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Get object propertiy from other table

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jas_esp

Post subject: Get object propertiy from other table

Posted: Thu Jun 22, 2006 10:38 am

Newbie

Joined: Thu Jun 22, 2006 7:19 am
Posts: 14

I have the followin db schema:

Table Person:
- person_id: int
- name: string
- ...

Table fingerprint
- fingerprint_id: int
- person_id: int
- fingerprint_data: string

-----------------------------------------------------------------------------
And these are the class definitions:

public class Person {

private int person_id;
private String name;
...
private Fingerprint fingerprint;

}

public class Fingerprint {

private int fingerprint_id;
private Person person;
private String fingerprintData;

}

-----------------------------------------------------------------------------
So a person can have 0 or 1 fingerprint, and one fingerprint belongs to only one person. How dshould be declared this situation in the XML mapping file?

Thanks.

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mojarrell

Post subject:

Posted: Thu Jun 22, 2006 6:44 pm

Beginner

Joined: Mon Jul 26, 2004 4:29 pm
Posts: 45
Location: TX, USA

Hi,

I set up your tables in HsqlDb using the following SQL:

Code:

-- drop tables (child table first)
drop table fingerprint;
drop table person;

-- person table
create table person (
   person_id identity primary key
  ,name varchar(30) );
  
-- fingerprint table

create table fingerprint (
   fingerprint_id identity primary key
  ,person_id int
  ,fingerprint_data varchar(50)
  ,foreign key (person_id) references person ( person_id ));
  

insert into person values (null, 'John Doe');
*person_id ~
call identity();

insert into fingerprint values (null, *{person_id}, 'Here is John''s fingerprint data.');

insert into person values (null, 'Sally Smith');
*person_id ~
call identity();

insert into fingerprint values (null, *{person_id}, 'Here is Sally''s fingerprint data.');

-- now a person without a fingerprint
insert into person values (null, 'Hank Williams');

Then I set up the following classes based on yours:

Code:

package hg;

public class Person {
    private int personId;
    private String name;
    private Fingerprint fingerprint;
    
    // getters and setters below
}

package hg;

Code:

public class Fingerprint {
    private int fingerprintId;
    private int personId;
    private String fingerprintData;
    private Person person;

    // getters and setters below
}

Here is the xml mapping:

Code:

   <class name="Person" table="PERSON">
        <id name="personId" column="PERSON_ID">
            <generator class="native"/>
        </id>
        <one-to-one name="fingerprint" class="Fingerprint" />
        <property name="name" />
    </class>

    <class name="Fingerprint" table="FINGERPRINT">
        <id name="fingerprintId" column="FINGERPRINT_ID">
            <generator class="native"/>
        </id>
        <one-to-one name="person"  class="Person" />
        <property name="fingerprintData" column="FINGERPRINT_DATA" />
    </class>

I then wrote some code to retrieve objects from either side of the association and this is the relavent output:

INFO Main - Let's list Persons then Fingerprints
INFO Main - Person is: hg.Person@12884e0[id=0,name=John Doe]
INFO Main - Associated fingerprint is: hg.Fingerprint@1b5340c[id=0,data=Here is John's fingerprint data.]
INFO Main - Person is: hg.Person@1ce3fc5[id=1,name=Sally Smith]
INFO Main - Associated fingerprint is: hg.Fingerprint@b307f0[id=1,data=Here is Sally's fingerprint data.]
INFO Main - Person is: hg.Person@cf66b[id=2,name=Hank Williams]
INFO Main - Associated fingerprint is: null
INFO Main - Let's list Fingerprints then Persons
INFO Main - Fingerprint is: hg.Fingerprint@814013[id=0,data=Here is John's fingerprint data.]
INFO Main - Associated person is: hg.Person@1884174[id=0,name=John Doe]
INFO Main - Fingerprint is: hg.Fingerprint@1d2b01b[id=1,data=Here is Sally's fingerprint data.]
INFO Main - Associated person is: hg.Person@195d4fe[id=1,name=Sally Smith]

Notice that the person without a fingerprint (Hank Williams) comes back with a null for the associated fingerprint object.

