I'm running Tomcat 7.0.22 and I wrote a simple servlet that connects to a SQL Anywhere 12.0 database. When I run the servlet I get java.lang.ClassCastException: org.apache.tomcat.dbcp.dbcp.BasicDataSource cannot be cast to org.apache.tomcat.jdbc.pool.DataSource. My ./META-INF/content.xml file looks like the following:
<Context> <Resource name="jdbc/FUDB" auth="Container" type="javax.sql.DataSource" username="dba" password="sql" driverClassName="sybase.jdbc.sqlanywhere.IDriver" factory="org.apache.tomcat.jdbc.pool.DataSourceFactory"
url="jdbc:sqlanywhere:uid=dba;pwd=sql;eng=BTH476331A_FedUtilization;" accessToUnderlyingConnectionAllowed="true" maxActive="8" maxIdle="4" />
My webapp web.xml looks like this:
<?xml version="1.0" encoding="ISO-8859-1"?> <web-app 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_3_0.xsd" version="3.0" metadata-complete="true"> <display-name>FedUtilization</display-name> <servlet> <servlet-name>Report1</servlet-name> <display-name>Report1</display-name> <servlet-class>com.sapgss.ps.servlet.Report1</servlet-class>
Report1 /Report1
SQL Anywhere 12.0.1 server jdbc3 jdbc/FUDB javax.sql.DataSource Container
The servlet code is as follows:
import java.io.*; import java.sql.*; import javax.servlet.*; import javax.servlet.http.*; import javax.naming.*; import org.apache.catalina.core.StandardContext.*; import org.apache.tomcat.jdbc.pool.*; import com.sapgss.ps.dbutil.*; import org.apache.tomcat.dbcp.dbcp.BasicDataSource; public class Report1 extends HttpServlet { public void doGet(HttpServletRequest request,
HttpServletResponse response) throws IOException, ServletException { try { response.setContentType("text/html"); PrintWriter out = response.getWriter(); out.println(""); out.println(""); out.println("Hello Elaine!"); out.println(""); out.println(""); out.println("
Hello Elaine!
");
// This is how to code access to the database in Java Context initCtx = new InitialContext(); Context envCtx = (Context) initCtx.lookup("java:comp/env"); DataSource ds = (DataSource) envCtx.lookup("jdbc/FUDB"); Connection conn = ds.getConnection(); . . .
} }
The error happens when I try to get a DataSource at this line: DataSource ds = (DataSource) envCtx.lookup("jdbc/FUDB");
Thanks in advance I'm pulling my hair out.
In my case I just forgot to put:
factory="org.apache.tomcat.jdbc.pool.DataSourceFactory"
in my /tomcat7/conf/context.xml
. Just added and all worked fine.
My context.xml:
<Context>
<Resource name="jdbc/gestrel" auth="Container"
type="javax.sql.DataSource"
driverClassName="org.postgresql.Driver"
url="jdbc:postgresql://127.0.0.1:5432/g...."
username="postgres"
password="....." maxActive="20" maxIdle="10"
factory="org.apache.tomcat.jdbc.pool.DataSourceFactory"
maxWait="-1"/>
</Context>
Today I've spent a half of the day trying to deal with the similar issue. I have tomcat server.xml file defining context like this:
<Context docBase="app" path="/my_context_path">
</Context>
Then I tried to add jdbc pool support using org.apache.tomcat.jdbc.pool.DataSource.
Just added resource definition to my server.xml context definition (see above). And of cause I defined resource-ref in my web.xml.
But there was always org.apache.tomcat.dbcp.dbcp.BasicDataSource returned. I spent time debugging tomcat and finally got to the following:
If you specify resource-ref in web.xml for bad cases 1,3 - tomcat will return org.apache.tomcat.dbcp.dbcp.BasicDataSource, cause as I can see it is some kind of default. BUT using such data source returned will cause something like this:
org.apache.tomcat.dbcp.dbcp.SQLNestedException: Cannot create JDBC driver of class '' for connect URL 'null'
at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createConnectionFactory(BasicDataSource.java:1452)
at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDataSource.java:1371)
at org.apache.tomcat.dbcp.dbcp.BasicDataSource.getConnection(BasicDataSource.java:1044)
If not specify resource-ref in web.xml then you will get an exception telling that resource with such name could not be find.
Also I noticed that for good cases 2,4 specifying resource-ref in web.xml is not necessary (works with and without resource-ref).
Try some of the cases I described. I hope something will help.
I would try to define resource in tomcat global context.xml file.
Good luck!
P.s. I run 7.0.22 version as well.
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