Merhabalar. Bir süredir bir tutorial üzerinden spring-mvc projesi ile uğraşmaktayım. Ancak anlam veremediğim bir şekilde org.springframework.beans.factory.BeanDefinitionStoreException: Unexpected exception parsing XML document from ServletContext resource [/WEB-INF/mvc-config.xml]; nested exception is java.lang.NoClassDefFoundError: org/springframework/transaction/interceptor/TransactionInterceptor hatası ile karşılaşıyorum. Classpath içine baktığımda spring-tx jar dosyası mevcut durumda. designnpatcher dosyası aşağıdaki şekilde:
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:jdbc="http://www.springframework.org/schema/jdbc" xmlns:tx="http://www.springframework.org/schema/tx" xmlns:context="http://www.springframework.org/schema/context" xmlns:mvc="http://www.springframework.org/schema/mvc" xsi:schemalocation="http://www.springframework.org/schema/beans <a href=" http:="" www.springframework.org="" schema="" beans="" spring-beans-4.1.xsd"="">http://www.springframework.org/schema/beans/spring-beans-4.1.xsd http://www.springframework.org/schema/aophttp://www.springframework.org/schema/aop/spring-aop-4.1.xsdhttp://www.springframework.org/schema/txhttp://www.springframework.org/schema/tx/spring-tx-4.1.xsdhttp://www.springframework.org/schema/contexthttp://www.springframework.org/schema/context/spring-context-4.1.xsdhttp://www.springframework.org/schema/mvchttp://www.springframework.org/schema/mvc/spring-mvc-4.1.xsdhttp://www.springframework.org/schema/jdbchttp://www.springframework.org/schema/jdbc/spring-jdbc-3.0.xsd "> <context:component-scan base-package="orh"/><!--context:annotation-config/-->
<mvc:annotation-driven/>
<tx:annotation-driven transaction-manager="transactionManager"/>
<mvc:resources mapping="/statics/**" location="/resource/"/>
<!--
Most controllers will use the ControllerClassNameHandlerMapping above, but
for the index controller we are using ParameterizableViewController, so we must
define an explicit mapping for it.
<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="/index">indexController</prop>
</props>
</property>
</bean>-->
<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver" p:prefix="/WEB-INF/views/jsp/" p:suffix=".jsp" />
<!--
The index controller.
-->
<bean id="indexController"
class="orh.web.IndexController"
/>
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer" p:location="/WEB-INF/database.properties"/> <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource" p:driverclassname="${jdbc.driverClassName}" p:url="${jdbc.url}" p:username="${jdbc.username}" p:password="${jdbc.password}"/>
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" />
</property>
<property name="jpaProperties">
<props>
<prop key="hibernate.show-sql">true</prop>
<prop key="hibernate.format-sql">true</prop>
<prop key="hibernate.hbm2ddl.auto">create</prop>
<prop key="hibernate.hbm2ddl.show">true</prop>
</props>
</property>
<property name="dataSource" ref="dataSource"/>
</bean>
<bean class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor" />
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
</beans>
Netbeans kullanıyorum ve projem bir maven ya da gradle projesi değil.
Çok teşekkürler.