM Enrichment e INNER JOIN e.links l WHERE l.link in (:links)") })
public class EnrichmentImpl {
#Id
#Column(name = "enrichmentId")
#GeneratedValue(strategy = GenerationType.AUTO)
private long id;
#ManyToMany
#JoinTable(name = "mh_EnrichmentLinks", joinColumns = { #JoinColumn(name = "EnrichmentId",
referencedColumnName = "enrichmentId") }, inverseJoinColumns = { #JoinColumn(name = "Link",
referencedColumnName = "link") })
private List<Link> links;
}
#Entity(name = "Link")
#Table(name = "mh_enrichment_link")
public class LinksImpl {
#Id
#Column(name = "link", length = 1024)
private String link;
}
Upon running the query with a String value in production I get:
Internal Exception: org.postgresql.util.PSQLException: ERROR: operator does not exist: character varying = bigint
Hinweis: No operator matches the given name and argument type(s). You might need to add explicit type casts.
Position: 215
Error Code: 0
Call: SELECT t1.enrichmentId FROM mh_enrichment_link t0, mh_EnrichmentLinks t2, mh_Enrichment t1 WHERE ((t0.link IN (?)) AND ((t2.EnrichmentId = t1.enrichmentId) AND (t0.link = t2.Link)))
Any ideas what's wrong? It is the query, isn't it?
The query is supposed to retrieve a list of Enrichments that are related to the given link.
Update #1
As requested: the tables in the DB look as follows:
For entity Link
CREATE TABLE mh_enrichment_link
(
link character varying(1024) NOT NULL,
CONSTRAINT mh_enrichment_link_pkey PRIMARY KEY (link)
)
For entity Enrichment
CREATE TABLE mh_enrichment
(
enrichmentid bigint NOT NULL,
CONSTRAINT mh_enrichment_pkey PRIMARY KEY (enrichmentid)
)
For the relation (See answer, this was where it went wrong)
CREATE TABLE mh_enrichmentlinks
(
link character varying(1024) NOT NULL,
CONSTRAINT mh_enrichment_link_pkey PRIMARY KEY (link)
)
The issue was fixed by dropping all related tables and having JPA regenerate them. Table definitions didn't match Entity definitions.
Thats also the quite obviously the reason why the test worked and the production didn't. In testing the tables are generated on runtime, in production they existed already (with an outdated definition).
Side note: The query is correct and does what it should.

JPA Error : The entity has no primary key attribute defined

I am using JPA in my application. In one of the table, I have not used primary key (I know its a bad design).
Now the generated entity is as mentioned below :
#Entity
#Table(name="INTI_SCHEME_TOKEN")
public class IntiSchemeToken implements Serializable {
private static final long serialVersionUID = 1L;
#Column(name="CREATED_BY")
private String createdBy;
#Temporal( TemporalType.DATE)
#Column(name="CREATED_ON")
private Date createdOn;
#Column(name="SCH_ID")
private BigDecimal schId;
#Column(name="TOKEN_ID")
private BigDecimal tokenId;
public IntiSchemeToken() {
}
public String getCreatedBy() {
return this.createdBy;
}
public void setCreatedBy(String createdBy) {
this.createdBy = createdBy;
}
public Date getCreatedOn() {
return this.createdOn;
}
public void setCreatedOn(Date createdOn) {
this.createdOn = createdOn;
}
public BigDecimal getSchId() {
return this.schId;
}
public void setSchId(BigDecimal schId) {
this.schId = schId;
}
public BigDecimal getTokenId() {
return this.tokenId;
}
public void setTokenId(BigDecimal tokenId) {
this.tokenId = tokenId;
}
}
Here In my project, eclipse IDE shows ERROR mark(RED colored cross) on this class and the error is "The entity has no primary key attribute defined".
Can anyone tell me, How to create an entity without primary key ?
Thanks.
