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View Full Version : ResourceBundleViewResolver/defaultParentView/Jasper problem



shoosh
May 20th, 2005, 12:36 PM
I have a ResourceBundleViewResolver for my Jasper reports :



<bean id="jasperViewResolver" class="org.springframework.web.servlet.view.ResourceBundl eViewResolver">
<property name="order"><value>0</value></property>
<property name="basename"><value>reports</value></property>
<property name="defaultParentView"><value>oaReportBase</value></property>
</bean>


Then i define some properties in the parent view to avoid repetition :



oaReportBase.class=com.kylietech.oaj.reports.OAJas perReportsPdfView
oaReportBase.reportDataKey=OAReportData


poSummary.url=/WEB-INF/reports/compiled/poSummary.jrxml

....


However, on startup now (after upgrading to 1.5) I get :

javax.servlet.ServletException: Error creating bean with name 'oaReportBase' defined in null: Initialization of bean failed; nested exception is java.lang.IllegalArgumentException: location is required
net.sf.acegisecurity.intercept.web.SecurityEnforce mentFilter.doFilter(SecurityEnforcementFilter.java :214)
net.sf.acegisecurity.util.FilterChainProxy$Virtual FilterChain.doFilter(FilterChainProxy.java:303)
net.sf.acegisecurity.ui.AbstractProcessingFilter.d oFilter(AbstractProcessingFilter.java:305)

It's trying to create a bean of type oaReportBase ?! Is this correct ??!!

Sherif

Deepak
Aug 29th, 2005, 10:52 AM
I think for a ResourceBundleViewResolver to work, you musl also specify the Viewname.url=.... property. That should resolve your problem.