Programmatically creating a JNDI DataSource for Spring
Asked Answered
U

1

6

I have an existing Spring web-based application that has datasources defined using JNDI, and I'm trying to create a standalone app to use the beans. How can I create the JNDI entry and database properties programmatically in the standalone application?

<bean id="myDataSource" class="org.springframework.jndi.JndiObjectFactoryBean">
    <property name="jndiName" value="java:comp/env/jdbc/MyDS" />
</bean>

    public static void main(String[] args) {

      // this throws an error since the JNDI lookup fails - can I programmatically define the database properties here?

    ClassPathXmlApplicationContext ctx = new  ClassPathXmlApplicationContext("applicationContext.xml");

    UserService userService = ctx.getBean(UserService.class);
    User user = userService.findUserById("jdoe");

    System.out.println("display name: " + user.getDisplayName());
}

EDIT:

I tried something like this, but am now getting the error "javax.naming.NoInitialContextException: Need to specify class name in environment or system property"

public static void main(String[] args) {
    setupJNDI();

    ClassPathXmlApplicationContext ctx = new  ClassPathXmlApplicationContext("applicationContext.xml");

    UserService userService = ctx.getBean(UserService.class);
    User user = userService.findUserById("jdoe");

    System.out.println("display name: " + user.getDisplayName());
}


private static void setupJNDI() {
    InitialContext ic;
    try {
        ic = new InitialContext();
        ic.createSubcontext("java:");
        ic.createSubcontext("java:/comp");
        ic.createSubcontext("java:/comp/env");
        ic.createSubcontext("java:/comp/env/jdbc");
        SQLServerConnectionPoolDataSource myDS = new SQLServerConnectionPoolDataSource();
        opaDS.setServerName("myserver");
        opaDS.setPortNumber(1433);
        opaDS.setUser("user");
        opaDS.setPassword("password");

        ic.bind("java:/comp/env/jdbc/MyDS", myDS);
    } catch (NamingException e) {
        e.printStackTrace();
    }
}
Unterwalden answered 20/2, 2013 at 14:21 Comment(0)
L
6

The org.springframework.test dependency has support for that via the SimpleNamingContextBuilder:

// First create the mock JNDI tree and bind the DS
SimpleNamingContextBuilder builder = new SimpleNamingContextBuilder();
DataSource ds = new ComboPooledDataSource();
ds.setDriverClass( ... ); // etc. for uid, password, url
builder.bind( "java:comp/env/jdbc/MyDS" , ds );
builder.activate();

// Then create the Spring context, which should now be able 
// to resolve the JNDI datasource
ApplicationContext context = new ClassPathXmlApplicationContext( "..." );

That should work.

Cheers,

Lafrance answered 20/2, 2013 at 14:29 Comment(3)
I tried this but am getting the error 'Exception in thread "main" org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'myDataSource' defined in class path resource [applicationContext.xml]: Invocation of init method failed; nested exception is javax.naming.NoInitialContextException: Need to specify class name in environment or system property, or as an applet parameter, or in an application resource file: java.naming.factory.initial'Unterwalden
My bad, the JNDI tree should of course be created and activated before trying to create the Spring context - I've edited the answer accordingly.Lafrance
Could you give me a reasonable explanation why this solution gives me an error using jtaDataSource?Detumescence

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