Question

Quick question based on a spring web app i am creating.

How do you go about setting up the context of the application so that you do not need to set the datasource parameters for simpleJDBC all the time and can call getSimpleJDBCTemplate().queryfor…. and it be set up with the datasource.

This is how i currently have it and it seems to go against inversion of control that spring is meant to provide as this is in every dao!

 ClassPathXmlApplicationContext ac = new ClassPathXmlApplicationContext("classpath:ApplicationContext.xml");
    DataSource dataSource = (DataSource) ac.getBean("dataSource");
    SimpleJdbcTemplate simpleJdbcTemplate = new SimpleJdbcTemplate(dataSource);

ApplicationContext

<beans xmlns="http://www.springframework.org/schema/beans"
          xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
          xmlns:context="http://www.springframework.org/schema/context"
          xsi:schemaLocation="
   http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
   http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-2.5.xsd">

    <bean id="placeholderConfig" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"> 
<property name="location" value="classpath:properties.properties"/>
</bean>

    <bean id="dataSource" destroy-method="close" class="org.apache.commons.dbcp.BasicDataSource">
        <property name="driverClassName" value="${jdbc.driverClassName}"/>
        <property name="url" value="${jdbc.url}"/>
        <property name="username" value="${username}"/>
        <property name="password" value="${password}"/>
    </bean>

 <bean name="SimpleJdbcTemplate" class="org.springframework.jdbc.core.simple.SimpleJdbcTemplate">
    <constructor-arg><ref bean="dataSource"/></constructor-arg>
</bean>


      <context:annotation-config/>
   </beans>

Latest Stack Trace from Tomcat log

13-Jan-2011 20:15:18 com.sun.jersey.api.core.PackagesResourceConfig init
INFO: Scanning for root resource and provider classes in the packages:
  ptc.jersey.spring
13-Jan-2011 20:15:18 com.sun.jersey.api.core.ScanningResourceConfig logClasses
INFO: Root resource classes found:
  class ptc.jersey.spring.resources.LoginResource
13-Jan-2011 20:15:18 com.sun.jersey.api.core.ScanningResourceConfig init
INFO: No provider classes found.
13-Jan-2011 20:15:19 com.sun.jersey.spi.spring.container.servlet.SpringServlet getContext
INFO: Using default applicationContext
13-Jan-2011 20:15:19 com.sun.jersey.server.impl.application.WebApplicationImpl _initiate
INFO: Initiating Jersey application, version 'Jersey: 1.4 09/11/2010 10:30 PM'
13-Jan-2011 20:15:21 com.sun.jersey.spi.container.ContainerResponse mapMappableContainerException
SEVERE: The RuntimeException could not be mapped to a response, re-throwing to the HTTP container
java.lang.NullPointerException
    at ptc.jersey.spring.daoImpl.UserDaoImpl.getUser(UserDaoImpl.java:43)

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

    <listener>
        <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
    </listener>

    <listener>
    <listener-class>
        org.springframework.web.context.ContextLoaderListener
    </listener-class>
</listener>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:ApplicationContext.xml</param-value>
</context-param>


    <servlet>
        <servlet-name>Jersey Spring Web Application</servlet-name>
        <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
  <param-name>com.sun.jersey.config.property.packages</param-name>
  <param-value>ptc.jersey.spring</param-value>
</init-param>
    </servlet>

    <servlet-mapping>
        <servlet-name>Jersey Spring Web Application</servlet-name>
        <url-pattern>/resources/*</url-pattern>
    </servlet-mapping>
</web-app>

Any help would be great
Thanks
Chris

Answer 1

Why don’t you declare the SimpleJdbcTemplate instance as well in the application context file?

For example with these bean declarations

<bean name="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
    <property name="driverClassName" value="org.gjt.mm.mysql.Driver"/>
    <property name="url" value="dburl"/>
    <property name="username" value="dbusername"/>
    <property name="password" value="password"/>
</bean>

<bean name="jdbcTemplate" class="org.springframework.jdbc.core.simple.SimpleJdbcTemplate">
    <constructor-arg><ref bean="dataSource"/></constructor-arg>
</bean>

And in your web.xml to load the application context

<listener>
    <listener-class>
        org.springframework.web.context.ContextLoaderListener
    </listener-class>
</listener>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:ApplicationContext.xml</param-value>
</context-param>