You can't. An entity MUST have a unique, immutable ID. It doesn't have to be defined as a primary key in the database, but the field or set of fields must uniquely identify the row, and its value may not change.
So, if one field in your entity, or one set of fields in your entity, satisfies these criteria, make it (or them) the ID of the entity. For example, if there is no way that a user can create two instances in the same day, you could make [createdOn, createdBy] the ID of the entity.
Of course this is a bad solution, and you should really change your schema and add an autogenerated, single-column ID in the entity.
If your Primary Key(PK) is a managed super class which is inherited in an entity class then you will have to include the mapped super class name in the persistence.xml file.
Look at the bug report:
https://bugs.eclipse.org/bugs/show_bug.cgi?id=361042
If you need to define a class without primary key, then you should mark that class as an Embeddable class. Otherwise you should give the primary key for all entities you are defining.
You can turn off (change) validation that was added.
Go to workspace preferences 'Java Persistence->JPA->Errors/Warnings' next 'Type' and change 'Entity has no primary key' to 'Warnning'.
In addition to http://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing#No_Primary_Key you can use some build-in columns like ROWID in Oracle:
Oracle legacy table without good PK: How to Hibernate?
but with care:
http://www.orafaq.com/wiki/ROWID
Entity frameworks doesn't work for all kind of data (like statistical data which was used for analysis not for querying).
Another solution without Hibernate
If
- you don't have PK on the table
- there is a logical combination of columns that could be PK (not necessary if you can use some kind of rowid)
-- but some of the columns are NULLable so you really can't create PK because of DB limitation
- and you can't modify the table structure (would break insert/select statements with no explicitly listed columns at legacy code)
then you can try the following trick
- create view at database with virtual column that has value of concatenated logical key columns ('A='||a||'B='||'C='c..) or rowid
- create your JPA entity class by this view
- mark the virtual column with #Id annotation
That's it. Update/delete data operations are also possible (not insert) but I wouldn't use them if the virtual key column is not made of rowid (to avoid full scan searches by the DB table)
P.S. The same idea is partly described at the linked question.
You need to create primary key ,If not found any eligible field then create auto increment Id.
CREATE TABLE fin_home_loan (
ID int NOT NULL AUTO_INCREMENT,
PRIMARY KEY (ID));
Just add fake id field.
In Postgres:
#Id
#Column(name="ctid")
String id;
In Oracle:
#Id
#Column(name="ROWID")
String rowid;

Java EE 6 JPA 2 ManyToOne Relation Creates Invalid Foreign Key

I am trying to create two entities where both entities have embeddedIds. One of the entities have 2 references to the other entity, where both of those references are related as ManyToOne.
Example codes are written below;
#Embeddable
public class ItemPK {
#Column(nullable = false, length = 100)
private String itemId;
#Column(name = "item_client_id", nullable = false)
private int clientId;
...
}
#Entity
#Table(name = "item")
public class Item {
#EmbeddedId
private ItemPK id;
#ManyToOne
#JoinColumn(name = "item_client_id")
private Client client;
#OneToMany(mappedBy="item", cascade = CascadeType.ALL, orphanRemoval = true)
private Set<RelatedItem> relatedItems;
#OneToMany(mappedBy="relatedItem", cascade = CascadeType.ALL, orphanRemoval = true)
private Set<RelatedItem> relatedItemsRHS;
...
}
#Embeddable
public class RelatedItemPK {
#Column(name = "itemId", length = 100, nullable = false)
private String itemId;
#Column(name = "item_client_id", nullable = false)
private int clientId;
#Column(name = "relatedItemId", length = 100, nullable = false)
private String relatedItemId;
#Column(name = "related_item_client_id", nullable = false)
private int relatedItemClientId;
...
