java.lang.ClassCastException:org.apache.tomcat.dbcp.dbcp.BasicDataSource无法强制转换为org.apache.tomcat.jdbc.pool.DataSource

Question

我正在运行Tomcat 7.0.22,我编写了一个连接到SQL Anywhere 12.0数据库的简单servlet.当我运行servlet时,我得到了java.lang.ClassCastException:org.apache.tomcat.dbcp.dbcp.BasicDataSource无法强制转换为org.apache.tomcat.jdbc.pool.DataSource.我的./META-INF/content.xml文件如下所示:

<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中:UID = DBA; PWD = SQL; ENG = BTH476331A_FedUtilization;" accessToUnderlyingConnectionAllowed ="true"maxActive ="8"maxIdle ="4"/>

我的webapp web.xml如下所示:

<?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服务器jdbc3 jdbc/FUDB javax.sql.DataSource容器

servlet代码如下:

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响应)抛出IOException,ServletException {try {response.setContentType("text/html"); PrintWriter out = response.getWriter(); 通过out.println( ""); 通过out.println( ""); out.println("Hello Elaine!"); 通过out.println( ""); 通过out.println( ""); 通过out.println("

你好Elaine!

");
//这是如何在Java上编写对数据库的访问权限上下文initCtx = new InitialContext(); Context envCtx =(Context)initCtx.lookup("java:comp/env"); DataSource ds =(DataSource)envCtx .lookup("jdbc/FUDB"); Connection conn = ds.getConnection(); ...
}}

当我尝试在此行获取DataSource时发生错误:DataSource ds =(DataSource)envCtx.lookup("jdbc/FUDB");

提前谢谢我把头发拉了出来.

Answer 1

在我的情况下,我只是忘了把:

factory="org.apache.tomcat.jdbc.pool.DataSourceFactory"

在我的/tomcat7/conf/context.xml.刚添加,一切正常.

我的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>