I can't seem to find any good examples of this which is strange since
it is a very common situation. Try adding this to your file (after the
last cmp-field-mapping):
<key-generator xmlns="http://www.openejb.org/xml/ns/
pkgen-2.1">
<uuid/>
</key-generator>
As I mentioned in another email, the actual contents of the key-
generator element is ignored, but when we see the element we add
@javax.persistence.GeneratedValue(strategy=IDENTITY) to the primary
key field. This causes OpenJPA to use the database generated identity
for the primary key of the entity.
-dain
On May 14, 2008, at 7:32 AM, SK Leung wrote:
I have created the openejb-jar.xml file as below.
<?xml version="1.0" encoding="UTF-8"?>
<openejb-jar xmlns="http://www.openejb.org/xml/ns/openejb-jar-2.1";
xmlns:nam="http://geronimo.apache.org/xml/ns/naming-1.1";
xmlns:pkgen="http//www.openejb.org/xml/ns/pkgen-2.0"
xmlns:sec="http://geronimo.apache.org/xml/ns/security-1.1";
xmlns:sys="http://geronimo.apache.org/xml/ns/deployment-1.1";>
<cmp-connection-factory>
<resource-link>jdbc/OracleCMTadmuser</resource-link>
</cmp-connection-factory>
<enterprise-beans>
<entity>
<ejb-name>ProductEJB</ejb-name>
<table-name>PRODUCTS</table-name>
<cmp-field-mapping>
<cmp-field-name>productID</cmp-field-name>
<table-column>PRODUCTID</table-column>
</cmp-field-mapping>
<cmp-field-mapping>
<cmp-field-name>name</cmp-field-name>
<table-column>NAME</table-column>
</cmp-field-mapping>
<cmp-field-mapping>
<cmp-field-name>description</cmp-field-name>
<table-column>DESCRIPTION</table-column>
</cmp-field-mapping>
<cmp-field-mapping>
<cmp-field-name>basePrice</cmp-field-name>
<table-column>BASEPRICE</table-column>
</cmp-field-mapping>
</entity>
</enterprise-beans>
</openejb-jar>
But the same error message is shown when creating the object. I have
the
following resource defined in openejb.xml.
<Resource id="jdbc/OracleCMTadmuser" type="DataSource">
JdbcDriver oracle.jdbc.driver.OracleDriver
JdbcUrl jdbc:oracle:thin:@localhost:1521:xe
UserName adm_user
Password password
</Resource>
I do not know whether the problem is in making connection to
database or
creating the record. If it is the latter one, then what is the cause
for
such error? I even try to insert all nulls into the table and it
accepts.
Please give help, much thanks.
SK
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