}
#Entity
#Table(name = "related_item")
public class RelatedItem {
#EmbeddedId
private RelatedItemPK id;
#ManyToOne(cascade = CascadeType.ALL, optional = false)
#JoinColumns({
#JoinColumn(name="itemId", referencedColumnName="itemId", insertable=false, updatable=false),
#JoinColumn(name="item_client_id", referencedColumnName="item_client_id", insertable=false, updatable=false)
})
private Item item;
#ManyToOne(cascade = CascadeType.ALL, optional = false)
#JoinColumns({
#JoinColumn(name="related_item_client_id", referencedColumnName="item_client_id", insertable=false, updatable=false),
#JoinColumn(name="relatedItemId", referencedColumnName="itemId", insertable=false, updatable=false)
})
private Item relatedItem;
...
}
The problem is while creating foreign keys for RelatedItem entity, I got an SQLException. It is the second ManyToOne relation that fails. The foreign key generation sql is below,
ALTER TABLE related_item ADD CONSTRAINT FK_related_item_related_item_client_id FOREIGN KEY (related_item_client_id, relatedItemId) REFERENCES item (item_client_id, itemId)
Since item table is indexed first by itemId then by item_client_id, this statement causes MySQL to produce an error.
I would like to switch the places of columns so that the SQL should look like the following,
ALTER TABLE related_item ADD CONSTRAINT FK_related_item_relatedItemId FOREIGN KEY (relatedItemId, related_item_client_id) REFERENCES item (itemId,item_client_id)
I tried changing the order of "JoinColumn"s but the result didn't change. I also tried renaming the fields to check if persistence provider choses the order by column name but again the result didn't change.
So, is there a way to enforce the column ordering?
p.s. I use following stuff:
MySQL 5.1
EclipseLink 2.0.0
Java EE 6
JPA 2
GlassFish v3
Edit: EclipseLink produces following SQL, which fails to run;
CREATE TABLE related_item (SIMILARITY DOUBLE, widget_id INTEGER NOT NULL, relatedItemId VARCHAR(100) NOT NULL, itemId VARCHAR(100) NOT NULL, related_item_client_id INTEGER NOT NULL, item_client_id INTEGER NOT NULL, PRIMARY KEY (widget_id, relatedItemId, itemId, related_item_client_id, item_client_id));
CREATE TABLE item (IMAGEURL VARCHAR(2048), STATUS VARCHAR(64), URL VARCHAR(2048), PRICE DOUBLE, STOCK INTEGER, DESCRIPTION TEXT(64000), NAME VARCHAR(255), ITEMID VARCHAR(100) NOT NULL, item_client_id INTEGER NOT NULL, PRIMARY KEY (ITEMID, item_client_id));
ALTER TABLE related_item ADD CONSTRAINT FK_related_item_itemId FOREIGN KEY (itemId, item_client_id) REFERENCES item (itemId, item_client_id);
ALTER TABLE related_item ADD CONSTRAINT FK_related_item_related_item_client_id FOREIGN KEY (related_item_client_id, relatedItemId) REFERENCES item (item_client_id, itemId);
ALTER TABLE item ADD CONSTRAINT FK_item_item_client_id FOREIGN KEY (item_client_id) REFERENCES client (ID);
Please include the stack trace. However, I strongly recommend you skip the #JoinColumn tags unless you have a VERY good reason for specifying the foreign keys yourself. By specifying the mappedBy attribute in one of the directions, JPA can figure out what to do by itself.
Java EE 6 and JPA put a lot of effort into enabling Convention over Configuration, which means that most of the time, things will work out of the box. It's desirable for you, the programmer because you have less boiler plate code to worry about, and it's desirable for the JPA and Jave EE container implementors because it gives them freedom to chose the best performing solutions. By declaring the foreign key relationships yourself, you rob both you and JPA of this advantage.
Edit: In fact, I suspect that both specifying mappedBy and specifying the #JoinTable could be the root cause of your problem. But I need to see the stack trace to tell for sure.
The order of the columns should not matter. If it does, then you could change the order in your index to match, or change the order you list your primary key in, or just use your scripts to generate your DDL.