我想配置连接池的Tomcat 7这里是代码创建资源实例不能在Tomcat7
<resource-ref>
<description>
Sample JNDI DataSource resource reference
</description>
<res-ref-name>jdbc/testDataSource</res-ref-name>
<res-type>java.sql.DatSource</res-type>
<res-auth>Container</res-auth>
</resource-ref>
和jsp页面的访问:
Context initialContext = new InitialContext();
Context envContext = (Context) initialContext.lookup("java:/comp/env");
conn = (Connection) envContext.lookup("jdbc/testDataSource");
但不幸的是,我得到异常:
javax.naming.NamingException: Cannot create resource instance
org.apache.naming.factory.ResourceFactory.getObjectInstance(ResourceFactory.java:146)
javax.naming.spi.NamingManager.getObjectInstance(NamingManager.java:321)
org.apache.naming.NamingContext.lookup(NamingContext.java:826)
org.apache.naming.NamingContext.lookup(NamingContext.java:145)
org.apache.naming.NamingContext.lookup(NamingContext.java:814)
org.apache.naming.NamingContext.lookup(NamingContext.java:159)
org.apache.jsp.index_jsp._jspService(index_jsp.java:91)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:433)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:389)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:333)
javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
我怎样才能解决这个问题? 谢谢。