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 Post subject: problem with find query in Hibernate
PostPosted: Fri Sep 30, 2005 4:27 pm 
Newbie

Joined: Fri Sep 30, 2005 3:17 pm
Posts: 1
When I start the sever, find query works fine for a while and then I keep getting this error. I do not know why it is causing this error. It is a very simple query to ACCESS_CODE table by passing CODE (string value) and returning the row. Instead of returning the row, I get error message. Any help will be appreciated. Here are the details:

Hibernate version:3.0.5

Mapping documents:<?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 package="xxx.xxx.xxx.xxxx">
<class name="AccessCode" table="ACCESS_CODE">
<id
column="CODE_ID"
name="id"
type="java.lang.Long"
>
<generator class="sequence">
<param name="sequence">seq_code_id</param>
</generator>
</id>
<property
column="CREATE_TIME"
length="26"
name="createTime"
not-null="false"
type="calendar"
/>
<property
column="DESCRIPTION"
length="1023"
name="description"
not-null="false"
type="string"
/>
<property
column="UPDATE_TIME"
length="26"
name="updateTime"
not-null="false"
type="calendar"
/>
<property
column="LDAP_VALUE"
length="128"
name="ldapValue"
not-null="true"
type="string"
/>
<property
column="CODE"
length="128"
name="code"
not-null="true"
type="string"
/>

</class>
</hibernate-mapping>


Code between sessionFactory.openSession() and session.close():

Full stack trace of any exception that occurs:org.hibernate.util.JDBCExceptionReporter logExceptions
SQL Error:-99999, SQLState: 22005

org.hibernate.util.JDBCExceptionReporter logExceptions
[IBM][CLI Driver] CLI0112E Error in assignment. SQLSTATW=22005


Name and version of the database you are using:DB2 v8.2

The generated SQL (show_sql=true):select accesscode0_.CODE_ID as
CODE1_, accesscode0_.CREATE_TIME as CREATE2_0_,
accesscode0_.DESCRIPTION as DESCRIPT3_0_, accesscode0_.UPDATE_TIME as UPDATE4_0_, accesscode0_.LDAP_VALUE as LDAP5_0_, accesscode0_.CODE as CODE0_ from ACCESS_CODE accesscode0_ where accesscode0_.CODE=?


Debug level Hibernate log excerpt:


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