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towler73
 Post subject: Composite Key with constant foreign key value
PostPosted: Thu Oct 27, 2005 6:19 pm 
Newbie

Joined: Thu Oct 27, 2005 5:57 pm
Posts: 3
I'm trying to create a mapping to a lagacy database and I can't figure out how to get this one situation to work.

Java 1.4.2
DB: Sybase ASA 9
Hibernate 3

Below is my mapping to my flags table. Pretty staight forward.

Code:
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
   <class name="com.imsi.shareholder.business.Flag" table="flags">
      <composite-id>
         <key-property name="tableName" column="table_name"/>
         <key-property name="flagName" column="flag_name"/>
         <key-property name="id" column="integer_value"/>
      </composite-id>
      <property name="description" type="string">
                         <column name="description" length="50" />
                </property>
                <property name="numericDescription" type="long">
         <column name="numeric_desc" precision="10" scale="0" />
      </property>
      <property name="sortOrder" type="long">
         <column name="sort_order" precision="3" scale="0" />
      </property>
      <property name="editable" type="boolean">
                        <column name="editable" precision="1" scale="0" />
                </property>
                <property name="inactive" type="boolean">
                        <column name="inactive" precision="1" scale="0" />
                </property>
   </class>
</hibernate-mapping>



In many other tables I want to make many-to-one references to flags. The problem is that the other tables only contain the integer_value key. The other two keys are basically constants. I was hoping the below mapping would work, but get a BadSqlGrammerException.

Code:
<many-to-one name="method">
    <formula>'FUND'</formula>
    <formula>'METHOD_ID'</formula>
    <column name="method_id"/>         
</many-to-one>


Code:
org.springframework.jdbc.BadSqlGrammarException: Bad SQL grammar [] in task 'Hibernate operation'; nested exception is com.sybase.jdbc2.jdbc.SybSQLException: ASA Error -131: Syntax error near 'FUND' on line 1


Is what I'm trying to do possilbe with the formula tags? Or some other way?

I looked into possibily a custom parameterized user type, but I figured I'd ask here first. If a custom user type is the way to go, can someone point me to a good example?

Thanks,
Brett
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snpesnpe
 Post subject:
PostPosted: Fri Oct 28, 2005 7:44 am 
Expert
Expert

Joined: Sat Jun 12, 2004 4:49 pm
Posts: 915
you can't make foreign key in database for this case, correct ?

you can only if integer_value is unique key - if you have this case then try set many-to-one with unique_key attribute or set that integer_value is id for flag
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towler73
 Post subject: Re: Composite Key with constant foreign key value
PostPosted: Fri Oct 28, 2005 9:33 am 
Newbie

Joined: Thu Oct 27, 2005 5:57 pm
Posts: 3
integer_value is only unique for a give table_name and flag_name. In other words all three are required for the key. The table(s) with the foreign key only contain the integer value. The other two key values are static values, for a given table, and field. My qusetion is how can I set those static keys?
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snpesnpe
 Post subject:
PostPosted: Fri Oct 28, 2005 10:07 am 
Expert
Expert

Joined: Sat Jun 12, 2004 4:49 pm
Posts: 915
Quote:
integer_value is only unique for a give table_name and flag_name. In other words all three are required for the key. The table(s) with the foreign key only contain the integer value. The other two key values are static values, for a given table, and field. My qusetion is how can I set those static keys?


what is static field (in relational database) ?

hibernate map relational database concepts to OO concepts - how you define it in your database ?
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towler73
 Post subject:
PostPosted: Fri Oct 28, 2005 11:02 am 
Newbie

Joined: Thu Oct 27, 2005 5:57 pm
Posts: 3
Static might be the wrong word, constant is probably better.

For my example above, I have a table with a column called "method_id" that corresponds to the integer value in the flags table. To complete the key I must specify the other two keys, which aren't found in the table that has the "method_id." Theses values are constants, in this case "table_name" = "FUND" and "flag_name" = "METHOD_ID".
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snpesnpe
 Post subject:
PostPosted: Fri Oct 28, 2005 12:08 pm 
Expert
Expert

Joined: Sat Jun 12, 2004 4:49 pm
Posts: 915
many-to-one is same foreign key in databases - you can't make foreign key for your case and you can't make many-to-one
I think that is better that you make helper method (in java class) or formula (in mapping)
you can't resolve this case with many-to-one - many-to-one must have primary/unique key in relation table

this is my opinion (i don't sure)
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