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[JVelilla]

I can obtain a XML bean factory independent of ApplicationContext? Example

ApplicationContext def..

beans>

<!-- Definition if bean ReportFactory -->
<bean id="reportFactory" class="tgn.util.tld.ReportFactory">
<constructor-arg>
<value>reports- spring.xml</value>
</constructor-arg>
</bean>


</beans>

my XML reports reports- spring.xml

<beans>

<!-- Definicion del bean StringParameters -->
<bean id="stringParamBean" class="report.impl.StringParameter" singleton="false">
<property name="name"><value>Motocompresor</value></property>
<property name="id"><value>planta</value></property>
<property name="value"><value>Algo que decir</value></property>
<property name="required"><value>true</value></property>
</bean>

</bean>


The file of reports is in the Web-inf directory, but the objective is to load this xml when it is needed not like part of the context

My Class ReportFactory

....
public ReportFactory(String path){
this.path = path;
ClassPathResource res = new ClassPathResource(path);

factory=new XmlBeanFactory(res);
}
...
}


But when loading the application I obtain one exception, of file nonfound in class path

It is possible to do this?

Thankz Javier

[nyname00]

Hi,
you have to move your reports file to the WEB-INF/classes dirctory in your webapp.
Or you just use FileSystemResource with the full path (or relative from TOMCAT_HOME/bin)

regards,
Mario