Hope this helps,

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jas_esp

Post subject:

Posted: Fri Jun 23, 2006 2:48 am

Newbie

Joined: Thu Jun 22, 2006 7:19 am
Posts: 14

Thank you very much! That is exactly what I need :)

¡Saludos desde España! / Greetings from Spain!

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jas_esp

Post subject:

Posted: Fri Jun 23, 2006 4:27 am

Newbie

Joined: Thu Jun 22, 2006 7:19 am
Posts: 14

I have talk too fast.

If I has this data:

Person person_id: 1
Person name: John

Fingerprint fingerprint_id: 9
Fingerprint person_id: 1
Fingerprint fingerprint_data: john data

(I am using Firebird db)

Then when i list the users, i see the user john, but without a fingerprint, because person_id and fingerprint_id are not equals. The field that indicates the person that owns a fingerprint is the column person_id on the fingerprint table, not the fingerprint_id.

I need to indicate in the one-to-one directive that the relation between the tables is that the person_id (not the fingerprint_id) on the fingerprint table corresponds to the person_id on the person table.

Any idea?

Thanks.

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mojarrell

Post subject:

Posted: Fri Jun 23, 2006 8:04 am

Beginner

Joined: Mon Jul 26, 2004 4:29 pm
Posts: 45
Location: TX, USA

You're exactly right. I didn't get the mapping quite right, but my data hid that fact. Give me a little while and I'll re-work the example.

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mojarrell

Post subject:

Posted: Fri Jun 23, 2006 11:05 am

Beginner

Joined: Mon Jul 26, 2004 4:29 pm
Posts: 45
Location: TX, USA

Here is new data with person_id's not equal to associated fingerprint_id's:

Code:

sql> select * from person;
PERSON_ID  NAME
---------  -------------
        0  John Doe
        1  Sally Smith
        2  Hank Williams

3 rows
sql> select * from fingerprint;
FINGERPRINT_ID  PERSON_ID  FINGERPRINT_DATA
--------------  ---------  ---------------------------------
             0          1  Here is Sally's fingerprint data.
             1          0  Here is John's fingerprint data.

2 rows

Here is the updated mapping:

Code:

    <class name="Person" table="PERSON">
        <id name="personId" column="PERSON_ID">
            <generator class="native"/>
        </id>
        <many-to-one name="fingerprint" column="PERSON_ID" unique="true"
        insert="false" update="false" not-found="ignore" />
        <property name="name" />
    </class>

    <class name="Fingerprint" table="FINGERPRINT">
        <id name="fingerprintId" column="FINGERPRINT_ID">
            <generator class="native" />
        </id>
        <one-to-one name="person" property-ref="fingerprint" />
        <property name="fingerprintData" column="FINGERPRINT_DATA" />
    </class>

Here is the output:

INFO Main - Let's list Persons then Fingerprints
INFO Main - Person is: hg.Person@13f210f[id=0,name=John Doe]
INFO Main - Associated fingerprint is: hg.Fingerprint@1fd5e2[fingerprintId=0,personId=1,data=Here is Sally's fingerprint data.]
INFO Main - Person is: hg.Person@c743eb[id=1,name=Sally Smith]
INFO Main - Associated fingerprint is: hg.Fingerprint@84cc09[fingerprintId=1,personId=0,data=Here is John's fingerprint data.]
INFO Main - Person is: hg.Person@3c342b[id=2,name=Hank Williams]
INFO Main - Associated fingerprint is: null
INFO Main - Let's list Fingerprints then Persons
INFO Main - Fingerprint is: hg.Fingerprint@25c828[fingerprintId=0,personId=1,data=Here is Sally's fingerprint data.]
INFO Main - Associated person is: hg.Person@1a12495[id=0,name=John Doe]
INFO Main - Fingerprint is: hg.Fingerprint@77ef83[fingerprintId=1,personId=0,data=Here is John's fingerprint data.]
INFO Main - Associated person is: hg.Person@955cd5[id=1,name=Sally Smith]

Now the correct fingerprint is associated with the correct person from both directions.

