java.lang.ClassCastException: org.apache.tomcat.dbcp.dbcp.BasicDataSource 无法转换为 org.apache.tomcat.jdbc.pool.DataSource [英] java.lang.ClassCastException: org.apache.tomcat.dbcp.dbcp.BasicDataSource cannot be cast to org.apache.tomcat.jdbc.pool.DataSource

问题描述

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 java.lang.ClassCastException: org.apache.tomcat.dbcp.dbcp.BasicDataSource cannot be cast to org.apache.tomcat.jdbc.pool.DataSource.我的 ./META-INF/content.xml 文件如下所示:

<块引用>

<资源名称="jdbc/FUDB"身份验证=容器"类型="javax.sql.DataSource"用户名 =数据库管理员"密码=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"最大活动=8"maxIdle="4"/>

我的 webapp web.xml 看起来像这样:

<块引用>

<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/javaeehttp://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"版本=3.0"元数据完成=真"><display-name>FedUtilization</display-name><小服务程序><servlet-name>Report1</servlet-name><display-name>Report1</display-name><servlet-class>com.sapgss.ps.servlet.Report1</servlet-class>

报告1/报告1
SQL Anywhere 12.0.1 服务器 jdbc3jdbc/FUDBjavax.sql.DataSource容器

servlet 代码如下:

<块引用>

import java.io.*;导入 java.sql.*;导入 javax.servlet.*;导入 javax.servlet.http.*;导入 javax.naming.*;导入 org.apache.catalina.core.StandardContext.*;导入 org.apache.tomcat.jdbc.pool.*;导入 com.sapgss.ps.dbutil.*;导入 org.apache.tomcat.dbcp.dbcp.BasicDataSource；公共类 Report1 扩展了 HttpServlet {public void doGet(HttpServletRequest request,

HttpServletResponse 响应)抛出 IOException, ServletException{尝试{response.setContentType("text/html");PrintWriter out = response.getWriter();out.println("");out.println("");out.println("你好伊莱恩！");out.println("");out.println("");out.println("

伊莱恩你好！

");
//这是如何在 Java Context 中编写对数据库的访问initCtx = 新的 InitialContext();上下文 envCtx = (上下文)initCtx.lookup("java:comp/env");数据源 ds = (数据源)envCtx.lookup("jdbc/FUDB");连接 conn = ds.getConnection();...
}}

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

提前谢谢我正在拔头发.

解决方案

就我而言，我只是忘记了:

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

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

我的 context.xml:

<资源名称="jdbc/gestrel" auth="容器"类型="javax.sql.DataSource"driverClassName="org.postgresql.Driver"url="jdbc:postgresql://127.0.0.1:5432/g...."用户名 =postgres"密码="....." maxActive="20" maxIdle="10"factory="org.apache.tomcat.jdbc.pool.DataSourceFactory"maxWait="-1"/>

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.

解决方案

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>

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