Hi,
Yeah, that sounds like something that you couldn't model directly. In
my initial reading, I hadn't noticed that you were creating an entity
for the join table.
Is it important that the B-A relation be unordered? If not, then you
could just model the A-B with ordering, and happen to get that
ordering also on the other side of the relation.
Also, is there any reason why you're modeling the BRef table as an
entity, vs. just using two many-to-many relationships?
-Patrick
On Jan 25, 2008 3:46 PM, Andy Schlaikjer <[EMAIL PROTECTED]> wrote:
> Hi Patrick,
>
> Thanks for your response. I did see in documentation the note on
> specifying the mapping of a bidirectional relation on only one side of
> the relation, but in my case the nature of the bidirectional
> relationship is a bit more constrained: the A to B relation is ordered,
> whereas the B to A relation is unordered. I'm not aware of any mechanism
> in JPA by which I can specify an "orderBy" column within a join table,
> so I'm at a loss here.
>
> Thanks,
> Andy
>
>
> Patrick Linskey wrote:
> > Hi,
> >
> > If you're trying to model a bidirectional relationship, you should
> > only specify mapping info for one side and use the mappedBy annotation
> > property on the other side of the relation.
> >
> > -Patrick
> >
> > On 1/25/08, Andy Schlaikjer <[EMAIL PROTECTED]> wrote:
> >> Hi all,
> >>
> >> I was curious if/how I might get into trouble by reusing a table to
> >> support mapping of an @Entity as well as a @JoinTable. Here's an example:
> >>
> >> I have the following entities A, BRef, and B where a unidirectional
> >> many-to-many relation between A and B is defined via BRef (acting here
> >> as an explicit join table):
> >>
> >> @Entity
> >> class A {
> >> @Id
> >> protected long id;
> >>
> >> @OneToMany(mappedBy = "a")
> >> @OrderBy("idx")
> >> protected List<BRef> brefs;
> >> }
> >>
> >> @Entity
> >> @Table(uniqueConstraints = @UniqueConstraint(columnNames = {"a_id",
> >> "b_id"}))
> >> class BRef {
> >> @Id
> >> protected long id;
> >>
> >> @ManyToOne(optional = false)
> >> protected A a;
> >>
> >> @ManyToOne(optional = false)
> >> protected B b;
> >>
> >> @Basic
> >> protected int idx;
> >> }
> >>
> >> @Entity
> >> class B {
> >> @Id
> >> protected long id;
> >> }
> >>
> >> With this mapping (using the mysql DBDictionary), the following CREATE
> >> TABLE statement reflects the BRef table:
> >>
> >> CREATE TABLE `BRef` (
> >> `id` bigint(20) NOT NULL,
> >> `idx` int(11) default NULL,
> >> `a_id` bigint(20) default NULL,
> >> `b_id` bigint(20) default NULL,
> >> PRIMARY KEY (`id`),
> >> UNIQUE KEY `a_id` (`a_id`,`b_id`),
> >> KEY `I_BREF_A` (`a_id`),
> >> KEY `I_BREF_B` (`b_id`)
> >> ) ENGINE=InnoDB DEFAULT CHARSET=utf8
> >>
> >> Now I'd like to extend the mapping to include the reverse many-to-many
> >> relation from B to A:
> >>
> >> @Entity
> >> class B {
> >> @Id
> >> protected long id;
> >>
> >> @OneToMany
> >> @JoinTable(
> >> name = "BRef",
> >> joinColumns = @JoinColumn(name = "b_id"),
> >> inverseJoinColumns = @JoinColumn(name = "a_id"),
> >> uniqueConstraints = @UniqueConstraint(columnNames = {"a_id", "b_id"})
> >> )
> >> protected Collection<A> as;
> >> }
> >>
> >> When I do this, the CREATE TABLE statement for table "BRef" changes
> >> slightly (ordering of columns, names of indices):
> >>
> >> CREATE TABLE `BRef` (
> >> `id` bigint(20) NOT NULL,
> >> `idx` int(11) default NULL,
> >> `b_id` bigint(20) default NULL,
> >> `a_id` bigint(20) default NULL,
> >> PRIMARY KEY (`id`),
> >> UNIQUE KEY `a_id` (`a_id`,`b_id`),
> >> KEY `I_BREF_B_ID` (`b_id`),
> >> KEY `I_BREF_ELEMENT` (`a_id`)
> >> ) ENGINE=InnoDB DEFAULT CHARSET=utf8
> >>
> >> So it is clear that the two separate mappings for table "BRef" are
> >> competing with one another. How else might the OpenJPA runtime be
> >> affected by this?
> >>
> >> Thanks,
> >> Andy
> >>
> >
> >
>
--
Patrick Linskey
202 669 5907