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mojarrell

Post subject:

Posted: Fri Jun 23, 2006 3:44 pm

Beginner

Joined: Mon Jul 26, 2004 4:29 pm
Posts: 45
Location: TX, USA

I was looking at my results and I realized that I made another mistake. The association from fingerprint to person wasn't pulling the right data. (Look at the end of the output above.) I tried to get several different one-to-one mappings to work with the tables you provided, but couldn't. Hibernate wants to map a one-to-one to a primary key, not another field. I tried to use the formula attribute to specify another field, but couldn't find the right incantation.

I finally re-wrote the example taking out the fingerprint_id altogether.

The sql now looks like this:

Code:

-- drop tables (child table first)
drop table fingerprint;
drop table person;

-- person table
create table person (
   person_id identity primary key
  ,name varchar(30) );
  
-- fingerprint table

create table fingerprint (
   person_id int primary key
  ,fingerprint_data varchar(50)
  ,foreign key (person_id) references person ( person_id ));
  

insert into person values (null, 'John Doe');
*john_id ~
call identity();


insert into person values (null, 'Sally Smith');
*sally_id ~
call identity();

insert into fingerprint values (*{sally_id}, 'Here is Sally''s fingerprint data.');
insert into fingerprint values (*{john_id}, 'Here is John''s fingerprint data.');

-- now a person without a fingerprint
insert into person values (null, 'Hank Williams');

The mappings now look like this:

Code:

    <class name="Person" table="PERSON">
        <id name="personId" column="PERSON_ID">
            <generator class="native"/>
        </id>
        <one-to-one name="fingerprint" />
        <property name="name" />
    </class>

    <class name="Fingerprint" table="FINGERPRINT">
        <id name="personId" column="PERSON_ID">
            <generator class="foreign">
                <param name="property">person</param>
            </generator>
        </id>
        <one-to-one name="person" constrained="true" lazy="false" />
        <property name="fingerprintData" column="FINGERPRINT_DATA" />
    </class>

And the output now appears to be correct from both sides:

INFO Main - Let's list Persons then Fingerprints
INFO Main - Person is: hg.Person@178460d[id=0,name=John Doe]
INFO Main - Associated fingerprint is: hg.Fingerprint@17f242c[personId=0,data=Here is John's fingerprint data.]
INFO Main - Person is: hg.Person@1938039[id=1,name=Sally Smith]
INFO Main - Associated fingerprint is: hg.Fingerprint@1b06041[personId=1,data=Here is Sally's fingerprint data.]
INFO Main - Person is: hg.Person@c743eb[id=2,name=Hank Williams]
INFO Main - Associated fingerprint is: null
INFO Main - Let's list Fingerprints then Persons
INFO Main - Fingerprint is: hg.Fingerprint@b1074a[personId=0,data=Here is John's fingerprint data.]
INFO Main - Associated person is: hg.Person@1eec35[id=0,name=John Doe]
INFO Main - Fingerprint is: hg.Fingerprint@55a338[personId=1,data=Here is Sally's fingerprint data.]
INFO Main - Associated person is: hg.Person@d85cc[id=1,name=Sally Smith]

I apologize for the wrong info. Maybe someone more experienced than me could show me how to take the original tables and map the one-to-one properly?

Cheers,

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jas_esp

Post subject:

Posted: Sat Jun 24, 2006 6:20 am

Newbie

Joined: Thu Jun 22, 2006 7:19 am
Posts: 14

OK. Many thanks for your effort and try to help me, even with my poor english. I'm going to try this.

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